By the end of this section, you will be able to:

- Describe the concept of electric charge
- Explain qualitatively the force electric charge creates

You are certainly familiar with electronic devices that you activate with the click of a switch, from computers to cell phones to television. And you have certainly seen electricity in a flash of lightning during a heavy thunderstorm. But you have also most likely experienced electrical effects in other ways, maybe without realizing that an electric force was involved. Let’s take a look at some of these activities and see what we can learn from them about electric charges and forces.

You have probably experienced the phenomenon of **static electricity**: When you first take clothes out of a dryer, many (not all) of them tend to stick together; for some fabrics, they can be very difficult to separate. Another example occurs if you take a woolen sweater off quickly—you can feel (and hear) the static electricity pulling on your clothes, and perhaps even your hair. If you comb your hair on a dry day and then put the comb close to a thin stream of water coming out of a faucet, you will find that the water stream bends toward (is attracted to) the comb (\ref{fig:1.1.1}).

Suppose you bring the comb close to some small strips of paper; the strips of paper are attracted to the comb and even cling to it (\ref{fig:1.1.2}). In the kitchen, quickly pull a length of plastic cling wrap off the roll; it will tend to cling to most any nonmetallic material (such as plastic, glass, or food). If you rub a balloon on a wall for a few seconds, it will stick to the wall. Probably the most annoying effect of static electricity is getting shocked by a doorknob (or a friend) after shuffling your feet on some types of carpeting.

\begin{gather}.\tag{Figure 1.1.2}\label{fig:1.1.2}\end{gather} Many of these phenomena have been known for centuries. The ancient Greek philosopher Thales of Miletus (624–546 BCE) recorded that whenThe English physicist William **Gilbert** (1544–1603) also studied this attractive force, using various substances. He worked with amber, and, in addition, he experimented with rock crystal and various precious and semi-precious gemstones. He also experimented with several metals. He found that the metals never exhibited this force, whereas the minerals did. Moreover, although an electrified amber rod would attract a piece of fur, it would repel another electrified amber rod; similarly, two electrified pieces of fur would repel each other.

This suggested there were two types of an electric property; this property eventually came to be called **electric charge**. The difference between the two types of electric charge is in the directions of the electric forces that each type of charge causes: These forces are repulsive when the same type of charge exists on two interacting objects and attractive when the charges are of opposite types. The SI unit of electric charge is the **coulomb** (C), after the French physicist Charles Augustine de **Coulomb** (1736–1806).

The most peculiar aspect of this new force is that it does not require physical contact between the two objects in order to cause an acceleration. This is an example of a so-called “long-range” force. (Or, as Albert Einstein later phrased it, “action at a distance.”) With the exception of gravity, all other forces we have discussed so far act only when the two interacting objects actually touch.

The American physicist and statesman BenjaminFranklin pointed out that the observed behavior could be explained by supposing that one of the two types of charge remained motionless, while the other type of charge flowed from one piece of foil to the other. He further suggested that an excess of what he called this “electrical fluid” be called “positive electricity” and the deficiency of it be called “negative electricity.” His suggestion, with some minor modifications, is the model we use today. (With the experiments that he was able to do, this was a pure guess; he had no way of actually determining the sign of the moving charge. Unfortunately, he guessed wrong; we now know that the charges that flow are the ones Franklin labeled negative, and the positive charges remain largely motionless. Fortunately, as we’ll see, it makes no practical or theoretical difference which choice we make, as long as we stay consistent with our choice.)

Let’s list the specific observations that we have of this **electric force**:

- The force acts without physical contact between the two objects.
- The force can be either attractive or repulsive: If two interacting objects carry the same sign of charge, the force is repulsive; if the charges are of opposite sign, the force is attractive. These interactions are referred to as
**electrostatic repulsion**and**electrostatic attraction**, respectively. - Not all objects are affected by this force.
- The magnitude of the force decreases (rapidly) with increasing separation distance between the objects.

To be more precise, we find experimentally that the magnitude of the force decreases as the square of the distance between the two interacting objects increases. Thus, for example, when the distance between two interacting objects is doubled, the force between them decreases to one fourth what it was in the original system. We can also observe that the surroundings of the charged objects affect the magnitude of the force. However, we will explore this issue in a later chapter.

In addition to the existence of two types of charge, several other properties of charge have been discovered.

**Charge is quantized.**This means that electric charge comes in discrete amounts, and there is a smallest possible amount of charge that an object can have. In the SI system, this smallest amount is $e\equiv1.602\times10^{-19}$. No free particle can have less charge than this, and, therefore, the charge on any object—the charge on all objects—must be an integer multiple of this amount. All macroscopic, charged objects have charge because electrons have either been added or taken away from them, resulting in a net charge.**The magnitude of the charge is independent of the type.**Phrased another way, the smallest possible positive charge (to four significant figures) is $+1.602\times10^{-19}~\mathrm{C}$, and the smallest possible negative charge is $-1.602\times10^{-19}~\mathrm{C}$; these values are exactly equal. This is simply how the laws of physics in our universe turned out.**Charge is conserved.**Charge can neither be created nor destroyed; it can only be transferred from place to place, from one object to another. Frequently, we speak of two charges “canceling”; this is verbal shorthand. It means that if two objects that have equal and opposite charges are physically close to each other, then the (oppositely directed) forces they apply on some other charged object cancel, for a net force of zero. It is important that you understand that the charges on the objects by no means disappear, however. The net charge of the universe is constant.**Charge is conserved in closed systems.**In principle, if a negative charge disappeared from your lab bench and reappeared on the Moon, conservation of charge would still hold. However, this never happens. If the total charge you have in your local system on your lab bench is changing, there will be a measurable flow of charge into or out of the system. Again, charges can and do move around, and their effects can and do cancel, but the net charge in your local environment (if closed) is conserved. The last two items are both referred to as the**law of conservation of charge**.

Once it became clear that all matter was composed of particles that came to be called atoms, it also quickly became clear that the constituents of the atom included both positively charged particles and negatively charged particles. The next question was, what are the physical properties of those electrically charged particles?

The negatively charged particle was the first one to be discovered. In 1897, the English physicist J. J. **Thomson** was studying what was then known as *cathode rays*. Some years before, the English physicist William Crookes had shown that these “rays” were negatively charged, but his experiments were unable to tell any more than that. (The fact that they carried a negative electric charge was strong evidence that these were not rays at all, but particles.) Thomson prepared a pure beam of these particles and sent them through crossed electric and magnetic fields, and adjusted the various field strengths until the net deflection of the beam was zero. With this experiment, he was able to determine the charge-to-mass ratio of the particle. This ratio showed that the mass of the particle was much smaller than that of any other previously known particle—1837 times smaller, in fact. Eventually, this particle came to be called the **electron**.

Since it was known that different atoms have different masses, and that ordinarily atoms are electrically neutral, it was natural to suppose that different atoms have different numbers of protons in their nucleus, with an equal number of negatively charged electrons orbiting about the positively charged nucleus, thus making the atoms overall electrically neutral. However, it was soon discovered that although the lightest atom, hydrogen, did indeed have a single proton as its nucleus, the next heaviest atom—helium—has twice the number of protons (two), but *four* times the mass of hydrogen.

This mystery was resolved in 1932 by the English physicist James **Chadwick**, with the discovery of the **neutron**. The neutron is, essentially, an electrically neutral twin of the proton, with no electric charge, but (nearly) identical mass to the proton. The helium nucleus therefore has two neutrons along with its two protons. (Later experiments were to show that although the neutron is electrically neutral overall, it does have an internal charge *structure*. Furthermore, although the masses of the neutron and the proton are *nearly* equal, they aren’t exactly equal: The neutron’s mass is very slightly larger than the mass of the proton. That slight mass excess turned out to be of great importance.)

The story of the atom does not stop there, however. In the latter part of the twentieth century, many more subatomic particles were discovered in the nucleus of the atom: pions, neutrinos, and quarks, among others. With the exception of the photon, none of these particles are directly relevant to the study of electromagnetism, so we will not discuss them further in this course.

As noted previously, electric charge is a property that an object can have. This is similar to how an object can have a property that we call mass, a property that we call density, a property that we call temperature, and so on. Technically, we should always say something like, “Suppose we have a particle that carries a charge of $3~\mu\mathrm{C}$.” However, it is very common to say instead, “Suppose we have a $3$-$\mu\mathrm{C}$ charge.” Similarly, we often say something like, “Six charges are located at the vertices of a regular hexagon.” A charge is not a particle; rather, it is a *property* of a particle. Nevertheless, this terminology is extremely common (and is frequently used in this book, as it is everywhere else). So, keep in the back of your mind what we really mean when we refer to a “charge.

By the end of this section, you will be able to:

- Explain what a conductor is
- Explain what an insulator is
- List the differences and similarities between conductors and insulators
- Describe the process of charging by induction

As discussed in the previous section, electrons surround the tiny nucleus in the form of a (comparatively) vast cloud of negative charge. However, this cloud does have a definite structure to it. Let’s consider an atom of the most commonly used conductor, copper.

There is an outermost electron that is only loosely bound to the atom’s nucleus. It can be easily dislodged; it then moves to a neighboring atom. In a large mass of copper atoms (such as a copper wire or a sheet of copper), these vast numbers of outermost electrons (one per atom) wander from atom to atom, and are the electrons that do the moving when electricity flows. These wandering, or “free,” electrons are called **conduction electrons**, and copper is therefore an excellent **conductor** (of electric charge). All conducting elements have a similar arrangement of their electrons, with one or two conduction electrons. This includes most metals.

**Insulators**, in contrast, are made from materials that lack conduction electrons; charge flows only with great difficulty, if at all. Even if excess charge is added to an insulating material, it cannot move, remaining indefinitely in place. This is why insulating materials exhibit the electrical attraction and repulsion forces described earlier, whereas conductors do not; any excess charge placed on a conductor would instantly flow away (due to mutual repulsion from existing charges), leaving no excess charge around to create forces. Charge cannot flow along or through an **insulator**, so its electric forces remain for long periods of time. (Charge will dissipate from an insulator, given enough time.) As it happens, amber, fur, and most semi-precious gems are insulators, as are materials like wood, glass, and plastic.

Let’s examine in more detail what happens in a conductor when an electrically charged object is brought close to it. As mentioned, the conduction electrons in the conductor are able to move with nearly complete freedom. As a result, when a charged insulator (such as a positively charged glass rod) is brought close to the conductor, the (total) charge on the insulator exerts an electric force on the conduction electrons. Since the rod is positively charged, the conduction electrons (which themselves are negatively charged) are attracted, flowing toward the insulator to the near side of the conductor (\ref{fig:1.2.2}).

Now, the conductor is still overall electrically neutral; the conduction electrons have changed position, but they are still in the conducting material. However, the conductor now has a chargeThe result is the formation of what is called an electric **dipole**, from a Latin phrase meaning “two ends.” The presence of electric charges on the insulator—and the electric forces they apply to the conduction electrons—creates, or “induces,” the dipole in the conductor.

When a charged rod is brought near a neutral substance, an insulator in this case, the distribution of charge in atoms and molecules is shifted slightly. Opposite charge is attracted nearer the external charged rod, while like charge is repelled. Since the electrostatic force decreases with distance, the repulsion of like charges is weaker than the attraction of unlike charges, and so there is a net attraction. Thus, a positively charged glass rod attracts neutral pieces of paper, as will a negatively charged rubber rod. Some molecules, like water, are polar molecules. Polar molecules have a natural or inherent separation of charge, although they are neutral overall. Polar molecules are particularly affected by other charged objects and show greater polarization effects than molecules with naturally uniform charge distributions.

When the two ends of a dipole can be separated, this method ofBy the end of this section, you will be able to:

- Describe the electric force, both qualitatively and quantitatively
- Calculate the force that charges exert on each other
- Determine the direction of the electric force for different source charges
- Correctly describe and apply the superposition principle for multiple source charges

Experiments with electric charges have shown that if two objects each have electric charge, then they exert an electric force on each other. The magnitude of the force is linearly proportional to the net charge on each object and inversely proportional to the square of the distance between them. (Interestingly, the force does not depend on the mass of the objects.) The direction of the force vector is along the imaginary line joining the two objects and is dictated by the signs of the charges involved.

Let

- $q_1,~q_2=$ the net electric charges of the two objects;
- $\vec{\mathbf{r}}_{12}=$ the vector displacement from $q_1$ to $q_2$.

The electric force $\vec{\mathbf{F}}$ on one of the charges is proportional to the magnitude of its own charge and the magnitude of the other charge, and is inversely proportional to the square of the distance between them:

\[F\propto\frac{q_1q_2}{{r_{12}}^2}.\]

This proportionality becomes an equality with the introduction of a proportionality constant. For reasons that will become clear in a later chapter, the proportionality constant that we use is actually a collection of constants. (We discuss this constant shortly.)

The electric force (or **Coulomb force**) between two electrically charged particles is equal to

We use absolute value signs around the product $q_1q_2$ because one of the charges may be negative, but the magnitude of the force is always positive. The unit vector $\hat{\mathbf{r}}$ points directly from the charge $q_1$ toward $q_2$. If $q_1$ and $q_2$ have the same sign, the force vector on $q_2$ points away from $q_1$; if they have opposite signs, the force on $q_2$ points toward $q_1$ (\ref{fig:1.3.1}).

It is important to note that the electric force is not constant; it is a function of the separation distance between the two charges. If either the test charge or the source charge (or both) move, then $\vec{\mathbf{r}}$ changes, and therefore so does the force. An immediate consequence of this is that direct application of Newton’s laws with this force can be mathematically difficult, depending on the specific problem at hand. It can (usually) be done, but we almost always look for easier methods of calculating whatever physical quantity we are interested in. (Conservation of energy is the most common choice.)

Finally, the new constant $\varepsilon_0$ in Coulomb’s law is called the *permittivity of free space*, or (better) the **permittivity of vacuum**. It has a very important physical meaning that we will discuss in a later chapter; for now, it is simply an empirical proportionality constant. Its numerical value (to three significant figures) turns out to be

These units are required to give the force in Coulomb’s law the correct units of newtons. Note that in Coulomb’s law, the permittivity of vacuum is only part of the proportionality constant. For convenience, we often define a Coulomb’s constant:

\[k_e=\frac{1}{4\pi\varepsilon_0}=8.99\times10^9~\frac{\mathrm{N}\cdot\mathrm{m}^2}{\mathrm{C}^2}.\]

\[q_1=+e=+1.602\times10^{-19}~\mathrm{C}\]

\[q_2=-e=-1.602\times10^{-19}~\mathrm{C}\]

\[r=5.29\times10^{-11}~m.\]The magnitude of the force on the electron is

\[F=\frac{1}{4\pi\varepsilon_0}\frac{|e|^2}{r^2}=\frac{1}{4\pi\left(8.85\times10^{-12}~\frac{\mathrm{C}^2}{\mathrm{N}\cdot\mathrm{m}^2}\right)}\frac{\left(1.602\times10^{-19}~\mathrm{C}\right)^2}{\left(5.29\times10^{-11}~m\right)^2}=8.25\times10^{-8}~\mathrm{N}.\]

As for the direction, since the charges on the two particles are opposite, the force is attractive; the force on the electron points radially directly toward the proton, everywhere in the electron’s orbit. The force is thus expressed as

\[\vec{\mathbf{F}}=(8.25\times10^{-8}~\mathrm{N})~\mathbf{\hat{r}}.\]

What would be different if the electron also had a positive charge?

The analysis that we have done for two particles can be extended to an arbitrary number of particles; we simply repeat the analysis, two charges at a time. Specifically, we ask the question: Given $N$ charges (which we refer to as source charge), what is the net electric force that they exert on some other point charge (which we call the test charge)? Note that we use these terms because we can think of the test charge being used to test the strength of the force provided by the source charges.

Like all forces that we have seen up to now, the net electric force on our test charge is simply the vector sum of each individual electric force exerted on it by each of the individual test charges. Thus, we can calculate the net force on the test charge $Q$ by calculating the force on it from each source charge, taken one at a time, and then adding all those forces together (as vectors). This ability to simply add up individual forces in this way is referred to as the **principle of superposition**, and is one of the more important features of the electric force. In mathematical form, this becomes

\begin{equation}{\mathbf{F}}=\frac{1}{4\pi\varepsilon_0}Q\sum_{i=1}^{N}\frac{q_i}{r_i^2}\mathbf{\hat{r}}_i.\tag{1.3.2}\label{eq:1.3.2}\end{equation}

(Note that the force vector $\vec{\mathbf{F}}_i$ does not necessarily point in the same direction as the unit vector $\hat{\mathbf{r}}_i$; it may point in the opposite direction, $-\hat{\mathbf{r}}_i$. The signs of the source charge and test charge determine the direction of the force on the test charge.)

There is a complication, however. Just as the source charges each exert a force on the test charge, so too (by Newton’s third law) does the test charge exert an equal and opposite force on each of the source charges. As a consequence, each source charge would change position. However, by Equation \ref{eq:1.3.2}, the force on the test charge is a function of position; thus, as the positions of the source charges change, the net force on the test charge necessarily changes, which changes the force, which again changes the positions. Thus, the entire mathematical analysis quickly becomes intractable. Later, we will learn techniques for handling this situation, but for now, we make the simplifying assumption that the source charges are fixed in place somehow, so that their positions are constant in time. (The test charge is allowed to move.) With this restriction in place, the analysis of charges is known as **electrostatics**, where “statics” refers to the constant (that is, static) positions of the source charges and the force is referred to as an **electrostatic force**.

We can’t add these forces directly because they don’t point in the same direction: $\vec{\mathbf{F}}_{12}$ points only in the $-x$-direction, while $\vec{\mathbf{F}}_{13}$ points only in the $+y$-direction. The net force is obtained from applying the Pythagorean theorem to its $x-$ and $y$-components:

\[F=\sqrt{F_x^2+F_y^2}\]

where

\begin{eqnarray*} F_x&=&-F_{23}=-\frac{1}{4\pi\epsilon_0}\frac{q_2q_3}{r_{23}^2}\\&=&-\left(8.99\times10^{9}~\mathrm{N}\cdot\frac{\mathrm{m}^2}{\mathrm{C}^2}\right)\frac{(4.806\times10^{-19}~\mathrm{C})(8.01\times10^{-19}~\mathrm{C})}{(4.00\times10^{-7}~\mathrm{m})^2}\\&=&-2.16\times10^{-14}~\mathrm{N}\end{eqnarray*}

and

\begin{eqnarray*} F_y&=&F_{21}=\frac{1}{4\pi\epsilon_0}\frac{q_2q_1}{r_{21}^2}\\&=&-\left(8.99\times10^{9}~\mathrm{N}\cdot\frac{\mathrm{m}^2}{\mathrm{C}^2}\right)\frac{(4.806\times10^{-19}~\mathrm{C})(3.024\times10^{-19}~\mathrm{C})}{(2.00\times10^{-7}~\mathrm{m})^2}\\&=&3.46\times10^{-14}~\mathrm{N}.\end{eqnarray*}

We find that

\[F=\sqrt{F_x^2+F_y^2}=4.08\times10^{-14}~\mathrm{N}\]

at an angle of

\[\phi=\tan^{-1}\left(\frac{F_y}{F_x}\right)=\tan^{-1}\left(\frac{3.46\times10^{-14}~\mathrm{N}}{-2.16\times10^{-14}~\mathrm{N}}\right)=-58^{\circ},\]

that is, $58^{\circ}$ above the $-x$-axis, as shown in the diagram.

It’s also worth noting that the only new concept in this example is how to calculate the electric forces; everything else (getting the net force from its components, breaking the forces into their components, finding the direction of the net force) is the same as force problems you have done earlier.

What would be different if $q_1$ were negative?

By the end of this section, you will be able to:

- Explain the purpose of the electric field concept
- Describe the properties of the electric field
- Calculate the field of a collection of source charges of either sign

As we showed in the preceding section, the net electric force on a test charge is the vector sum of all the electric forces acting on it, from all of the various source charges, located at their various positions. But what if we use a different test charge, one with a different magnitude, or sign, or both? Or suppose we have a dozen different test charges we wish to try at the same location? We would have to calculate the sum of the forces from scratch. Fortunately, it is possible to define a quantity, called the **electric field**, which is independent of the test charge. It only depends on the configuration of the source charges, and once found, allows us to calculate the force on any test charge.

Suppose we have $N$ source charges $q_1,\ q_2,\ q_3,\ldots,\ q_N$ located at positions $\vec{\mathbf{r}}_1,\vec{\mathbf{r}}_2,\vec{\mathbf{r}}_3,\ldots,\vec{\mathbf{r}}_N$, applying $N$ electrostatic forces on a test charge $Q$. The net force on $Q$ is (see Equation 1.3.2)

\begin{eqnarray*}\vec{\mathbf{F}}&=&\vec{\mathbf{F}}_1+\vec{\mathbf{F}}_2+\vec{\mathbf{F}}_3+\ldots+\vec{\mathbf{F}}_N\\&=&\frac{1}{4\pi\epsilon_0}\left(\frac{Qq_1}{r_1^2}\hat{\mathbf{r}}_1+\frac{Qq_2}{r_2^2}\hat{\mathbf{r}}_2+\frac{Qq_3}{r_3^2}\hat{\mathbf{r}}_3+\ldots+\frac{Qq_N}{r_N^2}\hat{\mathbf{r}}_N\right)\\&=&Q\left[\frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{r_1^2}\hat{\mathbf{r}}_1+\frac{q_2}{r_2^2}\hat{\mathbf{r}}_2+\frac{q_3}{r_3^2}\hat{\mathbf{r}}_3+\ldots+\frac{q_N}{r_N^2}\hat{\mathbf{r}}_N\right)\right].\end{eqnarray*}

We can rewrite this as

\begin{equation}\vec{\mathbf{F}}=Q\vec{\mathbf{E}}\tag{1.4.1}\label{eq:1.4.1}\end{equation}

where

\[\vec{\mathbf{E}}\equiv\frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{r_1^2}\hat{\mathbf{r}}_1+\frac{q_2}{r_2^2}\hat{\mathbf{r}}_2+\frac{q_3}{r_3^2}\hat{\mathbf{r}}_3+\ldots+\frac{q_N}{r_N^2}\hat{\mathbf{r}}_N\right)\]

or, more compactly,

\begin{equation}\vec{\mathbf{E}}\equiv\frac{1}{4\pi\epsilon_0}\sum_{i=1}^{N}\frac{q_i}{r_i^2}\hat{\mathbf{r}}_i.\tag{1.4.2}\label{eq:1.4.2}\end{equation}

This expression is called the electric field at position $P=P(x,y,z)$ of the $N$ source charges. Here, $P$ is the location of the point in space where you are calculating the field and is relative to the positions $\vec{\mathbf{r}}_i$ of the source charges (\ref{fig:1.4.1}). Note that we have to impose a coordinate system to solve actual problems.

\begin{gather}.\tag{Figure 1.4.1}\label{fig:1.4.1}\end{gather}Notice that the calculation of the electric field makes no reference to the test charge. Thus, the physically useful approach is to calculate the electric field and then use it to calculate the force on some test charge later, if needed. Different test charges experience different forces (Equation \ref{eq:1.4.1}), but it is the same electric field (Equation \ref{eq:1.4.2}). That being said, recall that there is no fundamental difference between a test charge and a source charge; these are merely convenient labels for the system of interest. Any charge produces an electric field; however, just as Earth’s orbit is not affected by Earth’s own gravity, a charge is not subject to a force due to the electric field it generates. Charges are only subject to forces from the electric fields of other charges.

In this respect, the electric field $\vec{\mathbf{E}}$ of a point charge is similar to the gravitational field $\vec{\mathbf{g}}$ of Earth; once we have calculated the gravitational field at some point in space, we can use it any time we want to calculate the resulting force on any mass we choose to place at that point. In fact, this is exactly what we do when we say the gravitational field of Earth (near Earth’s surface) has a value of $9.81~\mathrm{m/s}^2$, and then we calculate the resulting force (i.e., weight) on different masses. Also, the general expression for calculating $\vec{\mathbf{g}}$ at arbitrary distances from the center of Earth (i.e., not just near Earth’s surface) is very similar to the expression for $\vec{\mathbf{E}}$: $\vec{\mathbf{g}}=G\frac{M}{r^2}\hat{\mathbf{r}}$, where $G$ is a proportionality constant, playing the same role for $\vec{\mathbf{g}}$ as $\frac{1}{4\pi\epsilon_0}$ does for$\vec{\mathbf{E}}$. The value of $\vec{\mathbf{g}}$ is calculated once and is then used in an endless number of problems.

To push the analogy further, notice the units of the electric field: From $F=QE$, the units of $E$ are newtons per coulomb, $\mathrm{N/C}$, that is, the electric field applies a force on each unit charge. Now notice the units of $g$: From $w=mg$, the units of $g$ are newtons per kilogram, $\mathrm{N/kg}$, that is, the gravitational field applies a force on each unit mass. We could say that the gravitational field of Earth, near Earth’s surface, has a value of $9.81~\mathrm{N/kg}$.

Recall from your studies of gravity that the word “field” in this context has a precise meaning. A field, in physics, is a physical quantity whose value depends on (is a function of) position, relative to the source of the field. In the case of the electric field, Equation \ref{eq:1.4.2} shows that the value of $\vec{\mathbf{E}}$ (both the magnitude and the direction) depends on where in space the point $P$ is located, measured from the locations $\vec{\mathbf{r}}_i$ of the source charges $q_i$.

In addition, since the electric field is a vector quantity, the electric field is referred to as a ** vector field**. (The gravitational field is also a vector field.) In contrast, a field that has only a magnitude at every point is a

Also, as you did with the gravitational field of an object with mass, you should picture the electric field of a charge-bearing object (the source charge) as a continuous, immaterial substance that surrounds the source charge, filling all of space—in principle, to $\pm\infty$ in all directions. The field exists at every physical point in space. To put it another way, the electric charge on an object alters the space around the charged object in such a way that all other electrically charged objects in space experience an electric force as a result of being in that field. The electric field, then, is the mechanism by which the electric properties of the source charge are transmitted to and through the rest of the universe. (Again, the range of the electric force is infinite.)

We will see in subsequent chapters that the speed at which electrical phenomena travel is the same as the speed of light. There is a deep connection between the electric field and light.

Yet another experimental fact about the field is that it obeys the superposition principle. In this context, that means that we can (in principle) calculate the total electric field of many source charges by calculating the electric field of only $q_1$ at position $P$, then calculate the field of $q_2$ at $P$, while—and this is the crucial idea—ignoring the field of, and indeed even the existence of, $q_1$. We can repeat this process, calculating the field of each individual source charge, independently of the existence of any of the other charges. The total electric field, then, is the vector sum of all these fields. That, in essence, is what Equation \ref{eq:1.4.2} says.

In the next section, we describe how to determine the shape of an electric field of a source charge distribution and how to sketch it.

Equation \ref{eq:1.4.2} enables us to determine the magnitude of the electric field, but we need the direction also. We use the convention that the direction of any electric field vector is the same as the direction of the electric force vector that the field would apply to a positive test charge placed in that field. Such a charge would be repelled by positive source charges (the force on it would point away from the positive source charge) but attracted to negative charges (the force points toward the negative source).

By convention, all electric fields $\vec{\mathbf{E}}$ point away from positive source charges and point toward negative source charges.

Add charges to the Electric Field of Dreams and see how they react to the electric field. Turn on a background electric field and adjust the direction and magnitude.

\[\vec{\mathbf{E}}=\frac{1}{4\pi\epsilon_0}\sum_{i=1}^{N}\frac{q_i}{r_i^2}\hat{\mathbf{r}}_i.\]

Since there is only one source charge (the nucleus), this expression simplifies to

\[\vec{\mathbf{E}}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{\mathbf{r}}.\]

Here $q=2e=2\left(1.6\times10^{-19}~\mathrm{C}\right)$ (since there are two protons) and $r$ is given; substituting gives

\[\vec{\mathbf{E}}=\frac{1}{4\pi\left(8.85\times10^{-12}~\frac{\mathrm{C}^2}{\mathrm{N}\cdot\mathrm{m}^2}\right)}\frac{2\left(1.6\times10^{-19}~\mathrm{C}\right)}{\left(26.5\times10^{-12}~\mathrm{m}\right)^2}\hat{\mathbf{r}}=4.1\times10^{12}~\frac{\mathrm{N}}{\mathrm{C}}\hat{\mathbf{r}}.\]

The direction of $\vec{\mathbf{E}}$ is radially away from the nucleus in all directions. Why? Because a positive test charge placed in this field would accelerate radially away from the nucleus (since it is also positively charged), and again, the convention is that the direction of the electric field vector is defined in terms of the direction of the force it would apply to positive test charges.

Since none of the other components survive, this is the entire electric field, and it points in the $\hat{\mathbf{k}}$-direction. Notice that this calculation uses the principle of **superposition**; we calculate the fields of the two charges independently and then add them together.

What we want to do now is replace the quantities in this expression that we don’t know (such as $r$), or can’t easily measure (such as $\cos\theta$) with quantities that we do know, or can measure. In this case, by geometry,
\[r^2=z^2+\left(\frac{d}{2}\right)^2\]
and
\[\cos\theta=\frac{z}{r}=\frac{z}{\left[z^2+\left(\frac{d}{2}\right)^2\right]^{1/2}}.\]

Thus, substituting,
\[\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\frac{2q}{\left[z^2+\left(\frac{d}{2}\right)^2\right]}\frac{z}{\left[z^2+\left(\frac{d}{2}\right)^2\right]^{1/2}}\hat{\mathbf{k}}.\]

Simplifying, the desired answer is
\begin{equation}\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\frac{2qz}{\left[z^2+\left(\frac{d}{2}\right)^2\right]^{3/2}}\hat{\mathbf{k}}.\tag{1.4.3}\label{eq:1.4.3}\end{equation}
(b) If the source charges are equal and opposite, the vertical components cancel because
\[E_z=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\cos\theta-\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\cos\theta=0.\]

and we get, for the horizontal component of$\vec{\mathbf{E}}$,
\begin{eqnarray*}\vec{\mathbf{E}}(z)&=&\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\sin\theta\hat{\mathbf{i}}-\frac{1}{4\pi\epsilon_0}\frac{-q}{r^2}\sin\theta\hat{\mathbf{i}}\\&=&\frac{1}{4\pi\epsilon_0}\frac{2q}{r^2}\sin\theta\hat{\mathbf{i}}\\&=&\frac{1}{4\pi\epsilon_0}\frac{2q}{\left[z^2+\left(\frac{d}{2}\right)^2\right]}\frac{\left(\frac{d}{2}\right)}{\left[z^2+\left(\frac{d}{2}\right)^2\right]^{1/2}}\hat{\mathbf{i}}.\end{eqnarray*}
This becomes
\begin{equation}\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\frac{qd}{\left[z^2+\left(\frac{d}{2}\right)^2\right]^{3/2}}\hat{\mathbf{i}}\tag{1.4.4}\label{eq:1.4.4}.\end{equation}

Let’s start with Equation \ref{eq:1.4.3}, the field of two identical charges. From far away (i.e., $z\gg d$), the two source charges should “merge” and we should then “see” the field of just one charge, of size $2q$. So, let $z\gg d$; then we can neglect $d^2$ in Equation \ref{eq:1.4.3} to obtain

\begin{eqnarray*}\lim_{d\rightarrow0}\vec{\mathbf{E}}&=&\frac{1}{4\pi\epsilon_0}\frac{2qz}{\left[z^2\right]^{3/2}}\hat{\mathbf{k}}\\&=&\frac{1}{4\pi\epsilon_0}\frac{2qz}{z^3}\hat{\mathbf{k}}\\&=&\frac{1}{4\pi\epsilon_0}\frac{2q}{z^2}\hat{\mathbf{k}},\end{eqnarray*}

which is the correct expression for a field at a distance $z$ away from a charge $2q$.

Next, we consider the field of equal and opposite charges, Equation \ref{eq:1.4.4}. It can be shown (via a Taylor expansion) that for $d\ll z\ll\infty$, this becomes

\begin{equation}\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\frac{qd}{z^3}\hat{\mathbf{i}}\tag{1.4.5}\label{eq:1.4.5},\end{equation}

which is the field of a dipole, a system that we will study in more detail later. (Note that the units of $\vec{\mathbf{E}}$ are still correct in this expression, since the units of $d$ in the numerator cancel the unit of the “extra” $z$ in the denominator.) If $z$ is very large $(z\rightarrow\infty)$, then $E\rightarrow0$, as it should; the two charges “merge” and so cancel out.

What is the electric field due to a single point particle?

Try this simulation of electric field hockey to get the charge in the goal by placing other charges on the field.

- Explain what a continuous source charge distribution is and how it is related to the concept of quantization of charge
- Describe line charges, surface charges, and volume charges
- Calculate the field of a continuous source charge distribution of either sign

The charge distributions we have seen so far have been discrete: made up of individual point particles. This is in contrast with a **continuous charge distribution**, which has at least one nonzero dimension. If a charge distribution is continuous rather than discrete, we can generalize the definition of the electric field. We simply divide the charge into infinitesimal pieces and treat each piece as a point charge.

Note that because charge is quantized, there is no such thing as a “truly” continuous charge distribution. However, in most practical cases, the total charge creating the field involves such a huge number of discrete charges that we can safely ignore the discrete nature of the charge and consider it to be continuous. This is exactly the kind of approximation we make when we deal with a bucket of water as a continuous fluid, rather than a collection of H2OH2O molecules.

Our first step is to define a charge density for a charge distribution along a line, across a surface, or within a volume, as shown in \ref{fig:1.5.1}. \begin{gather}.\tag{Figure 1.5.1}\label{fig:1.5.1}\end{gather}Definitions of charge density:

- $\lambda\equiv$ charge per unit length (
**linear charge density**); units are coulombs per metre ($\mathrm{C/m}$) - $\sigma\equiv$ charge per unit area (
**surface charge density**); units are coulombs per square metre ($\mathrm{C/m^2}$) - $\rho\equiv$ charge per unit volume (
**volume charge density**); units are coulombs per cubic metre ($\mathrm{C/m^3}$)

Then, for a line charge, a surface charge, and a volume charge, the summation in Equation 1.4.2 becomes an integral and $q_i$ is replaced by $dq=\lambda dl$, $\sigma dA$, or $\rho dV$ respectively:

\begin{equation}\mathrm{Point~charge:}&~~~~~&\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\sum_{i=1}^{N}\left(\frac{q_i}{r^2}\right)\hat{\mathbf{r}}\tag{1.5.1}\label{eq:1.5.1}\end{equation}

\begin{equation}\mathrm{Line~charge:}&~~~~~&\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{line}}\left(\frac{\lambda dl}{r^2}\right)\hat{\mathbf{r}}\tag{1.5.2}\label{eq:1.5.2}\end{equation}

\begin{equation}\mathrm{Surface~charge:}&~~~~~&\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{surface}}\left(\frac{\sigma dA}{r^2}\right)\hat{\mathbf{r}}\tag{1.5.3}\label{eq:1.5.3}\end{equation}

\begin{equation}\mathrm{Volume~charge:}&~~~~~&\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{volume}}\left(\frac{\rho dV}{r^2}\right)\hat{\mathbf{r}}\tag{1.5.4}\label{eq:1.5.4}\end{equation}

The integrals are generalizations of the expression for the field of a point charge. They implicitly include and assume the principle of superposition. The “trick” to using them is almost always in coming up with correct expressions for $dl$, $dA$, or $dV$ as the case may be, expressed in terms of $r$, and also expressing the charge density function appropriately. It may be constant; it might be dependent on location.

Note carefully the meaning of $r$ in these equations: It is the distance from the charge element $(q_i,\lambda dl,\sigma dA,\rho dV)$ to the location of interest, $P(x,y,z)$ (the point in space where you want to determine the field). However, don’t confuse this with the meaning of $\hat{\mathbf{r}}$; we are using it and the vector notation $\vec{\mathbf{E}}$ to write three integrals at once. That is, Equation \ref{eq:1.5.2} is actually

\[E_x(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{line}}\left(\frac{\lambda dl}{r^2}\right)_x,\] \[E_y(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{line}}\left(\frac{\lambda dl}{r^2}\right)_y,\] \[E_z(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{line}}\left(\frac{\lambda dl}{r^2}\right)_z.\]The electric field for a line charge is given by the general expression

\[\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{line}}\frac{\lambda dl}{r^2}\hat{\mathbf{r}}.\]The symmetry of the situation (our choice of the two identical differential pieces of charge) implies the horizontal ($x$)-components of the field cancel, so that the net field points in the $z$-direction. Let’s check this formally.

The total field $\vec{\mathbf{E}}(P)$ is the vector sum of the fields from each of the two charge elements (call them $\vec{\mathbf{E}}_1$ and $\vec{\mathbf{E}}_2$, for now):

\[\vec{\mathbf{E}}(P)=\vec{\mathbf{E}}_1+\vec{\mathbf{E}}_2=E_{ix}\hat{\mathbf{i}}+E_{1z}\hat{\mathbf{k}}+E_{2x}(-\hat{\mathbf{i}})+E_{2z}\hat{\mathbf{k}}.\]Because the two charge elements are identical and are the same distance away from the point $P$ where we want to calculate the field, $E_{1x}=E_{2x}$, so those components cancel. This leaves

\[\vec{\mathbf{E}}(P)=E_{1z}\hat{\mathbf{k}}+E_{2z}\hat{\mathbf{k}}=E_1\cos\theta\hat{\mathbf{k}}+E_2\cos\theta\hat{\mathbf{k}}.\]

These components are also equal, so we have

\begin{eqnarray*}\hat{\mathbf{E}}(P)&=&\frac{1}{4\pi\epsilon_0}\int\frac{\lambda dl}{r^2}\cos\theta\hat{\mathbf{k}}+\frac{1}{4\pi\epsilon_0}\int\frac{\lambda dl}{r^2}\cos\theta\hat{\mathbf{k}}\\&=&\int_0^{L/2}\frac{2\lambda dx}{r^2}\cos\theta\hat{\mathbf{k}}\end{eqnarray*}

where our differential line element $dl$ is $dx$, in this example, since we are integrating along a line of charge that lies on the $x$-axis. (The limits of integration are $0$ to $\frac{L}{2}$, not $-\frac{L}{2}$ to $+\frac{L}{2}$, because we have constructed the net field from two differential pieces of charge $dq$. If we integrated along the entire length, we would pick up an erroneous factor of $2$.)

In principle, this is complete. However, to actually calculate this integral, we need to eliminate all the variables that are not given. In this case, both $r$ and $\theta$ change as we integrate outward to the end of the line charge, so those are the variables to get rid of. We can do that the same way we did for the two point charges: by noticing that

\[r=\left(z^2+x^2\right)^{1/2}\]

and

\[\cos\theta=\frac{z}{r}=\frac{z}{\left(z^2+x^2\right)^{1/2}}.\]

Substituting, we obtain

\begin{eqnarray*}\vec{\mathbf{E}}(P)&=&\frac{1}{4\pi\epsilon_0}\int_0^{L/2}\frac{2\lambda dx}{\left(z^2+x^2\right)}\frac{z}{\left(z^2+x^2\right)^{1/2}}\hat{\mathbf{k}}\\&=&\frac{1}{4\pi\epsilon_0}\int_0^{L/2}\frac{2\lambda z}{\left(z^2+x^2\right)^{3/2}}dx\hat{\mathbf{k}}\\&=&\frac{2\lambda z}{4\pi\epsilon_0}\left.\left[\frac{x}{z^2\sqrt{z^2+x^2}}\right]\right|_0^{L/2}\hat{\mathbf{k}}\end{eqnarray*}

which simplifies to

\begin{equation}\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\frac{\lambda L}{z\sqrt{z^2+\frac{L^2}{4}}}\hat{\mathbf{k}}.\tag{1.5.5}\label{eq:1.5.5}\end{equation}How would the strategy used above change to calculate the electric field at a point a distance $z$ above one end of the finite line segment?

where our differential line element $dl$ is $dx$, in this example, since we are integrating along a line of charge that lies on the $x$-axis. Again,

\[\cos\theta=\frac{z}{r}=\frac{z}{\left(z^2+x^2\right)^{1/2}}.\]Substituting, we obtain

\begin{eqnarray*}\vec{\mathbf{E}}(P)&=&\frac{1}{4\pi\epsilon_0}\int_{-\infty}^{\infty}\frac{\lambda dx}{\left(z^2+x^2\right)}\frac{z}{\left(z^2+x^2\right)^{1/2}}\hat{\mathbf{k}}\\&=&\frac{1}{4\pi\epsilon_0}\int_{-\infty}^{\infty}\frac{\lambda z}{\left(z^2+x^2\right)^{3/2}}dx\hat{\mathbf{k}}\\&=&\frac{\lambda z}{4\pi\epsilon_0}\left.\left[\frac{x}{z^2\sqrt{z^2+x^2}}\right]\right|_{-\infty}^{\infty}\hat{\mathbf{k}},\end{eqnarray*}which simplifies to

\[\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\frac{2\lambda}{z}\hat{\mathbf{k}}.\]

In the case of a finite line of charge, note that for $z\gg L$, $z^2$ dominates the $L$ in the denominator, so that Equation \ref{eq:1.5.5} simplifies to

\[\vec{\mathbf{E}}\approx\frac{1}{4\pi\epsilon_0}\frac{\lambda L}{z^2}\hat{\mathbf{k}}.\] If you recall that $\lambda L=q$, the total charge on the wire, we have retrieved the expression for the field of a point charge, as expected.In the limit $L\rightarrow\infty$, on the other hand, we get the field of an **infinite straight wire**, which is a straight wire whose length is much, much greater than either of its other dimensions, and also much, much greater than the distance at which the field is to be calculated:

\begin{equation}\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\frac{2\lambda}{z}\hat{\mathbf{k}}.\tag{1.5.6}\label{eq:1.5.6}\end{equation}

An interesting artifact of this infinite limit is that we have lost the usual $1/r^2$ dependence that we are used to. This will become even more intriguing in the case of an infinite plane.

A general element of the arc between $\theta$ and $\theta+d\theta$ is of length $Rd\theta$ and therefore contains a charge equal to $\lambda Rd\theta$. The element is at a distance of $r=\sqrt{z^2+R^2}$ from $P$, the angle is $\cos\phi=\frac{z}{\sqrt{z^2+R^2}}$, and therefore the electric field is

\begin{eqnarray*}\vec{\mathbf{E}}(P)&=&\int_{\mathrm{line}}\frac{\lambda dl}{r^2}\hat{\mathbf{r}}=\frac{1}{4\pi\epsilon_0}\int_0^{2\pi}\frac{\lambda Rd\theta}{z^2+R^2}\frac{z}{\sqrt{z^2+R^2}}\hat{\mathbf{z}}\\&=&\frac{1}{4\pi\epsilon_0}\frac{\lambda Rz}{\left(z^2+R^2\right)^{3/2}}\hat{\mathbf{z}}\int_0^{2\pi}d\theta=\frac{1}{4\pi\epsilon_0}\frac{2\pi\lambda Rz}{\left(z^2+R^2\right)^{3/2}}\hat{\mathbf{z}}\\&=&=\frac{1}{4\pi\epsilon_0}\frac{q_{\mathrm{tot}}z}{\left(z^2+R^2\right)^{3/2}}\hat{\mathbf{z}}.\end{eqnarray*}
\[\vec{\mathbf{E}}\approx\frac{1}{4\pi\epsilon_0}\frac{q_{\mathrm{tot}}}{z^2}\hat{\mathbf{z}},\]

as we expect.

To solve surface charge problems, we break the surface into symmetrical differential “stripes” that match the shape of the surface; here, we’ll use rings, as shown in the figure. Again, by symmetry, the horizontal components cancel and the field is entirely in the vertical $\hat{\mathbf{k}}$-direction. The vertical component of the electric field is extracted by multiplying by $\cos\theta$, so

\[\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{surface}}\frac{\sigma dA}{r^2}\cos\theta\hat{\mathbf{k}}.\] As before, we need to rewrite the unknown factors in the integrand in terms of the given quantities. In this case,
\begin{eqnarray*}dA&=&2\pi r'dr'\\r^2&=&r'^2+z^2\\\cos\theta&=&\frac{z}{\left(r'^2+z^2\right)^{1/2}}.\end{eqnarray*}

(Please take note of the two different "$r$s" here; $r$ is the distance from the differential ring of charge to the point $P$ where we wish to determine the field, whereas $r'$ is the distance from the centre of the disk to the differential ring of charge.) Also, we already performed the polar angle integral in writing down $dA$.

\begin{eqnarray*}\vec{\mathbf{E}}(P)&=&\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\int_0^R\frac{\sigma(2\pi r'dr')z}{(r'^2+z^2)^{3/2}}\hat{\mathbf{k}}\\&=&\frac{1}{4\pi\epsilon_0}(2\pi\sigma z)\left(\frac{1}{z}-\frac{1}{\sqrt{R^2+z^2}}\right)\hat{\mathbf{k}}\end{eqnarray*}

or, more simply,

\begin{equation}\hat{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\left(2\pi\sigma-\frac{2\pi\sigma z}{\sqrt{R^2+z^2}}\right)\hat{\mathbf{k}}.\tag{1.5.7}\label{eq:1.5.7}\end{equation}
\[\vec{\mathbf{E}}(z)\approx\frac{1}{4\pi\epsilon_0}\frac{\sigma\pi R^2}{z^2}\hat{\mathbf{k}},\]

which is the expression for a point charge $Q=\sigma\pi R^2$.

How would the above limit change with a uniformly charged rectangle instead of a disk?

As $R\rightarrow\infty$, Equation \ref{eq:1.5.7} reduces to the field of an **infinite plane**, which is a flat sheet whose area is much, much greater than its thickness, and also much, much greater than the distance at which the field is to be calculated:

\begin{equation}\vec{\mathbf{E}}=\frac{\sigma}{2\epsilon_0}\hat{\mathbf{k}}\tag{1.5.8}\label{eq:1.5.8}\end{equation}

Note that this field is constant. This surprising result is, again, an artifact of our limit, although one that we will make use of repeatedly in the future. To understand why this happens, imagine being placed above an infinite plane of constant charge. Does the plane look any different if you vary your altitude? No—you still see the plane going off to infinity, no matter how far you are from it. It is important to note that Equation \ref{eq:1.5.8} is because we are above the plane. If we were below, the field would point in the $-\hat{\mathbf{k}}$-direction.

However, in the region between the planes, the electric fields add, and we get

\[\vec{\mathbf{E}}=\frac{\sigma}{\epsilon_0}\hat{\mathbf{i}}\] for the electric field. The $\hat{\mathbf{i}}$ is because in the figure, the field is pointing in the $+x$-direction.What would the electric field look like in a system with two parallel positively charged planes with equal charge densities?

- Explain the purpose of an electric field diagram
- Describe the relationship between a vector diagram and a field line diagram
- Explain the rules for creating a field diagram and why these rules make physical sense
- Sketch the field of an arbitrary source charge

Now that we have some experience calculating electric fields, let’s try to gain some insight into the geometry of electric fields. As mentioned earlier, our model is that the charge on an object (the source charge) alters space in the region around it in such a way that when another charged object (the test charge) is placed in that region of space, that test charge experiences an electric force. The concept of electric **field lines**, and of electric field line diagrams, enables us to visualize the way in which the space is altered, allowing us to visualize the field. The purpose of this section is to enable you to create sketches of this geometry, so we will list the specific steps and rules involved in creating an accurate and useful sketch of an electric field.

It is important to remember that electric fields are three-dimensional. Although in this book we include some pseudo-three-dimensional images, several of the diagrams that you’ll see (both here, and in subsequent chapters) will be two-dimensional projections, or cross-sections. Always keep in mind that in fact, you’re looking at a three-dimensional phenomenon.

Our starting point is the physical fact that the electric field of the source charge causes a test charge in that field to experience a force. By definition, electric field vectors point in the same direction as the electric force that a (hypothetical) positive test charge would experience, if placed in the field (\ref{fig:1.6.1}). \begin{gather}.\tag{Figure 1.6.1}\label{fig:1.6.1}\end{gather}We’ve plotted many field vectors in the figure, which are distributed uniformly around the source charge. Since the electric field is a vector, the arrows that we draw correspond at every point in space to both the magnitude and the direction of the field at that point. As always, the length of the arrow that we draw corresponds to the magnitude of the field vector at that point. For a point source charge, the length decreases by the square of the distance from the source charge. In addition, the direction of the field vector is radially away from the source charge, because the direction of the electric field is defined by the direction of the force that a positive test charge would experience in that field. (Again, keep in mind that the actual field is three-dimensional; there are also field lines pointing out of and into the page.)

This diagram is correct, but it becomes less useful as the source charge distribution becomes more complicated. For example, consider the vector field diagram of a dipole (\ref{fig:1.6.2}). \begin{gather}.\tag{Figure 1.6.2}\label{fig:1.6.2}\end{gather}Although it may not be obvious at first glance, these field diagrams convey the same information about the electric field as do the vector diagrams. First, the direction of the field at every point is simply the direction of the field vector at that same point. In other words, at any point in space, the field vector at each point is tangent to the field line at that same point. The arrowhead placed on a field line indicates its direction.

As for the magnitude of the field, that is indicated by theIn \ref{fig:1.6.4}, the same number of field lines passes through both surfaces ($S$ and $S'$), but the surface $S$ is larger than surface $S'$. Therefore, the density of field lines (number of lines per unit area) is larger at the location of $S'$, indicating that the electric field is stronger at the location of $S'$ than at $S$. The rules for creating an electric field diagram are as follows.

- Electric field lines either originate on positive charges or come in from infinity, and either terminate on negative charges or extend out to infinity.
- The number of field lines originating or terminating at a charge is proportional to the magnitude of that charge. A charge of $2q$ will have twice as many lines as a charge of $q$.
- At every point in space, the field vector at that point is tangent to the field line at that same point.
- The field line density at any point in space is proportional to (and therefore is representative of) the magnitude of the field at that point in space.
- Field lines can never cross. Since a field line represents the direction of the field at a given point, if two field lines crossed at some point, that would imply that the electric field was pointing in two different directions at a single point. This in turn would suggest that the (net) force on a test charge placed at that point would point in two different directions. Since this is obviously impossible, it follows that field lines must never cross.

Always keep in mind that field lines serve only as a convenient way to visualize the electric field; they are not physical entities. Although the direction and relative intensity of the electric field can be deduced from a set of field lines, the lines can also be misleading. For example, the field lines drawn to represent the electric field in a region must, by necessity, be discrete. However, the actual electric field in that region exists at every point in space.

Field lines for three groups of discrete charges are shown in \ref{fig:1.6.5}. Since the charges in parts (a) and (b) have the same magnitude, the same number of field lines are shown starting from or terminating on each charge. In (c), however, we draw three times as many field lines leaving the $+3q$ charge as entering the $-q$. The field lines that do not terminate at $-q$ emanate outward from the charge configuration, to infinity. \begin{gather}.\tag{Figure 1.6.5}\label{fig:1.6.5}\end{gather}The ability to construct an accurate electric field diagram is an important, useful skill; it makes it much easier to estimate, predict, and therefore calculate the electric field of a source charge. The best way to develop this skill is with software that allows you to place source charges and then will draw the net field upon request. We strongly urge you to search the Internet for a program. Once you’ve found one you like, run several simulations to get the essential ideas of field diagram construction. Then practice drawing field diagrams, and checking your predictions with the computer-drawn diagrams.

One example of a field-line drawing program is from the PhET “Charges and Fields” simulation.

Coulomb’s law | \[\vec{\mathbf{F}}_{12}(r)=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}^2}\hat{\mathbf{r}}_{12}\] |

Superposition of electric forces | \[\vec{\mathbf{F}}(r)=\frac{1}{4\pi\epsilon_0}Q\sum_{i=1}^{N}\frac{q_i}{r_{i}^2}\hat{\mathbf{r}}_{i}\] |

Electric force due to an electric field | \[\vec{\mathbf{F}}=Q\vec{\mathbf{E}}\] |

Electric field at point $P$ | \[\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\sum_{i=1}^{N}\frac{q_i}{r_{i}^2}\hat{\mathbf{r}}_{i}\] |

Field of an infinite wire | \[\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\frac{2\lambda}{z}\hat{\mathbf{k}}\] |

Field of an infinite plane | \[\vec{\mathbf{E}}=\frac{\sigma}{2\epsilon_0}\hat{\mathbf{k}}\] |

Dipole moment | \[\vec{\mathbf{p}}=q\vec{\mathbf{d}}\] |

Torque on dipole in external $E$-field | \[\vec{\mathbf{\tau}}=\vec{\mathbf{p}}\times\vec{\mathbf{E}}\] |

- There are only two types of charge, which we call positive and negative. Like charges repel, unlike charges attract, and the force between charges decreases with the square of the distance.
- The vast majority of positive charge in nature is carried by protons, whereas the vast majority of negative charge is carried by electrons. The electric charge of one electron is equal in magnitude and opposite in sign to the charge of one proton.
- An ion is an atom or molecule that has nonzero total charge due to having unequal numbers of electrons and protons.
- The SI unit for charge is the coulomb ($\mathrm{C}$), with protons and electrons having charges of opposite sign but equal magnitude; the magnitude of this basic charge is $e=1.602\times10^{-19}~\mathrm{C}$
- Both positive and negative charges exist in neutral objects and can be separated by bringing the two objects into physical contact; rubbing the objects together can remove electrons from the bonds in one object and place them on the other object, increasing the charge separation.
- For macroscopic objects, negatively charged means an excess of electrons and positively charged means a depletion of electrons.
- The law of conservation of charge states that the net charge of a closed system is constant.

- A conductor is a substance that allows charge to flow freely through its atomic structure.
- An insulator holds charge fixed in place.
- Polarization is the separation of positive and negative charges in a neutral object. Polarized objects have their positive and negative charges concentrated in different areas, giving them a charge distribution.

- Coulomb’s law gives the magnitude of the force between point charges. It is \[\vec{\mathbf{F}}_{12}(r)=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}^2}\hat{\mathbf{r}}_{12}\] where $q_1$ and $q_2$ are two point charges separated by a distance $r$. This Coulomb force is extremely basic, since most charges are due to point-like particles. It is responsible for all electrostatic effects and underlies most macroscopic forces.

- The electric field is an alteration of space caused by the presence of an electric charge. The electric field mediates the electric force between a source charge and a test charge.
- The electric field, like the electric force, obeys the superposition principle
- The field is a vector; by definition, it points away from positive charges and toward negative charges.

- A very large number of charges can be treated as a continuous charge distribution, where the calculation of the field requires integration. Common cases are:
- one-dimensional (like a wire); uses a line charge density $\lambda$
- two-dimensional (metal plate); uses surface charge density $\sigma$
- three-dimensional (metal sphere); uses volume charge density $\rho$

- The “source charge” is a differential amount of charge $dq$. Calculating $dq$ depends on the type of source charge distribution: \[dq=\lambda dl;~~dq=\sigma dA;~~dq=\rho dV.\]
- Symmetry of the charge distribution is usually key.
- Important special cases are the field of an “infinite” wire and the field of an “infinite” plane.

- Electric field diagrams assist in visualizing the field of a source charge.
- The magnitude of the field is proportional to the field line density.
- Field vectors are everywhere tangent to field lines.

- If a permanent dipole is placed in an external electric field, it results in a torque that aligns it with the external field.
- If a nonpolar atom (or molecule) is placed in an external field, it gains an induced dipole that is aligned with the external field.
- The net field is the vector sum of the external field plus the field of the dipole (physical or induced).
- The strength of the polarization is described by the dipole moment of the dipole, $\vec{\mathbf{p}}=q\vec{\mathbf{d}}$.

1.1 The force would point outward.

1.2 The net force would point $58^{\circ}$ below the $-x$-axis. 1.3 $\vec{\mathbf{E}}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$ 1.4 We will no longer be able to take advantage of symmetry. Instead, we will need to calculate each of the two components of the electric field with their own integral. 1.5 The point charge would be $Q=\sigma ab$ where $a$ and $b$ are the sides of the rectangle but otherwise identical. 1.6 The electric field would be zero in between, and have magnitude $\frac{\sigma}{\epsilon_0}$ everywhere else.
1. There are very large numbers of charged particles in most objects. Why, then, don’t most objects exhibit static electricity?

2. Why do most objects tend to contain nearly equal numbers of positive and negative charges?

3. A positively charged rod attracts a small piece of cork. (a) Can we conclude that the cork is negatively charged? (b) The rod repels another small piece of cork. Can we conclude that this piece is positively charged?

4. Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.

5. How would you determine whether the charge on a particular rod is positive or negative?

6. An eccentric inventor attempts to levitate a cork ball by wrapping it with foil and placing a large negative charge on the ball and then putting a large positive charge on the ceiling of his workshop. Instead, while attempting to place a large negative charge on the ball, the foil flies off. Explain.

7. When a glass rod is rubbed with silk, it becomes positive and the silk becomes negative—yet both attract dust. Does the dust have a third type of charge that is attracted to both positive and negative? Explain.

8. Why does a car always attract dust right after it is polished? (Note that car wax and car tires are insulators.)

9. Does the uncharged conductor shown below experience a net electric force?

10. While walking on a rug, a person frequently becomes charged because of the rubbing between his shoes and the rug. This charge then causes a spark and a slight shock when the person gets close to a metal object. Why are these shocks so much more common on a dry day?

11. Compare charging by conduction to charging by induction.

12. Small pieces of tissue are attracted to a charged comb. Soon after sticking to the comb, the pieces of tissue are repelled from it. Explain.

13. Trucks that carry gasoline often have chains dangling from their undercarriages and brushing the ground. Why?

14. Why do electrostatic experiments work so poorly in humid weather?

15. Why do some clothes cling together after being removed from the clothes dryer? Does this happen if they’re still damp?

16. Can induction be used to produce charge on an insulator?

17. Suppose someone tells you that rubbing quartz with cotton cloth produces a third kind of charge on the quartz. Describe what you might do to test this claim.

18. A handheld copper rod does not acquire a charge when you rub it with a cloth. Explain why.

19. Suppose you place a charge $q$ near a large metal plate. (a) If $q$ is attracted to the plate, is the plate necessarily charged? (b) If $q$ is repelled by the plate, is the plate necessarily charged?

20. Would defining the charge on an electron to be positive have any effect on Coulomb’s law?

21. An atomic nucleus contains positively charged protons and uncharged neutrons. Since nuclei do stay together, what must we conclude about the forces between these nuclear particles?

22. Is the force between two fixed charges influenced by the presence of other charges?

23. When measuring an electric field, could we use a negative rather than a positive test charge?

24. During fair weather, the electric field due to the net charge on Earth points downward. Is Earth charged positively or negatively?

25. If the electric field at a point on the line between two charges is zero, what do you know about the charges?

26. Two charges lie along the $x$-axis. Is it true that the net electric field always vanishes at some point (other than infinity) along the $x$-axis?

27. Give a plausible argument as to why the electric field outside an infinite charged sheet is constant.

28. Compare the electric fields of an infinite sheet of charge, an infinite, charged conducting plate, and infinite, oppositely charged parallel plates.

29. Describe the electric fields of an infinite charged plate and of two infinite, charged parallel plates in terms of the electric field of an infinite sheet of charge.

30. A negative charge is placed at the center of a ring of uniform positive charge. What is the motion (if any) of the charge? What if the charge were placed at a point on the axis of the ring other than the center?

31. If a point charge is released from rest in a uniform electric field, will it follow a field line? Will it do so if the electric field is not uniform?

32. Under what conditions, if any, will the trajectory of a charged particle not follow a field line?

33. How would you experimentally distinguish an electric field from a gravitational field?

34. A representation of an electric field shows 10 field lines perpendicular to a square plate. How many field lines should pass perpendicularly through the plate to depict a field with twice the magnitude?

35. What is the ratio of the number of electric field lines leaving a charge 10$q$ and a charge $q$?
## 1.7 Electric Dipoles

36. What are the stable orientation(s) for a dipole in an external electric field? What happens if the dipole is slightly perturbed from these orientations?

37. Common static electricity involves charges ranging from nanocoulombs to microcoulombs. (a) How many electrons are needed to form a charge of $−2.00~\mathrm{nC}$? (b) How many electrons must be removed from a neutral object to leave a net charge of $0.500~\mu\mathrm{C}$?

38. If $1.80\times10^{20}$ electrons move through a pocket calculator during a full day’s operation, how many coulombs of charge moved through it?

39. To start a car engine, the car battery moves $3.75\times10^{21}$ electrons through the starter motor. How many coulombs of charge were moved?

40. A certain lightning bolt moves $40.0~\mathrm{C}$ of charge. How many fundamental units of charge is this?

41. A $2.5$-$\mathrm{g}$ copper penny is given a charge of $-2.0\times10^{-9}~\mathrm{C}$. (a) How many excess electrons are on the penny? (b) By what percent do the excess electrons change the mass of the penny?

42. A $2.5$-$\mathrm{g}$ copper penny is given a charge of $4.0\times10^{-9}~\mathrm{C}$. (a) How many electrons are removed from the penny? (b) If no more than one electron is removed from an atom, what percent of the atoms are ionized by this charging process?

43. Suppose a speck of dust in an electrostatic precipitator has $1.0000\times10^{12}$ protons in it and has a net charge of $−5.00~\mathrm{nC}$ (a very large charge for a small speck). How many electrons does it have?

44. An amoeba has $1.00\times{10}^{16}$ protons and a net charge of $0.300~\mathrm{pC}$. (a) How many fewer electrons are there than protons? (b) If you paired them up, what fraction of the protons would have no electrons?

45. A $50.0$-$\mathrm{g}$ ball of copper has a net charge of $2.00~\mu\mathrm{C}$. What fraction of the copper’s electrons has been removed? (Each copper atom has $29$ protons, and copper has an atomic mass of $63.5~\mathrm{u}$.)

46. What net charge would you place on a $100$-$\mathrm{g}$ piece of sulfur if you put an extra electron on $1$ in $10^{12}$ of its atoms? (Sulfur has an atomic mass of $32.1~\mathrm{u}$.)

47. How many coulombs of positive charge are there in $4.00\mathrm{kg}$ of plutonium, given its atomic mass is $244$ and that each plutonium atom has $94$ protons?

48. Two point particles with charges $+3~\mu\mathrm{C}$ and $+5~\mu\mathrm{C}$ are held in place by $3$-$\mathrm{N}$ forces on each charge in appropriate directions. (a) Draw a free-body diagram for each particle. (b) Find the distance between the charges.

49. Two charges $+3~\mu\mathrm{C}$ and $+12~\mu\mathrm{C}$ are fixed $1~\mathrm{m}$ apart, with the second one to the right. Find the magnitude and direction of the net force on a $-2$-$\mathrm{nC}$ charge when placed at the following locations: (a) halfway between the two (b) half a meter to the left of the $+3~\mu\mathrm{C}$ charge (c) half a meter above the $+12~\mu\mathrm{C}$ charge in a direction perpendicular to the line joining the two fixed charges.

50. In a salt crystal, the distance between adjacent sodium and chloride ions is $2.82\times10^{-10}~\mathrm{m}$. What is the force of attraction between the two singly charged ions?

51. Protons in an atomic nucleus are typically $10^{-15}~\mathrm{m}$ apart. What is the electric force of repulsion between nuclear protons?

52. Suppose Earth and the Moon each carried a net negative charge $-Q$. Approximate both bodies as point masses and point charges. (a) What value of $Q$ is required to balance the gravitational attraction between Earth and the Moon? (b) Does the distance between Earth and the Moon affect your answer? Explain. (c) How many electrons would be needed to produce this charge?

53. Point charges $q_1=50~\mu\mathrm{C}$ and $q_2=-25~\mu\mathrm{C}$ are placed $1.0~\mathrm{m}$ apart. What is the force on a third charge $q_3=10~\mu\mathrm{C}$ placed midway between $q_1$ and $q_2$?

54. Where must $q_3$ of the preceding problem be placed so that the net force on it is zero?

55. Two small balls, each of mass $5.0~\mathrm{g}$, are attached to silk threads $50\mathrm{cm}$ long, which are in turn tied to the same point on the ceiling, as shown below. When the balls are given the same charge $Q$, the threads hang at $5.0^{\circ}$ to the vertical, as shown below. What is the magnitude of $Q$? What are the signs of the two charges?

56. Point charges $Q_1=2.0~\mu\mathrm{C}$ and $Q_2=4.0~\mu\mathrm{C}$ are located at $\vec{\mathbf{r}}_1=(4.0\hat{\mathbf{i}}-2.0\hat{\mathbf{j}}+5.0\hat{\mathbf{k}})~\mathrm{m}$ and $\vec{\mathbf{r}}_2=(8.0\hat{\mathbf{i}}+5.0\hat{\mathbf{j}}-9.0\hat{\mathbf{k}})~\mathrm{m}$. What is the force of $Q_2$ on $Q_1$?

57. The net excess charge on two small spheres (small enough to be treated as point charges) is $Q$. Show that the force of repulsion between the spheres is greatest when each sphere has an excess charge $Q/2$. Assume that the distance between the spheres is so large compared with their radii that the spheres can be treated as point charges.

58. Two small, identical conducting spheres repel each other with a force of $0.050~\mathrm{N}$ when they are $0.25~\mathrm{m}$ apart. After a conducting wire is connected between the spheres and then removed, they repel each other with a force of $0.060~\mathrm{N}$. What is the original charge on each sphere?

59. A charge $q=2.0~\mu\mathrm{C}$ is placed at the point $P$ shown below. What is the force on $q$?

60. What is the net electric force on the charge located at the lower right-hand corner of the triangle shown here?

61. Two fixed particles, each of charge $5.0\times10^{-6}~\mathrm{C}$, are $24~\mathrm{cm}$ apart. What force do they exert on a third particle of charge $-2.5\times10^{-6}~\mathrm{C}$ that is $13~\mathrm{cm}$ from each of them?

62. The charges $q_1=2.0\times10^{-7}~\mathrm{C}$, $q_2=-4.0\times10^{-7}~\mathrm{C}$, and $q_3=-1.0\times10^{-7}~\mathrm{C}$ are placed at the corners of the triangle shown below. What is the force on $q_1$?

63. What is the force on the charge $q$ at the lower-right-hand corner of the square shown here?

64. Point charges $q_1=10~\mu\mathrm{C}$ and $q_2=-30~\mu\mathrm{C}$ are fixed at $\vec{\mathbf{r}}_1=(3.0\hat{\mathbf{i}}-4.0\hat{\mathbf{j}})~\mathrm{m}$ and$\vec{\mathbf{r}}_1=(9.0\hat{\mathbf{i}}+6.0\hat{\mathbf{j}})~\mathrm{m}$. What is the force of $q_2$ on $q_1$?

65. A particle of charge $2.0\times10^{-8}~\mathrm{C}$ experiences an upward force of magnitude $4.0\times10^{-6}~\mathrm{N}$ when it is placed in a particular point in an electric field. (a) What is the electric field at that point? (b) If a charge $q=-1.0\times10^{-8}~\mathrm{C}$ is placed there, what is the force on it?

66. On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately $100~\mathrm{N/C}$. Compare the gravitational and electric forces on a small dust particle of mass $2.0\times10^{-15}~\mathrm{g}$ that carries a single electron charge. What is the acceleration (both magnitude and direction) of the dust particle?

67. Consider an electron that is $10^{-10}~\mathrm{m}$ from an alpha particle ($q=3.2\times10^{-19}~\mathrm{C}$). (a) What is the electric field due to the alpha particle at the location of the electron? (b) What is the electric field due to the electron at the location of the alpha particle? (c) What is the electric force on the alpha particle? On the electron?

68. Each the balls shown below carries a charge $q$ and has a mass $m$. The length of each thread is $l$, and at equilibrium, the balls are separated by an angle $2\theta$. How does $\theta$ vary with $q$ and $l$? Show that $\theta$ satisfies $\sin^{2}\theta\tan\theta=\frac{q^2}{16\pi\epsilon_0gl^2m}$.

69. What is the electric field at a point where the force on a charge $q=-2.0\times10^{-6}~\mathrm{C}$ is $\left(4.0\hat{\mathbf{i}}-6.0\hat{\mathbf{j}}\right)\times10^{-6}~\mathrm{N}$?

70. A proton is suspended in the air by an electric field at the surface of Earth. What is the strength of this electric field?

71. The electric field in a particular thundercloud is $2.0\times10^{5}~\mathrm{N/C}$. What is the acceleration of an electron in this field?

72. A small piece of cork whose mass is $2.0~\mathrm{g}$ is given a charge of $5.0\times10^{-7}~\mathrm{C}$. What electric field is needed to place the cork in equilibrium under the combined electric and gravitational forces?

73. If the electric field is $100~\mathrm{N/C}$ at a distance of $50~\mathrm{cm}$ from a point charge $q$, what is the value of $q$?

74. What is the electric field of a proton at the first Bohr orbit for hydrogen ($r=5.29\times10^{-11}~\mathrm{m}$)? What is the force on the electron in that orbit?

75. (a) What is the electric field of an oxygen nucleus at a point that is $10^{-10}~\mathrm{m}$ from the nucleus? (b) What is the force this electric field exerts on a second oxygen nucleus placed at that point?

76. Two point charges, $q_1=2.0\times10^{-7}~\mathrm{C}$ and $q_2=-6.0\times10^{-8}~\mathrm{C}$, are held $25.0~\mathrm{cm}$ apart. (a) What is the electric field at a point $5.0~\mathrm{cm}$ from the negative charge and along the line between the two charges? (b)What is the force on an electron placed at that point?

77. Point charges $q_1=50~\mu\mathrm{C}$ and $q_1=-25~\mu\mathrm{C}$ are placed $1.0~\mathrm{m}$ apart. (a) What is the electric field at a point midway between them? (b) What is the force on a charge $q_3=20~\mu\mathrm{C}$ situated there?

78. Can you arrange the two point charges $q_1=-2.0\times10^{-6}~\mathrm{C}$ and $q_2=4.0\times10^{-6}~\mathrm{C}$ along the $x$-axis so that $E=0$ at the origin?

79. Point charges $q_1=q_2=4.0\times10^{-6}~\mathrm{C}$ are fixed on the $x$-axis at $x=-3.0~\mathrm{m}$ and $x=3.0~\mathrm{m}$. What charge $q$ must be placed at the origin so that the electric field vanishes at $x=0$, $y=3.0~\mathrm{m}$

80. A thin conducting plate $1.0~\mathrm{m}$ on the side is given a charge of $-2.0\times10^{-6}~\mathrm{C}$. An electron is placed $1.0~\mathrm{cm}$ above the centre of the plate. What is the acceleration of the electron?

81. Calculate the magnitude and direction of the electric field 2.0 m from a long wire that is charged uniformly at $\lambda=4.0\times10^{-6}~\mathrm{C/m}$.

82. Two thin conducting plates, each $25.0~\mathrm{cm}$ on a side, are situated parallel to one another and $5.0~\mathrm{mm}$ apart. If $10^{11}$ electrons are moved from one plate to the other, what is the electric field between the plates?

83. The charge per unit length on the thin rod shown below is $\lambda$. What is the electric field at the point $P$? (*Hint*: Solve this problem by first considering the electric field $d\vec{\mathbf{E}}$ at $P$ due to a small segment $dx$ of the rod, which contains charge $dq=\lambda dx$. Then find the net field by integrating $d\vec{\mathbf{E}}$ over the length of the rod.)

84. The charge per unit length on the thin semicircular wire shown below is $\lambda$. What is the electric field at the point $P$?

85. Two thin parallel conducting plates are placed $2.0~\mathrm{cm}$ apart. Each plate is $2.0~\mathrm{cm}$ on a side; one plate carries a net charge of $8.0~\mu\mathrm{C},$ and the other plate carries a net charge of $-8.0~\mu\mathrm{C}.$ What is the charge density on the inside surface of each plate? What is the electric field between the plates?

86. A thin conducing plate $2.0~\mathrm{m}$ on a side is given a total charge of $-10.0~\mu\mathrm{C}.$ (a) What is the electric field $1.0~\mathrm{cm}$ above the plate? (b) What is the force on an electron at this point? (c) Repeat these calculations for a point $2.0~\mathrm{cm}$ above the plate. (d) When the electron moves from $1.0$ to $2.0~\mathrm{cm}$ above the plate, how much work is done on it by the electric field?

87. A total charge $q$ is distributed uniformly along a thin, straight rod of length $L$ (see below). What is the electric field at $P_1$? At $P_2$?

88. Charge is distributed along the entire $x$-axis with uniform density $\lambda$. How much work does the electric field of this charge distribution do on an electron that moves along the $y$-axis from $y=a$ to $y=b$?

89. Charge is distributed along the entire $x$-axis with uniform density $\lambda_x$ and along the entire $y$-axis with uniform density $\lambda_y$. Calculate the resulting electric field at (a) $\vec{\mathbf{r}}=a\hat{\mathbf{i}}+b\hat{\mathbf{j}}$ and (b)$\vec{\mathbf{r}}=c\hat{\mathbf{k}}$

90. A rod bent into the arc of a circle subtends an angle $2\theta$ at the centre $P$ of the circle (see below). If the rod is charged uniformly with a total charge $Q$, what is the electric field at $P$?

91. A proton moves in the electric field $\vec{\mathbf{E}}=100\hat{\mathbf{i}}~\mathrm{N/C}$. (a) What are the force on and the acceleration of the proton? (b) Do the same calculation for an electron moving in this field.

92. An electron and a proton, each starting from rest, are accelerated by the same uniform electric field of $200~\mathrm{N/C}$. Determine the distance and time for each particle to acquire a kinetic energy of $3.2\times10^{-16}~\mathrm{J}$.

93. A spherical water droplet of radius $25~\mu\mathrm{m}$ carries an excess $250$ electrons. What vertical electric field is needed to balance the gravitational force on the droplet at the surface of the earth?

94. A proton enters the uniform electric field produced by the two charged plates shown below. The magnitude of the electric field is $4.0\times10^{5}~\mathrm{N/C},$ and the speed of the proton when it enters is $1.5\times10^7~\mathrm{m/s}.$ What distance $d$ has the proton been deflected downward when it leaves the plates?

95. Shown below is a small sphere of mass $0.25~\mathrm{g}$ that carries a charge of $9.0\times10^{-10}~\mathrm{C}.$ The sphere is attached to one end of a very thin silk string $5.0~\mathrm{cm}$ long. The other end of the string is attached to a large vertical conducting plate that has a charge density of $30\times10^{-6}~\mathrm{C/m}^2.$ What is the angle that the string makes with the vertical?

96. Two infinite rods, each carrying a uniform charge density $\lambda,$ are parallel to one another and perpendicular to the plane of the page. (See below.) What is the electrical field at $P_1$? At $P_2$?

97. Positive charge is distributed with a uniform density $\lambda$ along the positive $x$-axis from $r$ to $\infty$, along the positive $y$-axis from $r$ to $\infty$, and along a $90^{\circ}$ arc of a circle of radius $r,$ as shown below. What is the electric field at $O$?

98. From a distance of $10\mathrm{cm}$, a proton is projected with a speed of $v=4.0\times10^{6}~\mathrm{m/s}$ directly at a large, positively charged plate whose charge density is $\sigma=2.0\times10^{-5}~\mathrm{C/m}^2$. (See below.) (a) Does the proton reach the plate? (b) If not, how far from the plate does it turn around?

99. A particle of mass $m$ and charge $-q$ moves along a straight line away from a fixed particle of charge $Q$. When the distance between the two particles is $r_0,$ $-q$ is moving with a speed $v_0$. (a) Use the work-energy theorem to calculate the maximum separation of the charges. (b) What do you have to assume about v0v0 to make this calculation? (c) What is the minimum value of $v_0$ such that $-q$ escapes from $Q$?

100. Which of the following electric field lines are incorrect for point charges? Explain why.

101. In this exercise, you will practice drawing electric field lines. Make sure you represent both the magnitude and direction of the electric field adequately. Note that the number of lines into or out of charges is proportional to the charges. (a) Draw the electric field lines map for two charges $+20~\mu\mathrm{C}$ and $-20~\mu\mathrm{C}$ situated $5~\mathrm{cm}$ from each other. (b) Draw the electric field lines map for two charges $+20~\mu\mathrm{C}$ and $+20~\mu\mathrm{C}$ situated $5~\mathrm{cm}$ from each other. (c) Draw the electric field lines map for two charges $+20~\mu\mathrm{C}$ and $-30~\mu\mathrm{C}$ situated $5~\mathrm{cm}$ from each other.

102. Draw the electric field for a system of three particles of charges $+1~\mu\mathrm{C},$ $+2~\mu\mathrm{C},$ and $-3~\mu\mathrm{C}$ fixed at the corners of an equilateral triangle of side $2~\mathrm{cm}$.

103. Two charges of equal magnitude but opposite sign make up an electric dipole. A quadrupole consists of two electric dipoles are placed anti-parallel at two edges of a square as shown.

Draw the electric field of the charge distribution.

104. Suppose the electric field of an isolated point charge decreased with distance as $1/r^{2+\delta}$ rather than as $1/r^2$. Show that it is then impossible to draw continuous field lines so that their number per unit area is proportional to $E$.
## 1.7 Electric Dipoles

107. A water molecule consists of two hydrogen atoms bonded with one oxygen atom. The bond angle between the two hydrogen atoms is $104^{\circ}$ (see below). Calculate the net dipole moment of a water molecule that is placed in a uniform, horizontal electric field of magnitude $2.3\times10^{-8}~\mathrm{N/C}$. (You are missing some information for solving this problem; you will need to determine what information you need, and look it up.)

105. Consider the equal and opposite charges shown below. (a) Show that at all points on the $x$-axis for which $|x|\gg a$, $E\approx Qa/2\pi\epsilon_0x^3$. (b) Show that at all points on the $y$-axis for which $|y|\gg a$, $E\approx Qa/\pi\epsilon_0y^3$.

106. (a) What is the dipole moment of the configuration shown above? If $Q=4.0~\mu\mathrm{C}$, (b) what is the torque on this dipole with an electric field of $4.0\times10^5~\mathrm{N/C}\hat{\mathbf{i}}$? (c) What is the torque on this dipole with an electric field of$-4.0\times10^5~\mathrm{N/C}\hat{\mathbf{i}}$? (d) What is the torque on this dipole with an electric field of $\pm4.0\times10^5~\mathrm{N/C}\hat{\mathbf{j}}$?
108. Point charges $q_1=2.0~\mu\mathrm{C}$ and $q_2=4.0~\mu\mathrm{C}$ are located at $\vec{\mathbf{r}}_1=\left(4.0\hat{\mathbf{i}}-2.0\hat{\mathbf{j}}+2.0\hat{\mathbf{j}}\right)~\mathrm{m}$ and $\vec{\mathbf{r}}_2=\left(4.0\hat{\mathbf{i}}-2.0\hat{\mathbf{j}}+2.0\hat{\mathbf{j}}\right)~\mathrm{m}$. What is the force of $q_2$ on $q_1$?

109. What is the force on the $5.0$-$\mu\mathrm{C}$ charge shown below?

110. What is the force on the $2.0$-$\mu\mathrm{C}$ charge placed at the centre of the square shown below?

111. Four charged particles are positioned at the corners of a parallelogram as shown below. If $q=5.0~\mu\mathrm{C}$ and $Q=8.0~\mu\mathrm{C},$ what is the net force on $q$?

112. A charge $Q$ is fixed at the origin and a second charge $q$ moves along the $x$-axis, as shown below. How much work is done on $q$ by the electric force when $q$ moves from $x_1$ to $x_2?$

113. A charge $q=-2.0~\mu\mathrm{C}$ is released from rest when it is $2.0~\mathrm{m}$ from a fixed charge $Q=6.0~\mu\mathrm{C}.$ What is the kinetic energy of $q$ when it is $1.0~\mathrm{m}$ from $Q$?

114. What is the electric field at the midpoint $M$ of the hypotenuse of the triangle shown below?

115. Find the electric field at $P$ for the charge configurations shown below.

116. (a) What is the electric field at the lower-right-hand corner of the square shown below? (b) What is the force on a charge $q$ placed at that point?

117. Point charges are placed at the four corners of a rectangle as shown below: $q_1=2.0\times10^{-6}~\mathrm{C},$ $q_2=-2.0\times10^{-6}~\mathrm{C},$ $q_3=4.0\times10^{-6}~\mathrm{C},$ and $q_4=1.0\times10^{-6}~\mathrm{C}.$ What is the electric field at $P$?

118. Three charges are positioned at the corners of a parallelogram as shown below. (a) If $Q=8.0~\mu\mathrm{C},$ what is the electric field at the unoccupied corner? (b) What is the force on a $5.0$-$\mu\mathrm{C}$ charge placed at this corner?

119. A positive charge $q$ is released from rest at the origin of a rectangular coordinate system and moves under the influence of the electric field $\vec{\mathrm{E}}=E_0(1+x/a)\hat{\mathbf{i}}.$ What is the kinetic energy of $q$ when it passes through $x=3a$?

120. A particle of charge $-q$ and mass $m$ is placed at the centre of a uniformly charged ring of total charge $Q$ and radius $R$. The particle is displaced a small distance along the axis perpendicular to the plane of the ring and released. Assuming that the particle is constrained to move along the axis, show that the particle oscillates in simple harmonic motion with a frequency $f=\frac{1}{2\pi}\sqrt{\frac{qQ}{4\pi\epsilon_0mR^3}}.$

121. Charge is distributed uniformly along the entire $y$-axis with a density $\lambda_y$ and along the positive $x$-axis from $x=a$ to $x=b$ with a density $\lambda_x$. What is the force between the two distributions?

122. The circular arc shown below carries a charge per unit length $\lambda=\lambda_0\cos\theta,$ where $\theta$ is measured from the $x$-axis. What is the electric field at the origin?

123. Calculate the electric field due to a uniformly charged rod of length $L$, aligned with the $x$-axis with one end at the origin; at a point $P$ on the $z$-axis.

124. The charge per unit length on the thin rod shown below is $\lambda.$ What is the electric force on the point charge $q$? Solve this problem by first considering the electric force $d\vec{\mathbf{F}}$ on $q$ due to a small segment $dx$ of the rod, which contains charge $\lambda dx.$ Then, find the net force by integrating $d\vec{\mathbf{F}}$ over the length of the rod.

125. The charge per unit length on the thin rod shown here is $\lambda.$ What is the electric force on the point charge $q$? (See the preceding problem.)

126. The charge per unit length on the thin semicircular wire shown below is $\lambda.$ What is the electric force on the point charge $q$? (See the preceding problems.)

- Describe a permanent dipole
- Describe an induced dipole
- Define and calculate an electric dipole moment
- Explain the physical meaning of the dipole moment

Earlier we discussed, and calculated, the electric field of a dipole: two equal and opposite charges that are “close” to each other. (In this context, “close” means that the distance $d$ between the two charges is much, much less than the distance of the field point $P$, the location where you are calculating the field.) Let’s now consider what happens to a dipole when it is placed in an external field $\vec{\mathbf{E}}$. We assume that the dipole is a **permanent dipole**; it exists without the field, and does not break apart in the external field.

For now, we deal with only the simplest case: The external field is uniform in space. Suppose we have the situation depicted in \ref{fig:1.7.1}, where we denote the distance between the charges as the vector $\vec{\mathbf{d}}$, pointing from the negative charge to the positive charge. The forces on the two charges are equal and opposite, so there is no net force on the dipole. However, there is a torque:

\begin{eqnarray*}\vec{\pmb{\uptau}}&=&\left(\frac{\vec{\mathbf{d}}}{2}\times\vec{\mathbf{F}}_+\right)+\left(-\frac{\vec{\mathbf{d}}}{2}\times\vec{\mathbf{F}}_-\right)\\&=&\left[\left(\frac{\vec{\mathbf{d}}}{2}\right)\times\left(+q\vec{\mathbf{E}}\right)+\left(-\frac{\vec{\mathbf{d}}}{2}\right)\times\left(-q\vec{\mathbf{E}}\right)\right]\\&=&q\vec{\mathbf{d}}\times\vec{\mathbf{E}}.\end{eqnarray*}\begin{gather}.\tag{Figure 1.7.1}\label{fig:1.7.1}\end{gather}

The quantity $q\vec{\mathbf{d}}$ (the magnitude of each charge multiplied by the vector distance between them) is a property of the dipole; its value, as you can see, determines the torque that the dipole experiences in the external field. It is useful, therefore, to define this product as the so-called **dipole moment** of the dipole:

\begin{equation}\vec{\mathbf{p}}\equiv q\vec{\mathbf{d}}.\tag{1.7.1}\label{eq:1.7.1}\end{equation}

We can therefore write

\begin{equation}\vec{\pmb{\uptau}}=\vec{\mathbf{p}}\times\vec{\mathbf{E}}.\tag{1.7.2}\label{eq:1.7.2}\end{equation}

Recall that a torque changes the angular velocity of an object, the dipole, in this case. In this situation, the effect is to rotate the dipole (that is, align the direction of $\vec{\mathbf{p}}$) so that it is parallel to the direction of the external field.

Neutral atoms are, by definition, electrically neutral; they have equal amounts of positive and negative charge. Furthermore, since they are spherically symmetrical, they do not have a “built-in” dipole moment the way most asymmetrical molecules do. They obtain one, however, when placed in an external electric field, because the external field causes oppositely directed forces on the positive nucleus of the atom versus the negative electrons that surround the nucleus. The result is a new charge distribution of the atom, and therefore, an **induced dipole** moment (\ref{fig:1.7.2}).

An important fact here is that, just as for a rotated polar molecule, the result is that the dipole moment ends up aligned parallel to the external electric field. Generally, the magnitude of an induced dipole is much smaller than that of an inherent dipole. For both kinds of dipoles, notice that once the alignment of the dipole (rotated or induced) is complete, the net effect is to decrease the total electric field $\vec{\mathbf{E}}_{\mathrm{total}}=\vec{\mathbf{E}}_{\mathrm{external}}+\vec{\mathbf{E}}_{\mathrm{dipole}}$ in the regions outside the dipole charges (\ref{fig:1.7.3}). By “outside” we mean further from the charges than they are from each other. This effect is crucial for capacitors, as you will see in Capacitance.

\begin{gather}.\tag{Figure 1.7.3}\label{fig:1.7.3}\end{gather}Recall that we found the electric field of a dipole in Equation 1.4.5. If we rewrite it in terms of the dipole moment we get:

\[\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\frac{\vec{\mathbf{p}}}{z^3}.\] The form of this field is shown in \ref{fig:1.7.3}. Notice that along the plane perpendicular to the axis of the dipole and midway between the charges, the direction of the electric field is opposite that of the dipole and gets weaker the further from the axis one goes. Similarly, on the axis of the dipole (but outside it), the field points in the same direction as the dipole, again getting weaker the further one gets from the charges.- Define the work done by an electric force
- Define electric potential energy
- Apply work and potential energy in systems with electric charges

When a free positive charge *q* is accelerated by an electric field, it is given kinetic energy (\ref{fig:3.1.1}). The process is analogous to an object being accelerated by a gravitational field, as if the charge were going down an electrical hill where its electric potential energy is converted into kinetic energy, although of course the sources of the forces are very different. Let us explore the work done on a charge *q* by the electric field in this process, so that we may develop a definition of electric potential energy.

The electrostatic or Coulomb force is conservative, which means that the work done on $q$ is independent of the path taken, as we will demonstrate later. This is exactly analogous to the gravitational force. When a force is conservative, it is possible to define a potential energy associated with the force. It is usually easier to work with the potential energy (because it depends only on position) than to calculate the work directly.

To show this explicitly, consider an electric charge $+q$ fixed at the origin and move another charge $+Q$ toward $q$ in such a manner that, at each instant, the applied force $\vec{\mathbf{F}}$ exactly balances the electric force $\vec{\mathbf{F}}_e$ on $Q$ (\ref{fig:3.1.2}). The work done by the applied force $\vec{\mathbf{F}}$ on the charge $Q$ changes the potential energy of $Q$. We call this potential energy the **electrical potential energy** of $Q$.

The work $W_{12}$ done by the applied force $\vec{\mathbf{F}}$ when the particle moves from $P_1$ to $P_2$ may be calculated by

\[W_{12}=\int_{P_1}^{P_2}\vec{\mathbf{F}}\cdot d\vec{\mathbf{l}}.\]

Since the applied force $\vec{\mathbf{F}}$ balances the electric force $\vec{\mathbf{F}}_e$ on $Q$, the two forces have equal magnitude and opposite directions. Therefore, the applied force is

\[\vec{\mathbf{F}}=-\vec{\mathbf{F}}_e=-\frac{kqQ}{r^2}\hat{\mathbf{r}},\]

where we have defined positive to be pointing away from the origin and $r$ is the distance from the origin. The directions of both the displacement and the applied force in the system in \ref{fig:3.1.2} are parallel, and thus the work done on the system is positive.

We use the letter $U$ to denote electric potential energy, which has units of joules ($\mathrm{J}$). When a conservative force does negative work, the system gains potential energy. When a conservative force does positive work, the system loses potential energy, $\Delta U=-W$. In the system in \ref{fig:3.1.2}, the Coulomb force acts in the opposite direction to the displacement; therefore, the work is negative. However, we have increased the potential energy in the two-charge system.

- What is the work done by the electric field between $r_1$ and $r_2$?
- How much kinetic energy does $Q$ have at $r_2$?

This is also the value of the kinetic energy at $r_2$.

If $Q$ has a mass of $4.00~\mu\mathrm{g}$ what is the speed of $Q$ at $r_2$?

In this example, the work $W$ done to accelerate a positive charge from rest is positive and results from a loss in $U$, or a negative $\Delta U$. A value for $U$ can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.

Work $W$ done to accelerate a positive charge from rest is positive and results from a loss in $U$, or a negative $\Delta U$. Mathematically,

\begin{equation}W=-\Delta U.\tag{3.1.1}\label{eq:3.1.1}\end{equation}Gravitational potential energy and electric potential energy are quite analogous. Potential energy accounts for work done by a conservative force and gives added insight regarding energy and energy transformation without the necessity of dealing with the force directly. It is much more common, for example, to use the concept of electric potential energy than to deal with the Coulomb force directly in real-world applications.

In polar coordinates with $q$ at the origin and $Q$ located at $r$, the displacement element vector is $d\vec{\mathbf{l}}=\hat{\mathbf{r}}dr$ and thus the work becomes

\[W_{12}=-kqQ\int_{r_1}^{r_2}\frac{1}{r^2}\hat{\mathbf{r}}\cdot\hat{\mathbf{r}}dr=kqQ\frac{1}{r_2}-kqQ\frac{1}{r_1}.\]Notice that this result only depends on the endpoints and is otherwise independent of the path taken. To explore this further, compare path $P_1$ to $P_2$ with path $P_1P_3P_4P_2$ in \ref{fig:3.1.4}.

\begin{gather}.\tag{Figure 3.1.4}\label{fig:3.1.4}\end{gather}The segments $P_1P_3$ and $P_4P_2$ are arcs of circles centred at $q$. Since the force on $Q$ points either toward or away from $q$, no work is done by a force balancing the electric force, because it is perpendicular to the displacement along these arcs. Therefore, the only work done is along segment $P_3P_4$, which is identical to $P_1P_2$

One implication of this work calculation is that if we were to go around the path $P_1P_3P_4P_2P_1$, the net work would be zero (\ref{fig:3.1.5}). Recall that this is how we determine whether a force is conservative or not. Hence, because the electric force is related to the electric field by $\vec{\mathbf{F}}=q\vec{\mathbf{E}}$, the electric field is itself conservative. That is,

\[\oint\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}=0.\]Note that $Q$ is a constant.

\begin{gather}.\tag{Figure 3.1.5}\label{fig:3.1.5}\end{gather}Another implication is that we may define an electric potential energy. Recall that the work done by a conservative force is also expressed as the difference in the potential energy corresponding to that force. Therefore, the work $W_{\mathrm{ref}}$ to bring a charge from a reference point to a point of interest may be written as

\[W_{\mathrm{ref}}=\int_{r_{\mathrm{ref}}}^{r}\vec{\mathbf{F}}\cdot d\vec{\mathbf{l}}\]and, by Equation \ref{eq:3.1.1}, the difference in potential energy ($U_2-U_1$) of the test charge $Q$ between the two points is

\[\Delta U=-\int_{r_{\mathrm{ref}}}^{r}\vec{\mathbf{F}}\cdot d\vec{\mathbf{l}}\]

Therefore, we can write a general expression for the potential energy of two point charges (in spherical coordinates):

\[\Delta U=-\int_{r_{\mathrm{ref}}}^{r}\frac{kqQ}{r^2}dr=-\left[-\frac{kqQ}{r}\right]_{r_{\mathrm{ref}}}^{r}=kqQ\left[\frac{1}{r}-\frac{1}{r_{\mathrm{ref}}}\right]\]

We may take the second term to be an arbitrary constant reference level, which serves as the zero reference:

\[U(r)=k\frac{qQ}{r}-U_{\mathrm{ref}}.\]

A convenient choice of reference that relies on our common sense is that when the two charges are infinitely far apart, there is no interaction between them. Taking the potential energy of this state to be zero removes the term $U_{\mathrm{ref}}$ from the equation (just like when we say the ground is zero potential energy in a gravitational potential energy problem), and the potential energy of $Q$ when it is separated from $q$ by a distance $r$* *assumes the form

\begin{equation}U(r)=k\frac{qQ}{r}~(\mathrm{zero~reference~at~}r=\infty.)\tag{3.1.2}\label{eq:3.1.2}\end{equation}

This formula is symmetrical with respect to $q$ and $Q$, so it is best described as the potential energy of the two-charge system.

What is the change in the potential energy of the two-charge system from $r_1$ to $r_2$?

\begin{eqnarray*}\Delta U_{12}&\!\!\!=\!\!\!&-\int_{r_1}^{r_2}\vec{\mathbf{F}}\vdot d\vec{\mathbf{r}}=-\int_{r_1}^{r_2}\frac{kqQ}{r^2}dr=-\left[-\frac{kqQ}{r^2}\right]_{r_1}^{r_2}=kqQ\left[\frac{1}{r_2}-\frac{1}{r_1}\right]\\&\!\!\!=\!\!\!&(8.99\times10^9~\mathrm{Nm}^2/\mathrm{C}^2)(5.0\times10^{-9}~\mathrm{C})(3.0\times10^{-9}~\mathrm{C})\small{\left[\frac{1}{0.15~\mathrm{m}}-\frac{1}{0.10~\mathrm{m}}\right]}\\&\!\!\!=\!\!\!&-4.5\times10^{-7}~\mathrm{J}.\end{eqnarray*}

What is the potential energy of $Q$ relative to the zero reference at infinity at $r_2$ in the above example?

Due to Coulomb’s law, the forces due to multiple charges on a test charge $Q$ superimpose; they may be calculated individually and then added. This implies that the work integrals and hence the resulting potential energies exhibit the same behaviour. To demonstrate this, we consider an example of assembling a system of four charges.

\[W_1=0.\]

Step 2. While keeping the $+2.0\mathrm{-}\mu\mathrm{C}$ charge fixed at the origin, bring the $+3.0\mathrm{-}\mu\mathrm{C}$ charge to $(x,y,z)=(1.0~\mathrm{cm},0,0)$ (\ref{fig:3.1.8}). Now, the applied force must do work against the force exerted by the $+2.0\mathrm{-}\mu\mathrm{C}$ charge fixed at the origin. The work done equals the change in the potential energy of the $+3.0\mathrm{-}\mu\mathrm{C}$ charge:

\[W_2=k\frac{q_1q_2}{r_{12}}=8.99\times10^9~\mathrm{Nm}^2\mathrm{C}^{-2}\cdot\frac{(2.0\times10^{-6}~\mathrm{C})(3.0\times10^{-6}~\mathrm{C})}{1.0\times10^{-2}~\mathrm{m}}=5.4~\mathrm{J}.\]
\begin{gather}.\tag{Figure 3.1.8}\label{fig:3.1.8}\end{gather}

Step 3. While keeping the charges of $+3.0~\mu\mathrm{C}$ and $+2.0~\mu\mathrm{C}$ fixed in their places, bring in the $+4.0\mathrm{-}\mu\mathrm{C}$ charge to $(x,y,z)=(1.0~\mathrm{cm},1.0~\mathrm{cm},0)$ (\ref{fig:3.1.9}). The work done in this step is

\begin{eqnarray*}W_3&\!\!\!=\!\!\!&k\frac{q_1q_3}{r_{13}}+k\frac{q_2q_3}{r_{23}}\\&\!\!\!=\!\!\!&8.99\times10^9~\mathrm{Nm}^2\mathrm{C}^{-2}\\&\!\!\!~\!\!\!&\times\left[\frac{(2.0\times10^{-6}~\mathrm{C})(4.0\times10^{-6}~\mathrm{C})}{\sqrt{2.0}\times10^{-2}~\mathrm{m}}+\frac{(3.0\times10^{-6}~\mathrm{C})(4.0\times10^{-6}~\mathrm{C})}{1.0\times10^{-2}~\mathrm{m}}\right]\\&\!\!\!=\!\!\!&15.9~\mathrm{J}.\end{eqnarray*}
\begin{gather}.\tag{Figure 3.1.9}\label{fig:3.1.9}\end{gather}

Step 4. Finally, while keeping the first three charges in their places, bring the $+5.0\mathrm{-}\mu\mathrm{C}$ charge to $(x,y,z)=(0,1.0~\mathrm{cm},0)$ (\ref{fig:3.1.10}). The work done here is

\begin{eqnarray*}W_4&\!\!\!=\!\!\!&kq_4\left[\frac{q_1}{r_{14}}+\frac{q_2}{r_{24}}+\frac{q_3}{r_{34}}\right]\\&\!\!\!=\!\!\!&(8.99\times10^9~\mathrm{Nm}^2\mathrm{C}^{-2})(5.0\times10^{-6}~\mathrm{C})\\&\!\!\!~\!\!\!&\times\left[\frac{2.0\times10^{-6}~\mathrm{C}}{1.0\times10^{-2}~\mathrm{m}}+\frac{3.0\times10^{-6}~\mathrm{C}}{\sqrt{2.0}\times10^{-2}~\mathrm{m}}+\frac{4.0\times10^{-6}~\mathrm{C}}{1.0\times10^{-2}~\mathrm{m}}\right]\\&\!\!\!=\!\!\!&36.5~\mathrm{J}.\end{eqnarray*}
\begin{gather}.\tag{Figure 3.1.10}\label{fig:3.1.10}\end{gather}

Hence, the total work done by the applied force in assembling the four charges is equal to the sum of the work in bringing each charge from infinity to its final position:

\[W_T=W_1+W_2+W_3+W_4=0+5.4~\mathrm{J}+15.9~\mathrm{J}+36.5~\mathrm{J}=57.8~\mathrm{J}\]

Is the electrical potential energy of two point charges positive or negative if the charges are of the same sign? Opposite signs? How does this relate to the work necessary to bring the charges into proximity from infinity?

Note that the electrical potential energy is positive if the two charges are of the same type, either positive or negative, and negative if the two charges are of opposite types. This makes sense if you think of the change in the potential energy $\Delta U$ as you bring the two charges closer or move them farther apart. Depending on the relative types of charges, you may have to work on the system or the system would do work on you, that is, your work is either positive or negative. If you have to do positive work on the system (actually push the charges closer), then the energy of the system should increase. If you bring two positive charges or two negative charges closer, you have to do positive work on the system, which raises their potential energy. Since potential energy is proportional to $1/r$, the potential energy goes up when $r$ goes down between two positive or two negative charges.

On the other hand, if you bring a positive and a negative charge nearer, you have to do negative work on the system (the charges are pulling you), which means that you take energy away from the system. This reduces the potential energy. Since potential energy is negative in the case of a positive and a negative charge pair, the increase in $1/r$ makes the potential energy more negative, which is the same as a reduction in potential energy.

The result from Example 3.1.1 may be extended to systems with any arbitrary number of charges. In this case, it is most convenient to write the formula as

\begin{equation}W_{12\ldots N}=\frac{k}{2}\sum_i^N\sum_j^N\frac{q_iq_j}{r_{ij}}~\mathrm{for}~i\neq j.\tag{3.1.3}\label{eq:3.1.3}\end{equation}

The factor of 1/2 accounts for adding each pair of charges twice.

]]>- Define electric potential, voltage, and potential difference
- Define the electron-volt
- Calculate electric potential and potential difference from potential energy and electric field
- Describe systems in which the electron-volt is a useful unit
- Apply conservation of energy to electric systems

The electric potential energy per unit charge is

\begin{equation}V=\frac{U}{q}.\tag{3.2.1}\label{eq:3.2.1}\end{equation}

The **electric potential difference** between points $A$ and $B$, $V_B-V_A$ is defined to be the change in potential energy of a charge $q$ moved from $A$ to $B$, divided by the charge. Units of potential difference are joules per coulomb, given the name **volt (V)** after Alessandro **Volta**.

\[1~\mathrm{V}=1~\mathrm{J/C}\]

The familiar term **voltage** is the common name for electric potential difference. Keep in mind that whenever a voltage is quoted, it is understood to be the potential difference between two points. For example, every battery has two terminals, and its voltage is the potential difference between them. More fundamentally, the point you choose to be zero volts is arbitrary. This is analogous to the fact that gravitational potential energy has an arbitrary zero, such as sea level or perhaps a lecture hall floor. It is worthwhile to emphasize the distinction between potential difference and electrical potential energy.

The relationship between potential difference (or voltage) and electrical potential energy is given by

\begin{equation}\Delta V=\frac{\Delta U}{q}~\mathrm{or}~\Delta U=q\Delta V.\tag{3.2.2}\label{eq:3.2.2}\end{equation}

Voltage is not the same as energy. Voltage is the energy per unit charge. Thus, a motorcycle battery and a car battery can both have the same voltage (more precisely, the same potential difference between battery terminals), yet one stores much more energy than the other because $\Delta U=q\Delta V$. The car battery can move more charge than the motorcycle battery, although both are $12\mathrm{-V}$ batteries.

\[\Delta U_{\mathrm{cycle}}=(5000~\mathrm{C})(12.0~\mathrm{V})=(5000~\mathrm{C})(12.0~\mathrm{J/C})=6.00\times10^4~\mathrm{J}.\]

Similarly, for the car battery, $q=60,000~\mathrm{C}$ and

\[\Delta U_{\mathrm{cycle}}=(60,000~\mathrm{C})(12.0~\mathrm{V})=7.20\times10^5~\mathrm{J}.\]

How much energy does a $1.5{\text -}\mathrm{V}$ AAA battery have that can move $100~\mathrm{C}$?

Note that the energies calculated in the previous example are absolute values. The change in potential energy for the battery is negative, since it loses energy. These batteries, like many electrical systems, actually move negative charge—electrons in particular. The batteries repel electrons from their negative terminals ($A$) through whatever circuitry is involved and attract them to their positive terminals ($B$), as shown in \ref{fig:3.2.1}. The change in potential is $\Delta V=V_B-V_A=+12~\mathrm{V}$ and the charge $q$ is negative, so that $\Delta U=q\Delta V$ is negative, meaning the potential energy of the battery has decreased when $q$ has moved from $A$ to $B$.

\begin{gather}.\tag{Figure 3.2.1}\label{fig:3.2.1}\end{gather}
\[q=\frac{\Delta U}{\Delta V}.\]

Entering the values for $\Delta U$ and $\Delta V$ we get

\[q=\frac{-30.0~\mathrm{J}}{+12.0~\mathrm{V}}=\frac{-30.0~\mathrm{J}}{+12.0~\mathrm{J/C}}=-2.50~\mahtrm{C}.\]

The number of electrons $n_e$ is the total charge divided by the charge per electron. That is,

\[n_e=\frac{-2.50~\mathrm{C}}{-1.60\times10^{-19}~\mathrm{C/e}^{-}}=1.56\times10^{19}~\mathrm{electrons}.\]

How many electrons would go through a $24.0{\text -}\mathrm{W}$ lamp?

The energy per electron is very small in macroscopic situations like that in the previous example—a tiny fraction of a joule. But on a submicroscopic scale, such energy per particle (electron, proton, or ion) can be of great importance. For example, even a tiny fraction of a joule can be great enough for these particles to destroy organic molecules and harm living tissue. The particle may do its damage by direct collision, or it may create harmful X-rays, which can also inflict damage. It is useful to have an energy unit related to submicroscopic effects.

\ref{fig:3.2.2} shows a situation related to the definition of such an energy unit. An electron is accelerated between two charged metal plates, as it might be in an old-model television tube or oscilloscope. The electron gains kinetic energy that is later converted into another form—light in the television tube, for example. (Note that in terms of energy, “downhill” for the electron is “uphill” for a positive charge.) Since energy is related to voltage by $\Delta U=q\Delta V$, we can think of the joule as a coulomb-volt.

\begin{gather}.\tag{Figure 3.2.2}\label{fig:3.2.2}\end{gather}On the submicroscopic scale, it is more convenient to define an energy unit called the **electron-volt (eV)**, which is the energy given to a fundamental charge accelerated through a potential difference of $1~\mathrm{V}$. In equation form,

An electron accelerated through a potential difference of $1~\mathrm{V}$ is given an energy of $1~\mathrm{eV}$. It follows that an electron accelerated through $50~\mathrm{V}$ gains $50~\mathrm{eV}$. A potential difference of $100,000~\mathrm{V}$ ($100~\mathrm{kV}$) gives an electron an energy of $100,000~\mathrm{eV}$ ($100~\mathrm{keV}$), and so on. Similarly, an ion with a double positive charge accelerated through $100~\mathrm{V}$ gains $200~\mathrm{eV}$ of energy. These simple relationships between accelerating voltage and particle charges make the electron-volt a simple and convenient energy unit in such circumstances.

The electron-volt is commonly employed in submicroscopic processes—chemical valence energies and molecular and nuclear binding energies are among the quantities often expressed in electron-volts. For example, about $5~\mathrm{eV}$ of energy is required to break up certain organic molecules. If a proton is accelerated from rest through a potential difference of $30~\mathrm{kV}$, it acquires an energy of $30~\mathrm{keV}$ ($30,000~\mathrm{eV}$) and can break up as many as $6000$ of these molecules ($30,000~\mathrm{eV}\div5~\mathrm{eV}$ per molecule $=6000$ molecules). Nuclear decay energies are on the order of $1~\mathrm{MeV}$ ($1,000,000~\mathrm{eV}$) per event and can thus produce significant biological damage.

The total energy of a system is conserved if there is no net addition (or subtraction) due to work or heat transfer. For conservative forces, such as the electrostatic force, conservation of energy states that mechanical energy is a constant.

Mechanical energy is the sum of the kinetic energy and potential energy of a system; that is, $K+U=\mathrm{constant}$. A loss of $U$ for a charged particle becomes an increase in its $K$. Conservation of energy is stated in equation form as

\[K+U=\mathrm{constant}\]or

\[K_i+U_i=K_f+U_f\]

where $i$ and $f$ stand for initial and final conditions. As we have found many times before, considering energy can give us insights and facilitate problem solving.

\[K_i+U_i=K_f+U_f.\]

Entering the forms identified above, we obtain

\[qV=\frac{mv^2}{2}.\]

We solve this for $v$:

\[v=\sqrt{\frac{2qV}{m}}.\]

Entering values for $q$, $V$, and $m$ gives

\[v=\sqrt{\frac{2(-1.60\times10^{-19}~\mathrm{C})(-100~\mathrm{J/C})}{9.11\times10^{-31}~\mathrm{kg}}}=5.93\times10^6~\mathrm{m/s}.\]

How would this example change with a positron? A positron is identical to an electron except the charge is positive.

So far, we have explored the relationship between voltage and energy. Now we want to explore the relationship between voltage and electric field. We will start with the general case for a non-uniform $\vec{\mathbf{E}}$ field. Recall that our general formula for the potential energy of a test charge $q$ at point $P$ relative to reference point $R$ is

\[U_P=-\int_R^P\vec{\mathbf{F}}\cdot d\vec{\mathbf{l}}.\]

When we substitute in the definition of electric field ($\vec{\mathbf{E}}=\vec{\mathbf{F}}/q$), this becomes

\[U_P=-q\int_R^P\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}.\]

Applying our definition of potential ($V=U/q$) to this potential energy, we find that, in general,

\begin{equation}V_P=-\int_R^P\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}.\tag{3.2.3}\label{eq:3.2.3}\end{equation}

From our previous discussion of the potential energy of a charge in an electric field, the result is independent of the path chosen, and hence we can pick the integral path that is most convenient.

Consider the special case of a positive point charge $q$ at the origin. To calculate the potential caused by $q$ at a distance $r$ from the origin relative to a reference of $0$ at infinity (recall that we did the same for potential energy), let $P=r$ and $R=\infty$, with $d\vec{\mathbf{l}}=d\vec{\mathbf{r}}=\hat{\mathbf{r}}dr$ and use $\vec{\mathbf{E}}=\frac{kq}{r^2}\hat{\mathbf{r}}$. When we evaluate the integral

\[V_P=-\int_R^P\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}\]

for this system, we have

\[V_r=-\int_{-\infty}^r\frac{kq}{r^2}\hat{\mathbf{r}}\cdot d\hat{\mathbf{r}}dr,\]

which simplifies to
\[V_r=-\int_{-\infty}^r\frac{kq}{r^2}dr=\frac{kq}{r}-\frac{kq}{\infty}=\frac{kq}{r}.\]

This result,

\[V_r=\frac{kq}{r}.\]is the standard form of the potential of a point charge. This will be explored further in the next section.

To examine another interesting special case, suppose a uniform electric field $\vec{\mathbf{E}}$ is produced by placing a potential difference (or voltage) $\Delta V$ across two parallel metal plates, labeled $A$ and $B$ (\ref{fig:3.2.3}). Examining this situation will tell us what voltage is needed to produce a certain electric field strength. It will also reveal a more fundamental relationship between electric potential and electric field.

\begin{gather}.\tag{Figure 3.2.3}\label{fig:3.2.3}\end{gather}From a physicist’s point of view, either $\Delta V$ or $\vec{\mathbf{E}}$ can be used to describe any interaction between charges. However, $\Delta V$ is a scalar quantity and has no direction, whereas $\vec{\mathbf{E}}$ is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field, a scalar quantity, is represented by $E$.) The relationship between $\Delta V$ and $\vec{\mathbf{E}}$ is revealed by calculating the work done by the electric force in moving a charge from point $A$ to point $B$. But, as noted earlier, arbitrary charge distributions require calculus. We therefore look at a uniform electric field as an interesting special case.

The work done by the electric field in \ref{fig:3.2.3} to move a positive charge $q$ from $A$, the positive plate, higher potential, to $B$, the negative plate, lower potential, is

\[W=-\Delta U=-q\Delta V.\]

The potential difference between points $A$ and $B$ is

\[-\Delta V=-(V_B-V_A)=V_A-V_B=V_{BA}.\]

Entering this into the expression for work yields

\[W=qV_{AB}.\]

Work is $W=\vec{\mathbf{F}}\cdot\vec{\mathbf{d}}=FD\cos\theta$; here $\cos\theta=1$, since the path is parallel to the field. Thus, $W=Fd$. Since $F=qE$, we see that $W=qEd$.

Substituting this expression for work into the previous equation gives

\[qEd=qV_{AB}.\]

The charge cancels, so we obtain for the voltage between points $A$ and $B$

\[\left.\begin{array}{l}V_{AB}=Ed\\E=\frac{V_{AB}}{d}\end{array}\right\}~(\mathrm{uniform}~E{\text -}\mathrm{field~only})\]where $d$ is the distance from $A$ to $B$, or the distance between the plates in \ref{fig:3.2.3}. Note that this equation implies that the units for electric field are volts per meter. We already know the units for electric field are newtons per coulomb; thus, the following relation among units is valid:

\[1~\mathrm{N/C}=1~\mathrm{V/m}.\]

Furthermore, we may extend this to the integral form. Substituting Equation \ref{eq:3.2.2} into our definition for the potential difference between points $A$ and $B$, we obtain

\[V_{AB}=V_B-V_A=-\int_R^B\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}+\int_R^A\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}\]which simplifies to

\[V_B-V_A=-\int_A^B\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}.\]As a demonstration, from this we may calculate the potential difference between two points ($A$ and $B$) equidistant from a point charge $q$ at the origin, as shown in \ref{fig:3.2.4}.

\begin{gather}.\tag{Figure 3.2.4}\label{fig:3.2.4}\end{gather}To do this, we integrate around an arc of the circle of constant radius $r$ between $A$ and $B$, which means we let $d\vec{\mathbf{l}}=r\hat{\phi}d\phi$, while using $\vec{\mathbf{E}}=\frac{kq}{r^2}\hat{\mathbf{r}}$. Thus,

\begin{equation}\Delta V_{AB}=V_B-V_A=-\int_A^B\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}\tag{3.2.4}\label{eq:3.2.4}\end{equation}

for this system becomes

\[V_B-V_A=-\int_A^B\frac{kq}{r^2}\hat{\mathbf{r}}\cdot r\hat{\pmb{\upphi}}d\phi.\]

However, $\hat{\mathbf{r}}\cdot\hat{\pmb{\upphi}}=0$ and therefore

\[V_B-V_A=0.\]

This result, that there is no difference in potential along a constant radius from a point charge, will come in handy when we map potentials.

Entering the given values for $E$ and $d$ gives

\[V_{AB}=(3.0\times10^6~\mathrm{V/m})(0.025~\mathrm{m})=7.5\times10^4~\mathrm{V}\]or

\[V_{AB}=75~\mathrm{kV}.\]

(The answer is quoted to only two digits, since the maximum field strength is approximate.)

- The expression for the magnitude of the electric field between two uniform metal plates is \[E=\frac{V_{AB}}{d}.\]Since the electron is a single charge and is given $25.0~\mathrm{keV}$ of energy, the potential difference must be $25.0~\mathrm{kV}$. Entering this value for $V_{AB}$ and the plate separation of $0.0400~\mathrm{m}$, we obtain \[E=\frac{25.0~\mathrm{kV}}{0.0400~\mathrm{m}}=6.25\times10^5~\mathrm{V/m}.\]
- The magnitude of the force on a charge in an electric field is obtained from the equation
\[F=qE.\]Substituting known values gives \[F=(0.500\times10^{-6}~\mathrm{C})(6.25\times10^5~\mathrm{V/m})=0.313~\mathrm{N}.\]

\begin{eqnarray*}\Delta V&=&-\int_a^b\frac{kq}{r^2}dr=kq\left[\frac{1}{a}-\frac{1}{b}\right]\\&=&(8.99\times10^{9}~\mathrm{Nm}^2/\mathrm{C}^2)(2.0\times10^{-9}~\mathrm{C})\left[\frac{1}{0.040~\mathrm{m}}-\frac{1}{0.12~\mathrm{m}}\right]\\&=&300~\mathrm{V}.\end{eqnarray*}

For the second step, $V_B-V_A=-\int_A^B\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}$ becomes$\Delta V=-\int_0^{24^{\circ}}\frac{kq}{r^2}\hat{\mathbf{r}}\cdot r\hat{\pmb{\upphi}}d\phi$, but $\hat{\mathbf{r}}\cdot\hat{\pmb{\upphi}}=0$ and therefore $\Delta V=0$. Adding the two parts together, we get $300~\mathrm{V}$.

From the examples, how does the energy of a lightning strike vary with the height of the clouds from the ground? Consider the cloud-ground system to be two parallel plates.

Before presenting problems involving electrostatics, we suggest a problem-solving strategy to follow for this topic.

- Examine the situation to determine if static electricity is involved; this may concern separated stationary charges, the forces among them, and the electric fields they create.
- Identify the system of interest. This includes noting the number, locations, and types of charges involved.
- Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. Determine whether the Coulomb force is to be considered directly—if so, it may be useful to draw a free-body diagram, using electric field lines.
- Make a list of what is given or can be inferred from the problem as stated (identify the knowns). It is important to distinguish the Coulomb force $F$ from the electric field $E$, for example.
- Solve the appropriate equation for the quantity to be determined (the unknown) or draw the field lines as requested.
- Examine the answer to see if it is reasonable: Does it make sense? Are units correct and the numbers involved reasonable?

- Calculate the potential due to a point charge
- Calculate the potential of a system of multiple point charges
- Describe an electric dipole
- Define dipole moment
- Calculate the potential of a continuous charge distribution

Point charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (such as charge on a metal sphere) create external electric fields exactly like a point charge. The electric potential due to a point charge is, thus, a case we need to consider.

We can use calculus to find the work needed to move a test charge $q$ from a large distance away to a distance of $r$ from a point charge $q$. Noting the connection between work and potential $W=-q\Delta V$, as in the last section, we can obtain the following result.

The electric potential $V$ of a point charge is given by

\begin{equation}V=\frac{kq}{r}~(\mathrm{point~charge})\tag{3.3.1}\label{eq:3.3.1}\end{equation}

where $k$ is a constant equal to $8.99\times10^{9}~\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2$.

The potential at infinity is chosen to be zero. Thus, $V$ for a point charge decreases with distance, whereas $\vec{\mathbf{E}}$ for a point charge decreases with distance squared:

\[E=\frac{F}{q_t}=\frac{kq}{r^2}.\]Recall that the electric potential $V$ is a scalar and has no direction, whereas the electric field $\vec{\mathbf{E}}$ is a vector. To find the voltage due to a combination of point charges, you add the individual voltages as numbers. To find the total electric field, you must add the individual fields as vectors, taking magnitude and direction into account. This is consistent with the fact that $V$ is closely associated with energy, a scalar, whereas $\vec{\mathbf{E}}$ is closely associated with force, a vector.

\[V=k\frac{q}{r}=8.99\times10^{9}~\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2\left(\frac{-3.00\times10^{-9}~\mathrm{C}}{5.00\times10^{-2}~\mathrm{m}}\right)=-539~\mathrm{V}.\]

\[V=\frac{kq}{r}.\]

\[q=\frac{rV}{k}=\frac{(0.125~\mathrm{m})(100\times10^3~\mathrm{V})}{8.99\times10^{9}~\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2}=1.39\times10^{-6}~\mathrm{C}=1.39~\mu\mathrm{C}.\]

What is the potential inside the metal sphere in Example 3.3.1?

The voltages in both of these examples could be measured with a meter that compares the measured potential with ground potential. Ground potential is often taken to be zero (instead of taking the potential at infinity to be zero). It is the potential difference between two points that is of importance, and very often there is a tacit assumption that some reference point, such as Earth or a very distant point, is at zero potential. As noted earlier, this is analogous to taking sea level as $h=0$ when considering gravitational potential energy $U_g=mgh$.

Just as the electric field obeys a superposition principle, so does the electric potential. Consider a system consisting of $N$ charges $q_1,q_2,\ldots,q_N$. What is the net electric potential $V$ at a space point $P$ from these charges? Each of these charges is a source charge that produces its own electric potential at point $P$, independent of whatever other changes may be doing. Let $V_1,V_2,\ldots,V_N$ be the electric potentials at $P$ produced by the charges $q_1,q_2,\ldots,q_N$ respectively. Then, the net electric potential $V_P$ at that point is equal to the sum of these individual electric potentials. You can easily show this by calculating the potential energy of a test charge when you bring the test charge from the reference point at infinity to point $P$:

\[V_P=V_1+V_2+\ldots+V_N=\sum_1^NV_i.\]

Note that electric potential follows the same principle of superposition as electric field and electric potential energy. To show this more explicitly, note that a test charge $q_i$ at the point $P$ in space has distances of $r_1,r_2,\ldots,r_N$ from the $N$ charges fixed in space above, as shown in \ref{fig:3.3.2}. Using our formula for the potential of a point charge for each of these (assumed to be point) charges, we find that

\begin{equation}V_P=\sum_1^Nk\frac{q_i}{r_i}.\tag{3.3.2}\label{eq:3.3.2}\end{equation}

Therefore, the electric potential energy of the test charge is

\[U_P=q_tV_P=q_tk\sum_1^Nk\frac{q_i}{r_i},\]

which is the same as the work to bring the test charge into the system, as found in the first section of the chapter.

\begin{gather}.\tag{Figure 3.3.2}\label{fig:3.3.2}\end{gather}An **electric dipole** is a system of two equal but opposite charges a fixed distance apart. This system is used to model many real-world systems, including atomic and molecular interactions. One of these systems is the water molecule, under certain circumstances. These circumstances are met inside a microwave oven, where electric fields with alternating directions make the water molecules change orientation. This vibration is the same as heat at the molecular level.

What is the potential on the $x$-axis? The $z$-axis?

Now let us consider the special case when the distance of the point $P$ from the dipole is much greater than the distance between the charges in the dipole, $r\gg d$; for example, when we are interested in the electric potential due to a polarized molecule such as a water molecule. This is not so far (infinity) that we can simply treat the potential as zero, but the distance is great enough that we can simplify our calculations relative to the previous example.

We start by noting that in \ref{fig:3.3.4} the potential is given by

\[V_P=V_++V_-=k\left(\frac{q}{r_+}-\frac{q}{r_-}\right)\]where

\[r_{\pm}=\sqrt{x^2+\left(z\mp\frac{d}{2}\right)^2}.\]
\begin{gather}.\tag{Figure 3.3.4}\label{fig:3.3.4}\end{gather}

This is still the exact formula. To take advantage of the fact that $r\gg d$, we rewrite the radii in terms of polar coordinates, with $x=r\sin\theta$ and $z=r\cos\theta$. This gives us

\[r_{\pm}=\sqrt{r^2\sin^2\theta+\left(r\cos\theta\mp\frac{d}{2}\right)^2}.\]

We can simplify this expression by pulling $r$ out of the root,

\[r_{\pm}=r\sqrt{\sin^2\theta\left(\cos\theta\mp\frac{d}{2r}\right)^2}\]

and then multiplying out the parentheses

\[r_{\pm}=\sqrt{\sin^2\theta+\cos^2\theta\mp\cos\theta\frac{d}{r}+\left(\frac{d}{2r}\right)^2}=r\sqrt{1\mp\cos\theta\frac{d}{r}+\left(\frac{d}{2r}\right)^2}.\]

The last term in the root is small enough to be negligible (remember $r\gg d$ and hence $(d/r)^2$ is extremely small, effectively zero to the level we will probably be measuring), leaving us with

\[r_{\pm}=r\sqrt{1\mp\cos\theta\frac{d}{r}}.\]

Using the binomial approximation (a standard result from the mathematics of series, when $\alpha$ is small)

\[\frac{1}{1\mp\alpha}\approx1\pm\frac{\alpha}{2}\]

and substituting this into our formula for $V_P$, we get

\[V_P=k\left[\frac{q}{r}\left(1+\frac{d\cos\theta}{2r}\right)-\frac{q}{r}\left(1-\frac{d\cos\theta}{2r}\right)\right]=k\frac{qd\cos\theta}{r^2}.\]

This may be written more conveniently if we define a new quantity, the **electric dipole moment**,

\begin{equation}\vec{\mathbf{p}}=q\vec{\mathbf{d}},\tag{3.3.3}\label{eq:3.3.3}\end{equation}

where these vectors point from the negative to the positive charge. Note that this has magnitude $qd$. This quantity allows us to write the potential at point $P$ due to a dipole at the origin as

\begin{equation}V_P=k\frac{\vec{\mathbf{p}}\cdot\hat{\mathbf{r}}}{r^2}.\tag{3.3.4}\label{eq:3.3.4}\end{equation}

A diagram of the application of this formula is shown in \ref{fig:3.3.5}.

\begin{gather}.\tag{Figure 3.3.5}\label{fig:3.3.5}\end{gather}There are also higher-order moments, for quadrupoles, octupoles, and so on. You will see these in future classes.

We have been working with point charges a great deal, but what about continuous charge distributions? Recall from Equation \ref{eq:3.3.2} that

\[V_P=k\sum\frac{q_i}{r_i}.\]

We may treat a continuous charge distribution as a collection of infinitesimally separated individual points. This yields the integral

\begin{equation}V_P=k\int\frac{dq}{r}\tag{3.3.5}\label{eq:3.3.5}\end{equation}

for the potential at a point $P$. Note that $r$ is the distance from each individual point in the charge distribution to the point $P$. As we saw in Electric Charges and Fields, the infinitesimal charges are given by

\[dq=\left\{\begin{array}{ll}\lambda dl&\mathrm{(one~dimension)}\\\sigma dA&\mathrm{(two~dimensions)}\\\rho dV&\mathrm{(three~dimensions)}\\\end{array}\right.\]where $\lambda$ is linear charge density, $\sigma$ is the charge per unit area, and $\rho$ is the charge per unit volume.

\begin{eqnarray*}V_P&\!\!\!=\!\!\!&k\int\frac{dq}{r}=k\int_{-L/2}^{L/2}\frac{\lambda dy}{\sqrt{x^2+y^2}}=k\lambda\left[\ln\left(y+\sqrt{y^2+x^2}\right)\right]_{-L/2}^{L/2}\\&\!\!\!=\!\!\!&k\lambda\left[\ln\left(\left(\frac{L}{2}\right)+\sqrt{\left(\frac{L}{2}\right)^2+x^2}\right)\right]-\ln\left(\left(-\frac{L}{2}\right)+\sqrt{\left(-\frac{L}{2}\right)^2+x^2}\right)\\&\!\!\!=\!\!\!&k\lambda\ln\left[\frac{L+\sqrt{L^2+4x^2}}{-L+\sqrt{L^2+4x^2}}\right].\end{eqnarray*}

where

\[dq=\sigma2\pi rdr.\] The superposition of potential of all the infinitesimal rings that make up the disk gives the net potential at point $P$. This is accomplished by integrating from $r=0$ to $r=R$:
\begin{eqnarray*}V_P&=&\int dV_P=k2\pi\sigma\int_0^R\frac{rdr}{\sqrt{z^2+r^2}},\\&=&k2\pi\sigma(\sqrt{z^2+R^2}-\sqrt{z^2}\end{eqnarray*}

However, this limit does not exist because the argument of the logarithm becomes $[2/0]$ as $L\rightarrow\infty$, so this way of finding $V$ of an infinite wire does not work. The reason for this problem may be traced to the fact that the charges are not localized in some space but continue to infinity in the direction of the wire. Hence, our (unspoken) assumption that zero potential must be an infinite distance from the wire is no longer valid.

To avoid this difficulty in calculating limits, let us use the definition of potential by integrating over the electric field from the previous section, and the value of the electric field from this charge configuration from the previous chapter.

\[V_P=-\int_R^P\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}\]

where $R$ is a finite distance from the line of charge, as shown in \ref{fig:3.3.9}.

\begin{gather}.\tag{Figure 3.3.9}\label{fig:3.3.9}\end{gather}With this setup, we use $\vec{\mathbf{E}}_P=2k\lambda\frac{1}{s}\hat{\mathbf{s}}$ and $d\vec{\mathbf{l}}=d\vec{\mathbf{s}}$ to obtain

\[V_P-V_R=-\int_R^P2k\lambda\frac{1}{s}ds=-2k\lambda\lm\frac{s_P}{s_R}.\]

Now, if we define the reference potential $V_R=0$ at $s_R=1$, this simplifies to

\[V_P=-2k\lambda\ln s_P.\]

Note that this form of the potential is quite usable; it is $0$ at $1~\mathrm{m}$ and is undefined at infinity, which is why we could not use the latter as a reference.

What is the potential on the axis of a nonuniform ring of charge, where the charge density is $\lambda(\theta)=\lambda\cos\theta$?

- Explain how to calculate the electric field in a system from the given potential
- Calculate the electric field in a given direction from a given potential
- Calculate the electric field throughout space from a given potential

Recall that we were able, in certain systems, to calculate the potential by integrating over the electric field. As you may already suspect, this means that we may calculate the electric field by taking derivatives of the potential, although going from a scalar to a vector quantity introduces some interesting wrinkles. We frequently need $\vec{\mathbf{E}}$ to calculate the force in a system; since it is often simpler to calculate the potential directly, there are systems in which it is useful to calculate $V$ and then derive $\vec{\mathbf{E}}$ from it.

In general, regardless of whether the electric field is uniform, it points in the direction of decreasing potential, because the force on a positive charge is in the direction of $\vec{\mathbf{E}}$ and also in the direction of lower potential $V$. Furthermore, the magnitude of $\vec{\mathbf{E}}$ equals the rate of decrease of $V$ with distance. The faster $V$ decreases over distance, the greater the electric field. This gives us the following result.

In equation form, the relationship between voltage and uniform electric field is

\[E=-\frac{\Delta V}{\Delta s}\]

where $\Delta s$ is the distance over which the change in potential $\Delta V$ takes place. The minus sign tells us that $E$ points in the direction of decreasing potential. The electric field is said to be the gradient (as in grade or slope) of the electric potential.

For continually changing potentials, $\Delta V$ and $\Delta s$ become infinitesimals, and we need differential calculus to determine the electric field. As shown in \ref{fig:3.4.1}, if we treat the distance $\Delta s$ as very small so that the electric field is essentially constant over it, we find that

\[E_s=-\frac{dV}{ds}\]
\begin{gather}.\tag{Figure 3.4.1}\label{fig:3.4.1}\end{gather}

Therefore, the electric field components in the Cartesian directions are given by

\begin{equation}E_x=-\frac{\partial V}{\partial x},\ E_y=-\frac{\partial V}{\partial y},\ E_z=-\frac{\partial V}{\partial z}.\tag{3.4.1}\label{eq:3.4.1}\end{equation}

This allows us to define the “grad” or “del” vector operator, which allows us to compute the gradient in one step. In Cartesian coordinates, it takes the form
\begin{equation}\vec{\nabla}=\hat{\mathbf{i}}\frac{\partial}{\partial x}+\hat{\mathbf{j}}\frac{\partial}{\partial y}+\hat{\mathbf{k}}\frac{\partial}{\partial z}.\tag{3.4.2}\label{eq:3.4.2}\end{equation}

With this notation, we can calculate the electric field from the potential with

\begin{equation}\vec{\mathbf{E}}=-\vec{\nabla}V,\tag{3.4.3}\label{eq:3.4.3}\end{equation}

a process we call calculating the **gradient** of the potential.

If we have a system with either cylindrical or spherical symmetry, we only need to use the del operator in the appropriate coordinates:

\begin{equation}\mathrm{Cylindrical:}~\vec{\nabla}=\hat{\mathbf{r}}\frac{\partial}{\partial r}+\hat{\pmb{\upphi}}\frac{\partial}{\partial\phi}+\hat{\mathbf{z}}\frac{\partial}{\partial z}\tag{3.4.4}\label{eq:3.4.4}\end{equation}

\begin{equation}\mathrm{Spherical:}~\vec{\nabla}=\hat{\mathbf{r}}\frac{\partial}{\partial r}+\hat{\pmb{\uptheta}}\frac{1}{r}\frac{\partial}{\partial\theta}+\hat{\pmb{\upphi}}\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}\tag{3.4.5}\label{eq:3.4.5}\end{equation}

\begin{eqnarray*}\vec{\mathbf{E}}&=&-\left(\hat{\mathbf{r}}\frac{\partial}{\partial r}+\hat{\pmb{\uptheta}}\frac{1}{r}\frac{\partial}{\partial\theta}+\hat{\pmb{\upphi}}\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}\right)k\frac{q}{r}\\&=&-kq\left(\hat{\mathbf{r}}\frac{\partial}{\partial r}\frac{1}{r}+\hat{\pmb{\uptheta}}\frac{1}{r}\frac{\partial}{\partial\theta}\frac{1}{r}+\hat{\pmb{\upphi}}\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}\frac{1}{r}\right).\end{eqnarray*}

This equation simplifies to

\[\vec{\mathbf{E}}=-kq\left(\hat{\mathbf{r}}\frac{-1}{r^2}+\hat{\pmb{\uptheta}}0+\hat{\pmb{\upphi}}0\right)=-k\frac{q}{r^2}\hat{\mathbf{r}},\]

as expected.

with the potential $V=k\frac{q_{\mathrm{tot}}}{\sqrt{z^2+R^2}}$ found previously.

Which coordinate system would you use to calculate the electric field of a dipole?

- Define equipotential surfaces and equipotential lines
- Explain the relationship between equipotential lines and electric field lines
- Map equipotential lines for one or two point charges
- Describe the potential of a conductor
- Compare and contrast equipotential lines and elevation lines on topographic maps

We can represent electric potentials (voltages) pictorially, just as we drew pictures to illustrate electric fields. This is not surprising, since the two concepts are related. Consider \ref{fig:3.5.1}, which shows an isolated positive point charge and its electric field lines, which radiate out from a positive charge and terminate on negative charges. We use blue arrows to represent the magnitude and direction of the electric field, and we use green lines to represent places where the electric potential is constant. These are called **equipotential surfaces** in three dimensions, or **equipotential lines** in two dimensions. The term *equipotential* is also used as a noun, referring to an equipotential line or surface. The potential for a point charge is the same anywhere on an imaginary sphere of radius *r*surrounding the charge. This is true because the potential for a point charge is given by $V=kq/r$ and thus has the same value at any point that is a given distance $r$ from the charge. An equipotential sphere is a circle in the two-dimensional view of \ref{fig:3.5.1}. Because the electric field lines point radially away from the charge, they are perpendicular to the equipotential lines.

It is important to note that *equipotential lines are always perpendicular to electric field lines*. No work is required to move a charge along an equipotential, since $\Delta V=0$. Thus, the work is

\[W=-\Delta U=-q\Delta V=0.\]

Work is zero if the direction of the force is perpendicular to the displacement. Force is in the same direction as $E$, so motion along an equipotential must be perpendicular to $E$. More precisely, work is related to the electric field by

\[W=\vec{\mathbf{F}}\cdot\vec{\mathbf{d}}=q\vec{\mathbf{E}}\cdot\vec{\mathbf{d}}=qEd\cos\theta=0.\]

Note that in this equation, $E$ and $F$ symbolize the magnitudes of the electric field and force, respectively. Neither $q$ nor $E$ is zero; $d$ is also not zero. So $\cos\theta$ must be $0$, meaning $\theta$ must be $90^{\circ}$. In other words, motion along an equipotential is perpendicular to $E$.

One of the rules for static electric fields and conductors is that the electric field must be perpendicular to the surface of any conductor. This implies that a *conductor is an equipotential surface in static situations*. There can be no voltage difference across the surface of a conductor, or charges will flow. One of the uses of this fact is that a conductor can be fixed at what we consider zero volts by connecting it to the earth with a good conductor—a process called **grounding**. Grounding can be a useful safety tool. For example, grounding the metal case of an electrical appliance ensures that it is at zero volts relative to Earth.

Because a conductor is an equipotential, it can replace any equipotential surface. For example, in \ref{fig:3.5.1}, a charged spherical conductor can replace the point charge, and the electric field and potential surfaces outside of it will be unchanged, confirming the contention that a spherical charge distribution is equivalent to a point charge at its centre.

\ref{fig:3.5.2} shows the electric field and equipotential lines for two equal and opposite charges. Given the electric field lines, the equipotential lines can be drawn simply by making them perpendicular to the electric field lines. Conversely, given the equipotential lines, as in \ref{fig:3.5.3}(a), the electric field lines can be drawn by making them perpendicular to the equipotentials, as in \ref{fig:3.5.3}(b).

\begin{gather}.\tag{Figure 3.5.2}\label{fig:3.5.2}\end{gather}
\begin{gather}.\tag{Figure 3.5.3}\label{fig:3.5.3}\end{gather}

To improve your intuition, we show a three-dimensional variant of the potential in a system with two opposing charges. \ref{fig:3.5.4} displays a three-dimensional map of electric potential, where lines on the map are for equipotential surfaces. The hill is at the positive charge, and the trough is at the negative charge. The potential is zero far away from the charges. Note that the cut off at a particular potential implies that the charges are on conducting spheres with a finite radius.

\begin{gather}.\tag{Figure 3.5.4}\label{fig:3.5.4}\end{gather}A two-dimensional map of the cross-sectional plane that contains both charges is shown in \ref{fig:3.5.5}. The line that is equidistant from the two opposite charges corresponds to zero potential, since at the points on the line, the positive potential from the positive charge cancels the negative potential from the negative charge. Equipotential lines in the cross-sectional plane are closed loops, which are not necessarily circles, since at each point, the net potential is the sum of the potentials from each charge.

\begin{gather}.\tag{Figure 3.5.5}\label{fig:3.5.5}\end{gather}View this simulation to observe and modify the equipotential surfaces and electric fields for many standard charge configurations. There’s a lot to explore.

One of the most important cases is that of the familiar parallel conducting plates shown in \ref{fig:3.5.6}. Between the plates, the equipotentials are evenly spaced and parallel. The same field could be maintained by placing conducting plates at the equipotential lines at the potentials shown.

\begin{gather}.\tag{Figure 3.5.6}\label{fig:3.5.6}\end{gather}Consider the parallel plates in Figure 3.1.1. These have equipotential lines that are parallel to the plates in the space between and evenly spaced. An example of this (with sample values) is given in \ref{fig:3.5.6}. We could draw a similar set of equipotential isolines for gravity on the hill shown in Figure 3.1.1. If the hill has any extent at the same slope, the isolines along that extent would be parallel to each other. Furthermore, in regions of constant slope, the isolines would be evenly spaced. An example of real topographic lines is shown in \ref{fig:3.5.7}.

\begin{gather}.\tag{Figure 3.5.7}\label{fig:3.5.7}\end{gather}(b) Use $\Delta V_{AB}=-\int_A^B\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}$.

(c) Since the electric field is constant, find the ratio of $100~\mathrm{V}$ to the total potential difference; then calculate this fraction of the distance.

What are the equipotential surfaces for an infinite line charge?

In Example 3.5.1 with a point charge, we found that the equipotential surfaces were in the form of spheres, with the point charge at the centre. Given that a conducting sphere in electrostatic equilibrium is a spherical equipotential surface, we should expect that we could replace one of the surfaces in Example 3.5.1 with a conducting sphere and have an identical solution outside the sphere. Inside will be rather different, however.

\begin{gather}.\tag{Figure 3.5.9}\label{fig:3.5.9}\end{gather}To investigate this, consider the isolated conducting sphere of \ref{fig:3.5.9} that has a radius $R$ and an excess charge $q$. To find the electric field both inside and outside the sphere, note that the sphere is isolated, so its surface change distribution and the electric field of that distribution are spherically symmetric. We can therefore represent the field as $\vec{\mathbf{E}}=E(r)\hat{\mathbf{r}}$. To calculate $E(r)$, we apply Gauss’s law over a closed spherical surface $S$ of radius $r$ that is concentric with the conducting sphere. Since $r$ is constant and $\hat{\mathbf{n}}=\hat{\mathbf{r}}$ on the sphere,

\[\oint_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}da=E(r)\oint da=E(r)4\pi r^2.\]

For $r<R$, $S$ is within the conductor, so recall from our previous study of Gauss’s law that $q_{\mathrm{enc}}=0$ and Gauss’s law gives $E(r)=0$, as expected inside a conductor at equilibrium. If $r>R$, $S$ encloses the conductor so$q_{\mathrm{enc}}=q$. From Gauss’s law,

\[E(r)4\pi r^2=\frac{q}{\epsilon_0}.\]

The electric field of the sphere may therefore be written as

\[\begin{array}{lr}E=0&(r<R),\\E=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{\mathbf{r}}&(r\geqR).\end{array}\]

As expected, in the region $r\geq R$, the electric field due to a charge $q$ placed on an isolated conducting sphere of radius $R$ is identical to the electric field of a point charge $q$ located at the centre of the sphere.

To find the electric potential inside and outside the sphere, note that for $r\geq R$ the potential must be the same as that of an isolated point charge $q$ located at $r=0$,

\[V(r)=\frac{1}{4\pi r\epsilon_0}\frac{q}{r}~(r\geq R)\]

simply due to the similarity of the electric field.

For $r<R$, $E=0$, so $V(r)$ is constant in this region. Since $V(R)=q/4\pi\epsilon_0R$,

\[V(r)=\frac{1}{4\pi r\epsilon_0}\frac{q}{R}~(r<R)\]

We will use this result to show that

\[\sigma_1R_1=\sigma_2R_2,\]

for two conducting spheres of radii $R_1$ and $R_2$, with surface charge densities $\sigma_1$ and $\sigma_2$ respectively, that are connected by a thin wire, as shown in \ref{fig:3.5.10}. The spheres are sufficiently separated so that each can be treated as if it were isolated (aside from the wire). Note that the connection by the wire means that this entire system must be an equipotential.

\begin{gather}.\tag{Figure 3.5.10}\label{fig:3.5.10}\end{gather}We have just seen that the electrical potential at the surface of an isolated, charged conducting sphere of radius $R$ is

\[V=\frac{1}{4\pi r\epsilon_0}\frac{q}{R}.\]

Now, the spheres are connected by a conductor and are therefore at the same potential; hence

\[\frac{1}{4\pi r\epsilon_0}\frac{q_1}{R_1}=\frac{1}{4\pi r\epsilon_0}\frac{q_2}{R_2},\]

and

\[\frac{q_1}{R_1}=\frac{q_2}{R_2}.\]

The net charge on a conducting sphere and its surface charge density are related by $q=\sigma(4\pi R^2)$. Substituting this equation into the previous one, we find

\[\sigma_1 R_1=\sigma_2 R_2.\]

Obviously, two spheres connected by a thin wire do not constitute a typical conductor with a variable radius of curvature. Nevertheless, this result does at least provide a qualitative idea of how charge density varies over the surface of a conductor. The equation indicates that where the radius of curvature is large (points $B$ and $D$ in \ref{fig:3.5.11}), $\sigma$ and $E$ are small.

Similarly, the charges tend to be denser where the curvature of the surface is greater, as demonstrated by the charge distribution on oddly shaped metal (\ref{fig:3.5.11}). The surface charge density is higher at locations with a small radius of curvature than at locations with a large radius of curvature.

\begin{gather}.\tag{Figure 3.5.11}\label{fig:3.5.11}\end{gather}A practical application of this phenomenon is the **lightning rod**, which is simply a grounded metal rod with a sharp end pointing upward. As positive charge accumulates in the ground due to a negatively charged cloud overhead, the electric field around the sharp point gets very large. When the field reaches a value of approximately $3.0\times10^6~\mathrm{N/C}$ (the ** dielectric strength** of the air), the free ions in the air are accelerated to such high energies that their collisions with air molecules actually ionize the molecules. The resulting free electrons in the air then flow through the rod to Earth, thereby neutralizing some of the positive charge. This keeps the electric field between the cloud and the ground from getting large enough to produce a lightning bolt in the region around the rod.

An important application of electric fields and equipotential lines involves the heart. The heart relies on electrical signals to maintain its rhythm. The movement of electrical signals causes the chambers of the heart to contract and relax. When a person has a heart attack, the movement of these electrical signals may be disturbed. An artificial pacemaker and a defibrillator can be used to initiate the rhythm of electrical signals. The equipotential lines around the heart, the thoracic region, and the axis of the heart are useful ways of monitoring the structure and functions of the heart. An electrocardiogram (ECG) measures the small electric signals being generated during the activity of the heart.

Play around with this simulation to move point charges around on the playing field and then view the electric field, voltages, equipotential lines, and more.

- Describe some of the many practical applications of electrostatics, including several printing technologies
- Relate these applications to Newton’s second law and the electric force

The study of electrostatics has proven useful in many areas. This module covers just a few of the many applications of electrostatics.

**Van de Graaff generators** (or Van de Graaffs) are not only spectacular devices used to demonstrate high voltage due to static electricity—they are also used for serious research. The first was built by Robert Van de Graaff in 1931 (based on original suggestions by Lord Kelvin) for use in nuclear physics research. \ref{fig:3.6.1} shows a schematic of a large research version. Van de Graaffs use both smooth and pointed surfaces, and conductors and insulators to generate large static charges and, hence, large voltages.

A very large excess charge can be deposited on the sphere because it moves quickly to the outer surface. Practical limits arise because the large electric fields polarize and eventually ionize surrounding materials, creating free charges that neutralize excess charge or allow it to escape. Nevertheless, voltages of 15 million volts are well within practical limits.

\begin{gather}.\tag{Figure 3.6.1}\label{fig:3.6.1}\end{gather}Most copy machines use an electrostatic process called **xerography**—a word coined from the Greek words *xeros* for dry and *graphos *for writing. The heart of the process is shown in simplified form in \ref{fig:3.6.2}.

A selenium-coated aluminum drum is sprayed with positive charge from points on a device called a corotron. Selenium is a substance with an interesting property—it is a **photoconductor**. That is, selenium is an insulator when in the dark and a conductor when exposed to light.

In the first stage of the xerography process, the conducting aluminum drum is grounded so that a negative charge is induced under the thin layer of uniformly positively charged selenium. In the second stage, the surface of the drum is exposed to the image of whatever is to be copied. In locations where the image is light, the selenium becomes conducting, and the positive charge is neutralized. In dark areas, the positive charge remains, so the image has been transferred to the drum.

The third stage takes a dry black powder, called toner, and sprays it with a negative charge so that it is attracted to the positive regions of the drum. Next, a blank piece of paper is given a greater positive charge than on the drum so that it will pull the toner from the drum. Finally, the paper and electrostatically held toner are passed through heated pressure rollers, which melt and permanently adhere the toner to the fibers of the paper.

Laser printers use the xerographic process to make high-quality images on paper, employing a laser to produce an image on the photoconducting drum as shown in \ref{fig:3.6.3}. In its most common application, the **laser printer** receives output from a computer, and it can achieve high-quality output because of the precision with which laser light can be controlled. Many laser printers do significant information processing, such as making sophisticated letters or fonts, and in the past may have contained a computer more powerful than the one giving them the raw data to be printed.

The **ink jet printer**, commonly used to print computer-generated text and graphics, also employs electrostatics. A nozzle makes a fine spray of tiny ink droplets, which are then given an electrostatic charge (\ref{fig:3.6.4}).

Once charged, the droplets can be directed, using pairs of charged plates, with great precision to form letters and images on paper. Ink jet printers can produce color images by using a black jet and three other jets with primary colors, usually cyan, magenta, and yellow, much as a color television produces color. (This is more difficult with xerography, requiring multiple drums and toners.)

\begin{gather}.\tag{Figure 3.6.4}\label{fig:3.6.4}\end{gather}Electrostatic painting employs electrostatic charge to spray paint onto oddly shaped surfaces. Mutual repulsion of like charges causes the paint to fly away from its source. Surface tension forms drops, which are then attracted by unlike charges to the surface to be painted. Electrostatic painting can reach hard-to-get-to places, applying an even coat in a controlled manner. If the object is a conductor, the electric field is perpendicular to the surface, tending to bring the drops in perpendicularly. Corners and points on conductors will receive extra paint. Felt can similarly be applied.

Another important application of electrostatics is found in air cleaners, both large and small. The electrostatic part of the process places excess (usually positive) charge on smoke, dust, pollen, and other particles in the air and then passes the air through an oppositely charged grid that attracts and retains the charged particles (\ref{fig:3.6.5})

Large **electrostatic precipitators** are used industrially to remove over $99\%$ of the particles from stack gas emissions associated with the burning of coal and oil. Home precipitators, often in conjunction with the home heating and air conditioning system, are very effective in removing polluting particles, irritants, and allergens.

Potential energy of a two-charge system | $U(r)=k\frac{qQ}{r}$ |

Work done to assemble a system of charges | $W_{12\ldots N}=\frac{k}{2}\sum_i^N\sum_j^N\frac{q_iq_j}{r_{ij}}~\mathrm{for}~i\neq j$ |

Potential difference | $\Delta V=\frac{\Delta U}{q}$ or $\Delta U=q\Delta V$ |

Electric potential | $V=\frac{U}{q}=-\int_R^P\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}$ |

Potential difference between two points | $\Delta V_{AB}=V_B-V_A=-\int_A^B\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}$ |

Electric potential of a point charge | $V=\frac{kq}{r}$ |

Electric potential of a system of point charges | $V_P=k\sum_1^N\frac{q_i}{r_i}$ |

Electric dipole moment | $\vec{\mathbf{p}}=q\vec{\mathbf{d}}$ |

Electric potential due to a dipole | $V_P=k\frac{\vec{\mathbf{p}}\cdot\hat{\mathbf{r}}}{r^2}$ |

Electric potential of a continuous charge distribution | $V_P=k\int\frac{dq}{r}$ |

Electric field components | $E_x=-\frac{\partial V}{\partial x},~E_y=-\frac{\partial V}{\partial y},~E_z=-\frac{\partial V}{\partial z}$ |

Del operator in Cartesian coordinates | $\vec{\nabla}=\hat{\mathbf{i}}\frac{\partial}{\partial x}+\hat{\mathbf{j}}\frac{\partial}{\partial y}+\hat{\mathbf{k}}\frac{\partial}{\partial z}$ |

Electric field as gradient of potential | $\vec{\mathbf{E}}=-\vec{\nabla}V$ |

Del operator in cylindrical coordinates | $\vec{\nabla}=\hat{\mathbf{r}}\frac{\partial}{\partial r}+\hat{\pmb{\upphi}}\frac{\partial}{\partial\phi}+\hat{\mathbf{z}}\frac{\partial}{\partial z}$ |

Del operator in spherical coordinates | $\vec{\nabla}=\hat{\mathbf{r}}\frac{\partial}{\partial r}+\hat{\pmb{\uptheta}}\frac{\partial}{\partial\theta}+\hat{\pmb{\upphi}}\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}$ |

- The work done to move a charge from point $A$ to $B$ in an electric field is path independent, and the work around a closed path is zero. Therefore, the electric field and electric force are conservative.
- We can define an electric potential energy, which between point charges is $U(r)=k\frac{qQ}{r}$, with the zero reference taken to be at infinity.
- The superposition principle holds for electric potential energy; the potential energy of a system of multiple charges is the sum of the potential energies of the individual pairs.

- Electric potential is potential energy per unit charge.
- The potential difference between points $A$ and $B$, $V_B-V_A$, that is, the change in potential of a charge $q$ moved from $A$ to $B$, is equal to the change in potential energy divided by the charge.
- Potential difference is commonly called voltage, represented by the symbol $\Delta V$: \[\Delta V=\frac{\Delta U}{q}~\mathrm{or}~\Delta U=q\Delta V.\]
- An electron-volt is the energy given to a fundamental charge accelerated through a potential difference of $1~\mathrm{V}$. In equation form, \[1~\mathrm{eV}=(1.60\times10^{-19}~\mathrm{C})(~\mathrm{V})=(1.60\times10^{-19}~\mathrm{C})(~\mathrm{J/C})=1.60\times10^{-19}~\mathrm{J}.\]

- Electric potential is a scalar whereas electric field is a vector.
- Addition of voltages as numbers gives the voltage due to a combination of point charges, allowing us to use the principle of superposition: $V_P=k\sum_1^N\frac{q_i}{r_i}$.
- An electric dipole consists of two equal and opposite charges a fixed distance apart, with a dipole moment $\vec{\mathbf{p}}=q\vec{\mathbf{d}}$.
- Continuous charge distributions may be calculated with $V_P=k\int\frac{dq}{r}$.

- Just as we may integrate over the electric field to calculate the potential, we may take the derivative of the potential to calculate the electric field.
- This may be done for individual components of the electric field, or we may calculate the entire electric field vector with the gradient operator.

- An equipotential surface is the collection of points in space that are all at the same potential. Equipotential lines are the two-dimensional representation of equipotential surfaces.
- Equipotential surfaces are always perpendicular to electric field lines.
- Conductors in static equilibrium are equipotential surfaces.
- Topographic maps may be thought of as showing gravitational equipotential lines.

- Electrostatics is the study of electric fields in static equilibrium.
- In addition to research using equipment such as a Van de Graaff generator, many practical applications of electrostatics exist, including photocopiers, laser printers, ink jet printers, and electrostatic air filters.

3.1. $K=\frac{1}{2}mv^2,$ $v=\sqrt{2\frac{K}{m}}=\sqrt{2\frac{4.5\times10^{-7}~\mathrm{J}}{4.00\times10^{-9}~\mathrm{kg}}}=15~\mathrm{m/s}$

3.2. It has kinetic energy of $4.5\times10^{-7}~\mathrm{J}$ at point $r_2$ and potential energy of $9.0\times10^{-7}~\mathrm{J},$ which means that as $Q$ approaches infinity, its kinetic energy totals three times the kinetic energy at $r_2,$ since all of the potential energy gets converted to kinetic.

3.3. positive, negative, and these quantities are the same as the work you would need to do to bring the charges in from infinity

3.4. $\Delta U=q\Delta V=(100~\mathrm{C})(1.5~\mathrm{V})=150~\mathrm{J}$

3.5. $-2.00~\mathrm{C},$ $n_e=1.25\times10^{19}$ electrons

3.6. It would be going in the opposite direction, with no effect on the calculations as presented.

3.7. Given a fixed maximum electric field strength, the potential at which a strike occurs increases with increasing height above the ground. Hence, each electron will carry more energy. Determining if there is an effect on the total number of electrons lies in the future.

3.8. $V=k\frac{q}{r}=(8.99\times10^9~\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2)\left(\frac{-3.00\times10^{-9}~\mathrm{C}}{5.00\times10^{-3}~\mathrm{m}}\right)=-5390~\mathrm{V}$ recall that the electric field inside a conductor is zero. Hence, any path from a point on the surface to any point in the interior will have an integrand of zero when calculating the change in potential, and thus the potential in the interior of the sphere is identical to that on the surface.

3.9. The x-axis the potential is zero, due to the equal and opposite charges the same distance from it. On the $z$-axis, we may superimpose the two potentials; we will find that for $z\gg d,$ again the potential goes to zero due to cancellation.

3.10. It will be zero, as at all points on the axis, there are equal and opposite charges equidistant from the point of interest. Note that this distribution will, in fact, have a dipole moment.

3.11. Any, but cylindrical is closest to the symmetry of a dipole.

3.12. infinite cylinders of constant radius, with the line charge as the axis

1. Would electric potential energy be meaningful if the electric field were not conservative?

2. Why do we need to be careful about work done *on* the system versus work done *by* the system in calculations?

3. Does the order in which we assemble a system of point charges affect the total work done?

4. Discuss how potential difference and electric field strength are related. Give an example.

5. What is the strength of the electric field in a region where the electric potential is constant?

6. If a proton is released from rest in an electric field, will it move in the direction of increasing or decreasing potential? Also answer this question for an electron and a neutron. Explain why.

7. Voltage is the common word for potential difference. Which term is more descriptive, voltage or potential difference?

8. If the voltage between two points is zero, can a test charge be moved between them with zero net work being done? Can this necessarily be done without exerting a force? Explain.

9. What is the relationship between voltage and energy? More precisely, what is the relationship between potential difference and electric potential energy?

10. Voltages are always measured between two points. Why?

11. How are units of volts and electron-volts related? How do they differ?

12. Can a particle move in a direction of increasing electric potential, yet have its electric potential energy decrease? Explain.

13. Compare the electric dipole moments of charges $\pm Q$ separated by a distance $d$ and charges $\pm Q/2$ separated by a distance $d/2$.

14. Would Gauss’s law be helpful for determining the electric field of a dipole? Why?

15. In what region of space is the potential due to a uniformly charged sphere the same as that of a point charge? In what region does it differ from that of a point charge?

16. Can the potential of a nonuniformly charged sphere be the same as that of a point charge? Explain.

17. If the electric field is zero throughout a region, must the electric potential also be zero in that region?

18. Explain why knowledge of $\vec{\mathbf{E}}(x,y,z)$ is not sufficient to determine $V(x,y,z)$ What about the other way around?

19. If two points are at the same potential, are there any electric field lines connecting them?

20. Suppose you have a map of equipotential surfaces spaced $1.0~\mathrm{V}$ apart. What do the distances between the surfaces in a particular region tell you about the strength of the $\vec{\mathbf{E}}$ in that region?

21. Is the electric potential necessarily constant over the surface of a conductor?

22. Under electrostatic conditions, the excess charge on a conductor resides on its surface. Does this mean that all of the conduction electrons in a conductor are on the surface?

23. Can a positively charged conductor be at a negative potential? Explain.

24. Can equipotential surfaces intersect?

25. Why are the metal support rods for satellite network dishes generally grounded?

26. (a) Why are fish reasonably safe in an electrical storm? (b) Why are swimmers nonetheless ordered to get out of the water in the same circumstance?

27. What are the similarities and differences between the processes in a photocopier and an electrostatic precipitator?

28. About what magnitude of potential is used to charge the drum of a photocopy machine? A web search for “xerography” may be of use.

29. Consider a charge $Q_1(+5.0~\mu\mathrm{C})$ fixed at a site with another charge $Q_2$ (charge $+3~\mu\mathrm{C},$ mass $6.0~\mu\mathrm{g}$) moving in the neighboring space. (a) Evaluate the potential energy of $Q_2$ when it is $4.0~\mathrm{cm}$ from $Q_1.$ (b) If $Q_2$ starts from rest from a point $4.0~\mathrm{cm}$ from $Q_1,$ what will be its speed when it is $8.0~\mathrm{cm}$ from $Q_1$? (Note: $Q_1$ is held fixed in its place.)

30. Two charges $Q_1(+2.00~\mu\mathrm{C})$ and $Q_2(+2.00~\mu\mathrm{C})$ are placed symmetrically along the $x$-axis at $x=\pm3.00~\mathrm{cm}.$ Consider a charge $Q_3=+4.00~\mu\mathrm{C}$ and mass $10.0~\mathrm{mg}$ moving along the $y$-axis. If $Q_3$ starts from rest at $y=2.00~\mathrm{cm}$ what is its speed when it reaches $y=4.00~\mathrm{cm}$?

31. To form a hydrogen atom, a proton is fixed at a point and an electron is brought from far away to a distance of $0.529\times10^{-10}~\mathrm{m},$ the average distance between proton and electron in a hydrogen atom. How much work is done?

32. (a) What is the average power output of a heart defibrillator that dissipates $400~\mathrm{J}$ of energy in $10.0~\mathrm{ms}$? (b) Considering the high-power output, why doesn’t the defibrillator produce serious burns?

33. Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same voltage, assuming non-relativistic final speeds. Take the mass of the hydrogen ion to be $1.67\times10^{-27}~\mathrm{kg}.$

34. An evacuated tube uses an accelerating voltage of $40~\mathrm{kV}$ to accelerate electrons to hit a copper plate and produce X-rays. Non-relativistically, what would be the maximum speed of these electrons?

35. Show that units of $\mathrm{V/m}$ and $\mathrm{N/C}$ for electric field strength are indeed equivalent.

36. What is the strength of the electric field between two parallel conducting plates separated by $1.00~\mathrm{cm}$ and having a potential difference (voltage) between them of $1.50\times10^{4}~\mathrm{V}$?

37. The electric field strength between two parallel conducting plates separated by $4.00~\mathrm{cm}$ is $7.50\times10^4~\mathrm{V}.$ (a) What is the potential difference between the plates? (b) The plate with the lowest potential is taken to be zero volts. What is the potential $1.00~\mathrm{cm}$ from that plate and $3.00~\mathrm{cm}$ from the other?

38. The voltage across a membrane forming a cell wall is $8.0~\mathrm{mV}$ and the membrane is $9.00~\mathrm{nm}$ thick. What is the electric field strength? (The value is surprisingly large, but correct.) You may assume a uniform electric field.

39. Two parallel conducting plates are separated by $10.0~\mathrm{cm}$, and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential $8.00~\mathrm{cm}$ from the zero volt plate (and $2.00~\mathrm{cm}$ from the other) is $450~\mathrm{V}$? (b) What is the voltage between the plates?

40. Find the maximum potential difference between two parallel conducting plates separated by $0.500~\mathrm{cm}$ of air, given the maximum sustainable electric field strength in air to be $3.0\times10^6~\mathrm{V/m}.$

41. An electron is to be accelerated in a uniform electric field having a strength of $2.00\times10^6~\mathrm{V/m}.$ (a) What energy in $\mathrm{keV}$ is given to the electron if it is accelerated through $0.400~\mathrm{M}$? (b) Over what distance would it have to be accelerated to increase its energy by $50.0~\mathrm{GeV}$?

42. Use the definition of potential difference in terms of electric field to deduce the formula for potential difference between $r=r_a$ and $r=r_b$ for a point charge located at the origin. Here $r$ is the spherical radial coordinate.

43. The electric field in a region is pointed away from the $z$-axis and the magnitude depends upon the distance $s$ from the axis. The magnitude of the electric field is given as $E=\frac{\alpha}{s}$ where $\alpha$ is a constant. Find the potential difference between points $P_1$ and $P_2,$ explicitly stating the path over which you conduct the integration for the line integral.

44. Singly charged gas ions are accelerated from rest through a voltage of $13.0~\mathrm{V}$. At what temperature will the average kinetic energy of gas molecules be the same as that given these ions?

45. A $0.500{\text -}\mathrm{cm}$-diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed $40.0{\text -}\mathrm{pC}$ charge on its surface. What is the potential near its surface?

46. How far from a $1.00{\text -}\mu\mathrm{C}$ point charge is the potential $100~\mathrm{V}$? At what distance is it $2.00\times10^2~\mathrm{V}$?

47. If the potential due to a point charge is $5.00\times10^2~\mathrm{V}$ at a distance of $15.0~\mathrm{m},$ what are the sign and magnitude of the charge?

48. In nuclear fission, a nucleus splits roughly in half. (a) What is the potential $2.00\times10^{-14}~\mathrm{m}$ from a fragment that has $46$ protons in it? (b) What is the potential energy in $\mathrm{MeV}$ of a similarly charged fragment at this distance?

49. A research Van de Graaff generator has a $2.00{\text -}\mathrm{m}$-diameter metal sphere with a charge of $5.00~\mathrm{mC}$ on it. (a) What is the potential near its surface? (b) At what distance from its center is the potential $1.00~\mathrm{MV}$? (c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its energy in $\mathrm{MeV}$ when the atom is at the distance found in part b?

50. An electrostatic paint sprayer has a $0.200{\text -}\mathrm{m}$-diameter metal sphere at a potential of $25.0~\mathrm{kV}$ that repels paint droplets onto a grounded object. (a) What charge is on the sphere? (b) What charge must a $0.100{\text -}\mathrm{mg}$ drop of paint have to arrive at the object with a speed of $10.0~\mathrm{m/s}$?

51. (a) What is the potential between two points situated $10~\mathrm{cm}$ and $20~\mathrm{cm}$ from a $3.0{\text -}\mu\mathrm{C}$ point charge? (b) To what location should the point at $20~\mathrm{cm}$ be moved to increase this potential difference by a factor of two?

52. Find the potential at points $P_1,$ $P_2,$ $P_3,$ and $P_4$ in the diagram due to the two given charges.

53. Two charges $-2.0~\mu\mathrm{C}$ and $+2.0~\mu\mathrm{C}$ are separated by $4.0~\mathrm{cm}$ on the $z$-axis symmetrically about origin, with the positive one uppermost. Two space points of interest $P_1$ and $P_2$ are located $3.0~\mathrm{cm}$ and $30~\mathrm{cm}$ from origin at an angle $30^{\circ}$ with respect to the $z$-axis. Evaluate electric potentials at $P_1$ and $P_2$ in two ways: (a) Using the exact formula for point charges, and (b) using the approximate dipole potential formula.

54. (a) Plot the potential of a uniformly charged $1{\text -}\mathrm{m}$ rod with $1~\mathrm{C/m}$ charge as a function of the perpendicular distance from the centre. Draw your graph from $s=0.1~\mathrm{m}$ to $s=1.0~\mathrm{m}.$ (b) On the same graph, plot the potential of a point charge with a $1{\text -}\mathrm{C}$ charge at the origin. (c) Which potential is stronger near the rod? (d) What happens to the difference as the distance increases? Interpret your result.

55. Throughout a region, equipotential surfaces are given by $z=\mathrm{constant}.$ The surfaces are equally spaced with $V=100~\mathrm{V}$ for $z=0.00~\mathrm{m},$ $V=200~\mathrm{V}$ for $z=0.50~\mathrm{m},$ $V=300~\mathrm{V}$ for $z=1.00~\mathrm{m}.$ What is the electric field in this region?

56. In a particular region, the electric potential is given by $V=-xy^2z+4xy.$ What is the electric field in this region?

57. Calculate the electric field of an infinite line charge, throughout space.

58. Two very large metal plates are placed $2.0~\mathrm{cm}$ apart, with a potential difference of $12~\mathrm{V}$ between them. Consider one plate to be at $12~\mathrm{V},$ and the other at $0~\mathrm{V}.$ (a) Sketch the equipotential surfaces for $0,$ $4,$ $8,$ and $12~\mathrm{V}.$ (b) Next sketch in some electric field lines, and confirm that they are perpendicular to the equipotential lines.

59. A very large sheet of insulating material has had an excess of electrons placed on it to a surface charge density of $-3.00~\mathrm{nC/m}^2.$ (a) As the distance from the sheet increases, does the potential increase or decrease? Can you explain why without any calculations? Does the location of your reference point matter? (b) What is the shape of the equipotential surfaces? (c) What is the spacing between surfaces that differ by $1.00~\mathrm{V}$?

60. A metallic sphere of radius $2.0~\mathrm{cm}$ is charged with $+5.00{\text -}\mu\mathrm{C}$ charge, which spreads on the surface of the sphere uniformly. The metallic sphere stands on an insulated stand and is surrounded by a larger metallic spherical shell, of inner radius $5.0~\mathrm{cm}$ and outer radius $6.0~\mathrm{cm}.$ Now, a charge of $-5.0{\text -}\mu\mathrm{C}$ is placed on the inside of the spherical shell, which spreads out uniformly on the inside surface of the shell. If potential is zero at infinity, what is the potential of (a) the spherical shell, (b) the sphere, (c) the space between the two, (d) inside the sphere, and (e) outside the shell?

61. Two large charged plates of charge density $\pm30~\mu\mathrm{C}/m^2$ face each other at a separation of $5.0~\mathrm{mm}.$ (a) Find the electric potential everywhere. (b) An electron is released from rest at the negative plate; with what speed will it strike the positive plate?

62. A long cylinder of aluminum of radius $R$ meters is charged so that it has a uniform charge per unit length on its surface of $\lambda.$ (a) Find the electric field inside and outside the cylinder. (b) Find the electric potential inside and outside the cylinder. (c) Plot electric field and electric potential as a function of distance from the centre of the rod.

63. Two parallel plates $10~\mathrm{cm}$ on a side are given equal and opposite charges of magnitude $5.0\times10^{-9}.$ The plates are $1.5~\mathrm{mm}$ apart. What is the potential difference between the plates?

64. The surface charge density on a long straight metallic pipe is $\sigma.$ What is the electric potential outside and inside the pipe? Assume the pipe has a diameter of $2a.$

65. Concentric conducting spherical shells carry charges $Q$ and $-Q,$ respectively. The inner shell has negligible thickness. What is the potential difference between the shells?

66. Shown below are two concentric spherical shells of negligible thicknesses and radii $R_1$ and $R_2.$ The inner and outer shell carry net charges $q_1$ and $q_2,$ respectively, where both $q_1$ and $q_2$ are positive. What is the electric potential in the regions (a) $r<R_1,$ (b) $R_1<r<R_2,$ and (c) $r>R_2$?

67. A solid cylindrical conductor of radius a is surrounded by a concentric cylindrical shell of inner radius $b$. The solid cylinder and the shell carry charges $Q$ and $-Q,$ respectively. Assuming that the length $L$ of both conductors is much greater than $a$ or $b,$ what is the potential difference between the two conductors?

68. (a) What is the electric field $5.00~\mathrm{m}$ from the centre of the terminal of a Van de Graaff with a $3.00{\text -}\mathrm{mC}$ charge, noting that the field is equivalent to that of a point charge at the centre of the terminal? (b) At this distance, what force does the field exert on a $2.00{\text -}\mu\mathrm{C}$ charge on the Van de Graaff’s belt?

69. (a) What is the direction and magnitude of an electric field that supports the weight of a free electron near the surface of Earth? (b) Discuss what the small value for this field implies regarding the relative strength of the gravitational and electrostatic forces.

70. A simple and common technique for accelerating electrons is shown in \ref{fig:3.7.1}, where there is a uniform electric field between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to continue moving. (a) Calculate the acceleration of the electron if the field strength is $2.50\times10^4~\mathrm{N/C}.$ (b) Explain why the electron will not be pulled back to the positive plate once it moves through the hole.
\begin{gather}.\tag{Figure 3.7.1}\label{fig:3.7.1}\end{gather}

71. In a Geiger counter, a thin metallic wire at the center of a metallic tube is kept at a high voltage with respect to the metal tube. Ionizing radiation entering the tube knocks electrons off gas molecules or sides of the tube that then accelerate towards the center wire, knocking off even more electrons. This process eventually leads to an avalanche that is detectable as a current. A particular Geiger counter has a tube of radius $R$ and the inner wire of radius $a$ is at a potential of $V_0$ volts with respect to the outer metal tube. Consider a point $P$ at a distance $s$ from the centre wire and far away from the ends. (a) Find a formula for the electric field at a point $P$ inside using the infinite wire approximation. (b) Find a formula for the electric potential at a point $P$ inside. (c) Use $V_0=900~\mathrm{V},$ $a=3.00~\mathrm{mm},$ $R=2.00~\mathrm{cm},$ and find the value of the electric field at a point $1.00~\mathrm{cm}$ from the centre.

72. The practical limit to an electric field in air is about $3.00\times10^6~\mathrm{N/C}.$ Above this strength, sparking takes place because air begins to ionize. (a) At this electric field strength, how far would a proton travel before hitting the speed of light (ignore relativistic effects)? (b) Is it practical to leave air in particle accelerators?

73. To form a helium atom, an alpha particle that contains two protons and two neutrons is fixed at one location, and two electrons are brought in from far away, one at a time. The first electron is placed at $0.600\times10^{-10}~\mathrm{m}$ from the alpha particle and held there while the second electron is brought to $0.600\times10^{-10}~\mathrm{m}$ from the alpha particle on the other side from the first electron. See the final configuration below. (a) How much work is done in each step? (b) What is the electrostatic energy of the alpha particle and two electrons in the final configuration?

74. Find the electrostatic energy of eight equal charges $(+3~\mu\mathrm{C})$ each fixed at the corners of a cube of side $2~\mathrm{cm}.$

75. The probability of fusion occurring is greatly enhanced when appropriate nuclei are brought close together, but mutual Coulomb repulsion must be overcome. This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei toward one another. (a) Calculate the potential energy of two singly charged nuclei separated by $1.00\times10^{-12}~\mathrm{m}$ (b) At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?

76. A bare helium nucleus has two positive charges and a mass of $6.64\times10^{-27}~\mathrm{kg}.$ (a) Calculate its kinetic energy in joules at $2.00\%$ of the speed of light. (b) What is this in electron-volts? (c) What voltage would be needed to obtain this energy?

77. An electron enters a region between two large parallel plates made of aluminum separated by a distance of $2.0~\mathrm{cm}$ and kept at a potential difference of $200~\mathrm{V}.$ The electron enters through a small hole in the negative plate and moves toward the positive plate. At the time the electron is near the negative plate, its speed is $4.0\times10^5~\mathrm{m/s}.$ Assume the electric field between the plates to be uniform, and find the speed of electron at (a) $0.10~\mathrm{cm},$ (b) $0.50~\mathrm{cm},$ (c) $1.0~\mathrm{cm},$ and (d) $1.5~\mathrm{cm}$ from the negative plate, and (e) immediately before it hits the positive plate.

78. How far apart are two conducting plates that have an electric field strength of $4.50\times10^3~\mathrm{V/m}$ between them, if their potential difference is $15.0~\mathrm{kV}$?

79. (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength of dry air, which is $3.00\times10^6~\mathrm{V/m},$ if the plates are separated by $2.00~\mathrm{mm}$ and a potential difference of $5.0\times10^3~\mathrm{V}$ is applied? (b) How close together can the plates be with this applied voltage?

80. Membrane walls of living cells have surprisingly large electric fields across them due to separation of ions. What is the voltage across an $8.00{\text -}\mathrm{nm}$-thick membrane if the electric field strength across it is $5.50~\mathrm{MV/m}$? You may assume a uniform electric field.

81. A double charged ion is accelerated to an energy of $32.0~\mathrm{keV}$ by the electric field between two parallel conducting plates separated by $2.00~\mathrm{cm}.$ What is the electric field strength between the plates?

82. The temperature near the centre of the Sun is thought to be $15$ million degrees Celsius $(1.5\times10^7~^{\circ}\mathrm{C})$ (or kelvin). Through what voltage must a singly charged ion be accelerated to have the same energy as the average kinetic energy of ions at this temperature?

83. A lightning bolt strikes a tree, moving $20.0~\mathrm{C}$ of charge through a potential difference of $1.00\times10^2~\mathrm{MV}$ (a) What energy was dissipated? (b) What mass of water could be raised from $15~^{\circ}\mathrm{C}$ to the boiling point and then boiled by this energy? (c) Discuss the damage that could be caused to the tree by the expansion of the boiling steam.

84. What is the potential $0.530\times10^{-10}~\mathrm{m}$ from a proton (the average distance between the proton and electron in a hydrogen atom)?

85. (a) A sphere has a surface uniformly charged with $1.00~\mathrm{C}$. At what distance from its centre is the potential $5.00~\mathrm{MV}$? (b) What does your answer imply about the practical aspect of isolating such a large charge?

86. What are the sign and magnitude of a point charge that produces a potential of $-2.00~\mathrm{V}$ at a distance of $1.00~\mathrm{mm}$?

87. In one of the classic nuclear physics experiments at the beginning of the twentieth century, an alpha particle was accelerated toward a gold nucleus, and its path was substantially deflected by the Coulomb interaction. If the energy of the doubly charged alpha nucleus was $5.00~\mathrm{MeV},$ how close to the gold nucleus ($79$ protons) could it come before being deflected?

88. A $12.0{\text -}\mathrm{V}$ battery-operated bottle warmer heats $50.0~\mathrm{g}$ of glass, $2.50\times10^2~\mathrm{g}$ of baby formula, and $2.00\times10^2~\mathrm{g}$ of aluminum from $20.0~^{\circ}\mathrm{C}$ to $90.0~^{\circ}\mathrm{C}.$ (a) How much charge is moved by the battery? (b) How many electrons per second flow if it takes $5.00~\mathrm{min}$ to warm the formula? (*Hint:* Assume that the specific heat of baby formula is about the same as the specific heat of water.)

89. A battery-operated car uses a $12.0{\text -}\mathrm{V}$ system. Find the charge the batteries must be able to move in order to accelerate the $750~\mathrm{kg}$ car from rest to $25.0~\mathrm{m/s},$ make it climb a $2.00\times10^2{\text -}\mathrm{m}$ high hill, and finally cause it to travel at a constant $25.0~\mathrm{m/s}$ while climbing with $5.00\times10^2{\text -}\mathrm{N}$ force for an hour.

90. (a) Find the voltage near a $10.0~\mathrm{cm}$ diameter metal sphere that has $8.00~\mathrm{C}$ of excess positive charge on it. (b) What is unreasonable about this result? (c) Which assumptions are responsible?

91. A uniformly charged ring of radius $10~\mathrm{cm}$ is placed on a nonconducting table. It is found that $3.0~\mathrm{cm}$ above the centre of the half-ring the potential is $-3.0~\mathrm{V}$ with respect to zero potential at infinity. How much charge is in the half-ring?

92. A glass ring of radius $5.0~\mathrm{cm}$ is painted with a charged paint such that the charge density around the ring varies continuously given by the following function of the polar angle $\theta,$ $\lambda=(3.0\times10^{-6}~\mathrm{C/m}).$ Find the potential at a point $15~\mathrm{cm}$ above the centre.

93. A compact disk of radius $(R=3.0~\mathrm{cm})$ is sprayed with a charged paint so that the charge varies continually with radial distance $r$ from the centre in the following manner: $\sigma=-(6.0~\mathrm{C/m})r/R.$ Find the potential at a point $4~\mathrm{cm}$ above the centre.

94. (a) What is the final speed of an electron accelerated from rest through a voltage of $25.0~\mathrm{MV}$ by a negatively charged Van de Graff terminal? (b) What is unreasonable about this result? (c) Which assumptions are responsible?

95. A large metal plate is charged uniformly to a density of $\sigma=2.0\times10^{-9}~\mathrm{C/m}^2.$ How far apart are the equipotential surfaces that represent a potential difference of $25~\mathrm{V}$?

96. Your friend gets really excited by the idea of making a lightning rod or maybe just a sparking toy by connecting two spheres as shown in Figure 3.5.10, and making $R_2$ so small that the electric field is greater than the dielectric strength of air, just from the usual $150~\mathrm{V/m}$ electric field near the surface of the Earth. If $R_1$ is $10~\mathrm{cm},$ how small does $R_2$ need to be, and does this seem practical?

97. (a) Find $x\gg L$ limit of the potential of a finite uniformly charged rod and show that it coincides with that of a point charge formula. (b) Why would you expect this result?

98. A small spherical pith ball of radius $0.50~\mathrm{cm}$ is painted with a silver paint and then $-10~\mu\mathrm{C}$ of charge is placed on it. The charged pith ball is put at the centre of a gold spherical shell of inner radius $2.0~\mathrm{cm}$ and outer radius $2.2~\mathrm{cm}.$ (a) Find the electric potential of the gold shell with respect to zero potential at infinity. (b) How much charge should you put on the gold shell if you want to make its potential $100~\mathrm{V}$?

99. Two parallel conducting plates, each of cross-sectional area $400~\mathrm{cm}^2,$ are $2.0~\mathrm{cm}$ apart and uncharged. If $1.0\times10^{12}$ electrons are transferred from one plate to the other, (a) what is the potential difference between the plates? (b) What is the potential difference between the positive plate and a point $1.25~\mathrm{cm}$ from it that is between the plates?

100. A point charge of $q=5.0\times10^{-8}~\mathrm{C}$ is placed at the centre of an uncharged spherical conducting shell of inner radius $6.0~\mathrm{cm}$ and outer radius $9.0~\mathrm{cm}.$ Find the electric potential at (a) $r=4.0~\mathrm{cm},$ (b) $r=8.0~\mathrm{cm},$ (c) $r=12.0~\mathrm{cm}.$

101. Earth has a net charge that produces an electric field of approximately $150~\mathrm{N/C}$ downward at its surface. (a) What is the magnitude and sign of the excess charge, noting the electric field of a conducting sphere is equivalent to a point charge at its centre? (b) What acceleration will the field produce on a free electron near Earth’s surface? (c) What mass object with a single extra electron will have its weight supported by this field?

102. Point charges of $25.0~\mu\mathrm{C}$ and $45.0~\mu\mathrm{C}$ are placed $0.500~\mathrm{m}$ apart. (a) At what point along the line between them is the electric field zero? (b) What is the electric field halfway between them?

103. What can you say about two charges $q_1$ and $q_2,$ if the electric field one-fourth of the way from $q_1$ to $q_2$ is zero?

104. Calculate the angular velocity $\omega$ of an electron orbiting a proton in the hydrogen atom, given the radius of the orbit is $0.530\times10^{-10}~\mathrm{m}.$ You may assume that the proton is stationary and the centripetal force is supplied by Coulomb attraction.

105. An electron has an initial velocity of $5.00\times10^6~\mathrm{m/s}$ in a uniform $2.00\times10^5{\text -}\mathrm{N/C}$ electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? (b) How far does the electron travel before coming to rest? (c) How long does it take the electron to come to rest? (d) What is the electron’s velocity when it returns to its starting point?

106. Three $\mathrm{Na}^+$ and three $\mathrm{Cl}^-$ ions are placed alternately and equally spaced around a circle of radius $50~\mathrm{nm}$. Find the electrostatic energy stored.

107. Look up (presumably online, or by dismantling an old device and making measurements) the magnitude of the potential deflection plates (and the space between them) in an ink jet printer. Then look up the speed with which the ink comes out the nozzle. Can you calculate the typical mass of an ink drop?

108. Use the electric field of a finite sphere with constant volume charge density to calculate the electric potential, throughout space. Then check your results by calculating the electric field from the potential.

109. Calculate the electric field of a dipole throughout space from the potential.

- Explain the concepts of a capacitor and its capacitance
- Describe how to evaluate the capacitance of a system of conductors

A **capacitor** is a device used to store electrical charge and electrical energy. It consists of at least two electrical conductors separated by a distance. (Note that such electrical conductors are sometimes referred to as “electrodes,” but more correctly, they are “capacitor plates.”) The space between capacitors may simply be a vacuum, and, in that case, a capacitor is then known as a “vacuum capacitor.” However, the space is usually filled with an insulating material known as a **dielectric**. (You will learn more about dielectrics in the sections on dielectrics later in this chapter.) The amount of storage in a capacitor is determined by a property called *capacitance*, which you will learn more about a bit later in this section.

Capacitors have applications ranging from filtering static from radio reception to energy storage in heart defibrillators. Typically, commercial capacitors have two conducting parts close to one another but not touching, such as those in \ref{fig:4.1.1}. Most of the time, a dielectric is used between the two plates. When battery terminals are connected to an initially uncharged capacitor, the battery potential moves a small amount of charge of magnitude $Q$ from the positive plate to the negative plate. The capacitor remains neutral overall, but with charges $+Q$ and $-Q$ residing on opposite plates.

\begin{gather}.\tag{Figure 4.1.1}\label{fig:4.1.1}\end{gather}A system composed of two identical parallel-conducting plates separated by a distance is called a **parallel-plate capacitor** (\ref{fig:4.1.2}). The magnitude of the electrical field in the space between the parallel plates is $E=\sigma/\epsilon_0$, where $\sigma$ denotes the surface charge density on one plate (recall that σσ is the charge $Q$ per the surface area $A$). Thus, the magnitude of the field is directly proportional to $Q$.

Capacitors with different physical characteristics (such as shape and size of their plates) store different amounts of charge for the same applied voltage $V$ across their plates. The **capacitance** $C$ of a capacitor is defined as the ratio of the maximum charge $Q$ that can be stored in a capacitor to the applied voltage $V$ across its plates. In other words, capacitance is the largest amount of charge per volt that can be stored on the device:

\begin{equation}C=\frac{Q}{V}.\tag{4.1.1}\label{eq:4.1.1}\end{equation}

The SI unit of capacitance is the **farad** ($\mathrm{F}$), named after Michael **Faraday** (1791–1867). Since capacitance is the charge per unit voltage, one farad is one coulomb per one volt, or

By definition, a $1.0{\text -}\mathrm{F}$ capacitor is able to store $1.0~\mathrm{C}$ of charge (a very large amount of charge) when the potential difference between its plates is only $1.0~\mathrm{V}$. One farad is therefore a very large capacitance. Typical capacitance values range from picofarads ($1~\mathrm{pF}=10^{-12}~\mathrm{F}$) to millifarads ($1~\mathrm{mF}=10^{-3}~\mathrm{F}$), which also includes microfarads ($1~\mu\mathrm{F}=10^{-6}~\mathrm{F}$). Capacitors can be produced in various shapes and sizes (\ref{fig:4.1.3}).

\begin{gather}.\tag{Figure 4.1.3}\label{fig:4.1.3}\end{gather}We can calculate the capacitance of a pair of conductors with the standard approach that follows.

- Assume that the capacitor has a charge $Q$.
- Determine the electrical field $\vec{\mathbf{E}}$ between the conductors. If symmetry is present in the arrangement of conductors, you may be able to use Gauss’s law for this calculation.
- Find the potential difference between the conductors from \begin{equation}V_B-V_A=-\int_A^B\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}},\tag{4.1.2}\label{eq:4.1.2}\end{equation} where the path of integration leads from one conductor to the other. The magnitude of the potential difference is then $V=|V_B-V_A|$.
- With $V$ known, obtain the capacitance directly from Equation \ref{eq:4.1.1}.

The parallel-plate capacitor (\ref{fig:4.1.4}) has two identical conducting plates, each having a surface area $A$, separated by a distance $d$. When a voltage $V$ is applied to the capacitor, it stores a charge $Q$, as shown. We can see how its capacitance may depend on $A$ and $d$* *by considering characteristics of the Coulomb force. We know that force between the charges increases with charge values and decreases with the distance between them. We should expect that the bigger the plates are, the more charge they can store. Thus, $C$* *should be greater for a larger value of $A$. Similarly, the closer the plates are together, the greater the attraction of the opposite charges on them. Therefore, $C$ should be greater for a smaller $d$.

We define the surface charge density σσ on the plates as

\[\sigma=\frac{Q}{A}.\]We know from previous chapters that when $d$ is small, the electrical field between the plates is fairly uniform (ignoring edge effects) and that its magnitude is given by

\[E=\frac{\sigma}{\epsilon_0},\] where the constant ε0ε0 is the permittivity of free space, $\epsilon_0=8.85\times10^{-12}~\mathrm{F/m}$. The SI unit of $\mathrm{F/m}$ is equivalent to $\mathrm{C}^2/(\mathrm{N}\cdot\mathrm{m}^2)$. Since the electrical field $\vec{\mathbf{E}}$ between the plates is uniform, the potential difference between the plates is
\[V=Ed=\frac{\sigma d}{\epsilon_0}=\frac{Qd}{\epsilon_0A}.\]

Therefore Equation \ref{eq:4.1.3} gives the capacitance of a parallel-plate capacitor as

\begin{equation}C=\frac{Q}{V}=\frac{Q}{Qd/\epsilon_0A}=\epsilon_0\frac{A}{d}.\tag{4.1.3}\label{eq:4.1.3}\end{equation}

Notice from this equation that capacitance is a function *only of the geometry* and what material fills the space between the plates (in this case, vacuum) of this capacitor. In fact, this is true not only for a parallel-plate capacitor, but for all capacitors: The capacitance is independent of $Q$ or $V$. If the charge changes, the potential changes correspondingly so that $Q/V$ remains constant.

\[Q=CV=(8.85\times10^{-9}~\mathrm{F})(3.00\times10^3~\mathrm{V})=26.6~\mu\mathrm{C}.\]

\[A=\frac{Cd}{\epsilon_0}=\frac{(1.0~\mathrm{F})(1.0\times10^{-3}~\mathrm{m})}{8.85\times10^{-12}~\mathrm{F/m}}=1.1\times10^8~\mathrm{m}^2.\]

Each square plate would have to be $10~\mathrm{km}$ across. It used to be a common prank to ask a student to go to the laboratory stockroom and request a $1{\text -}\mathrm{F}$ parallel-plate capacitor, until stockroom attendants got tired of the joke.

The capacitance of a parallel-plate capacitor is $2.0~\mathrm{pF}$. If the area of each plate is $2.4~\mathrm{cm}^2$, what is the plate separation?

Verify that $\sigma/V$ and $\epsilon_0/d$ have the same physical units.

A spherical capacitor is another set of conductors whose capacitance can be easily determined (\ref{fig:4.1.5}). It consists of two concentric conducting spherical shells of radii $R_1$ (inner shell) and $R_2$ (outer shell). The shells are given equal and opposite charges $+Q$ and $-Q$, respectively. From symmetry, the electrical field between the shells is directed radially outward. We can obtain the magnitude of the field by applying Gauss’s law over a spherical Gaussian surface of radius *r* concentric with the shells. The enclosed charge is $+Q$; therefore we have

Thus, the electrical field between the conductors is

\[\vec{\mathbf{E}}=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{\mathbf{r}}.\]

We substitute this $\vec{\mathbf{E}}$ into Equation \ref{eq:4.1.2} and integrate along a radial path between the shells:

\[V=\int_{R_1}^{R_2}\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}=\int_{R_1}^{R_2}\left(\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\hat{\mathbf{r}}\right)\cdot(\hat{\mathbf{r}}dr)=\frac{Q}{4\pi\epsilon_0}\int_{R_1}^{R_2}\frac{dr}{r^2}=\frac{Q}{4\pi\epsilon_0}\left(\frac{1}{R_1}-\frac{1}{R_2}\right).\]

In this equation, the potential difference between the plates is $V=-(V_2-V_1=V_1-V_2)$. We substitute this result into Equation \ref{eq:4.1.1} to find the capacitance of a spherical capacitor:

\begin{equation}C=\frac{Q}{V}=4\pi\epsilon_0\frac{R_1R_2}{R_2-R_1}.\tag{4.1.4}\label{eq:4.1.4}\end{equation}

\begin{gather}.\tag{Figure 4.1.5}\label{fig:4.1.5}\end{gather}

\[V=\int_{R_1}^{+\infty}\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}=\frac{Q}{4\pi\epsilon_0}\int_{R_1}^{+\infty}\frac{1}{r^2}\hat{\mathbf{r}}\cdot(\hat{\mathbf{r}}dr)=\frac{Q}{4\pi\epsilon_0}\int_{R_1}^{+\infty}\frac{dr}{r^2}=\frac{1}{4\pi\epsilon_0}\frac{Q}{R_1}.\]

The capacitance of an isolated sphere is therefore

\[C=\frac{Q}{V}=Q\frac{4\pi\epsilon_0R_1}{Q}=4\pi\epsilon_0R_1.\]

The radius of the outer sphere of a spherical capacitor is five times the radius of its inner shell. What are the dimensions of this capacitor if its capacitance is $5.00~\mathrm{pF}$?

A cylindrical capacitor consists of two concentric, conducting cylinders (\ref{fig:4.1.6}). The inner cylinder, of radius $R_1$, may either be a shell or be completely solid. The outer cylinder is a shell of inner radius $R_2$. We assume that the length of each cylinder is $l$ and that the excess charges $+Q$ and $-Q$ reside on the inner and outer cylinders, respectively.

\begin{gather}.\tag{Figure 4.1.6}\label{fig:4.1.6}\end{gather}With edge effects ignored, the electrical field between the conductors is directed radially outward from the common axis of the cylinders. Using the Gaussian surface shown in \ref{fig:4.1.6}, we have

\[\oint_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=E(2\pi rl)=\frac{Q}{\epsilon_0}.\]Therefore, the electrical field between the cylinders is

\begin{equation}\vec{\mathbf{E}}=\frac{1}{2\pi\epsilon_0}\frac{Q}{r\,l}\hat{\mathrm{r}}.\tag{4.1.5}\label{eq:4.1.5}\end{equation}

Here \hat{\mathrm{r}} is the unit radial vector along the radius of the cylinder. We can substitute into Equation \ref{eq:4.1.2} and find the potential difference between the cylinders:

\begin{eqnarray*}V&=&\int_{R_1}^{R_2}\vec{\mathbf{E}}\cdot d\vec{\mathbf{l}}_p=\frac{Q}{2\pi\epsilon_0\,l}\int_{R_1}^{R_2}\frac{1}{r}\hat{\mathbf{r}}\cdot(\hat{\mathbf{r}}dr)=\frac{Q}{2\pi\epsilon_0\,l}\int_{R_1}^{R_2}\frac{dr}{r}\hat{\mathbf{r}}\\&=&\frac{Q}{2\pi\epsilon_0\,l}\ln r|_{R_1}^{R_2}=\frac{Q}{2\pi\epsilon_0\,l}\ln\frac{R_2}{R_1}\end{eqnarray*}

Thus, the capacitance of a cylindrical capacitor is

\begin{equation}C=\frac{Q}{V}=\frac{2\pi\epsilon_0\,l}{\ln(R_2/R_1)}\tag{4.1.6}\label{eq:4.1.6}\end{equation}

As in other cases, this capacitance depends only on the geometry of the conductor arrangement. An important application of Equation \ref{eq:4.1.6} is the determination of the capacitance per unit length of a *coaxial cable*, which is commonly used to transmit time-varying electrical signals. A **coaxial cable** consists of two concentric, cylindrical conductors separated by an insulating material. (Here, we assume a vacuum between the conductors, but the physics is qualitatively almost the same when the space between the conductors is filled by a dielectric.) This configuration shields the electrical signal propagating down the inner conductor from stray electrical fields external to the cable. Current flows in opposite directions in the inner and the outer conductors, with the outer conductor usually grounded. Now, from Equation \ref{eq:4.1.6}, the capacitance per unit length of the coaxial cable is given by

\[\frac{C}{l}=\frac{2\pi\epsilon_0}{\ln(R_2/R_1)}.\]

In practical applications, it is important to select specific values of $C/l$. This can be accomplished with appropriate choices of radii of the conductors and of the insulating material between them.

When a cylindrical capacitor is given a charge of $0.500~\mathrm{nC}$, a potential difference of $20.0~\mathrm{V}$ is measured between the cylinders. (a) What is the capacitance of this system? (b) If the cylinders are $1.0~\mathrm{m}$ long, what is the ratio of their radii?

Several types of practical capacitors are shown in \ref{fig:4.1.3}. Common capacitors are often made of two small pieces of metal foil separated by two small pieces of insulation (see \ref{fig:4.1.1}(b)). The metal foil and insulation are encased in a protective coating, and two metal leads are used for connecting the foils to an external circuit. Some common insulating materials are mica, ceramic, paper, and Teflon™ non-stick coating.

Another popular type of capacitor is an **electrolytic capacitor**. It consists of an oxidized metal in a conducting paste. The main advantage of an electrolytic capacitor is its high capacitance relative to other common types of capacitors. For example, capacitance of one type of aluminum electrolytic capacitor can be as high as $1.0~\mathrm{F}$. However, you must be careful when using an electrolytic capacitor in a circuit, because it only functions correctly when the metal foil is at a higher potential than the conducting paste. When reverse polarization occurs, electrolytic action destroys the oxide film. This type of capacitor cannot be connected across an alternating current source, because half of the time, ac voltage would have the wrong polarity, as an alternating current reverses its polarity (see Alternating-Current Circuts on alternating-current circuits).

A **variable air capacitor** (\ref{fig:4.1.7}) has two sets of parallel plates. One set of plates is fixed (indicated as “stator”), and the other set of plates is attached to a shaft that can be rotated (indicated as “rotor”). By turning the shaft, the cross-sectional area in the overlap of the plates can be changed; therefore, the capacitance of this system can be tuned to a desired value. Capacitor tuning has applications in any type of radio transmission and in receiving radio signals from electronic devices. Any time you tune your car radio to your favorite station, think of capacitance.

The symbols shown in \ref{fig:4.1.8} are circuit representations of various types of capacitors. We generally use the symbol shown in \ref{fig:4.1.8}(a). The symbol in \ref{fig:4.1.8}(c) represents a variable-capacitance capacitor. Notice the similarity of these symbols to the symmetry of a parallel-plate capacitor. An electrolytic capacitor is represented by the symbol in part \ref{fig:4.1.8}(b), where the curved plate indicates the negative terminal.

\begin{gather}.\tag{Figure 4.1.8}\label{fig:4.1.8}\end{gather}An interesting applied example of a capacitor model comes from cell biology and deals with the electrical potential in the plasma membrane of a living cell (\ref{fig:4.1.9}). **Cell membranes** separate cells from their surroundings but allow some selected ions to pass in or out of the cell. The potential difference across a membrane is about $70~\mathrm{mV}$. The cell membrane may be $7$ to $10~\mathrm{nm}$ thick. Treating the cell membrane as a nano-sized capacitor, the estimate of the smallest electrical field strength across its ‘plates’ yields the value $E=\frac{V}{d}=\frac{70\times10^{-3}~\mathrm{V}}{10\times10^{-9}~\mathrm{m}}=7\times10^6~\mathrm{V/m}>3~\mathrm{MV/m}$.

This magnitude of electrical field is great enough to create an electrical spark in the air.

\begin{gather}.\tag{Figure 4.1.9}\label{fig:4.1.9}\end{gather}Visit the PhET Explorations: Capacitor Lab to explore how a capacitor works. Change the size of the plates and add a dielectric to see the effect on capacitance. Change the voltage and see charges built up on the plates. Observe the electrical field in the capacitor. Measure the voltage and the electrical field.

- Explain how to determine the equivalent capacitance of capacitors in series and in parallel combinations
- Compute the potential difference across the plates and the charge on the plates for a capacitor in a network and determine the net capacitance of a network of capacitors

Several capacitors can be connected together to be used in a variety of applications. Multiple connections of capacitors behave as a single equivalent capacitor. The total capacitance of this equivalent single capacitor depends both on the individual capacitors and how they are connected. Capacitors can be arranged in two simple and common types of connections, known as *series* and *parallel*, for which we can easily calculate the total capacitance. These two basic combinations, series and parallel, can also be used as part of more complex connections.

\ref{fig:4.2.1} illustrates a series combination of three capacitors, arranged in a row within the circuit. As for any capacitor, the capacitance of the combination is related to the charge and voltage by using Equation 4.1.1. When this series combination is connected to a battery with voltage *V*, each of the capacitors acquires an identical charge $Q$. To explain, first note that the charge on the plate connected to the positive terminal of the battery is $+Q$ and the charge on the plate connected to the negative terminal is $-Q$. Charges are then induced on the other plates so that the sum of the charges on all plates, and the sum of charges on any pair of capacitor plates, is zero. However, the potential drop $V_1=Q/C_1$ on one capacitor may be different from the potential drop $V_2=Q/C_2$ on another capacitor, because, generally, the capacitors may have different capacitances. The series combination of two or three capacitors resembles a single capacitor with a smaller capacitance. Generally, any number of capacitors connected in series is equivalent to one capacitor whose capacitance (called the *equivalent capacitance*) is smaller than the smallest of the capacitances in the series combination. Charge on this equivalent capacitor is the same as the charge on any capacitor in a series combination: That is, *all capacitors of a series combination have the same charge*. This occurs due to the conservation of charge in the circuit. When a charge $Q$ in a series circuit is removed from a plate of the first capacitor (which we denote as $-Q$), it must be placed on a plate of the second capacitor (which we denote as $+Q$), and so on.

We can find an expression for the total (equivalent) capacitance by considering the voltages across the individual capacitors. The potentials across capacitors $1$, $2$, and $3$ are, respectively, $V_1=Q/C_1$, $V_2=Q/C_2$, and$V_3=Q/C_3$,. These potentials must sum up to the voltage of the battery, giving the following potential balance:

\[V=V_1+V_2+V_3.\]

Potential $V$ is measured across an equivalent capacitor that holds charge $Q$ and has an equivalent capacitance $C_{\mathrm{S}}$. Entering the expressions for $V_1$, $V_2$, and $V_3$, we get

\[\frac{Q}{C_{\mathrm{S}}}=\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}.\]

Canceling the charge $Q$, we obtain an expression containing the equivalent capacitance, $C_{\mathrm{S}}$, of three capacitors connected in series:

\[\frac{1}{C_{\mathrm{S}}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}.\]

This expression can be generalized to any number of capacitors in a series network.

For capacitors connected in a **series combination**, the reciprocal of the equivalent capacitance is the sum of reciprocals of individual capacitances:

\begin{equation}\frac{1}{C_{\mathrm{S}}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\ldots.\tag{4.2.1}\label{eq:4.2.1}\end{equation}

\begin{eqnarray*}\frac{1}{C_{\mathrm{S}}}&=&\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\\&=&\frac{1}{1.000~\mu\mathrm{F}}+\frac{1}{5.000~\mu\mathrm{F}}+\frac{1}{8.000~\mu\mathrm{F}}\\&=&\frac{1.325}{\mu\mathrm{F}}.\end{eqnarray*}

Now we invert this result and obtain $C_{\mathrm{S}}=\frac{\mu\mathrm{F}}{1.325}=0.755~\mu\mathrm{F}$.

A parallel combination of three capacitors, with one plate of each capacitor connected to one side of the circuit and the other plate connected to the other side, is illustrated in \ref{fig:4.2.2}(a). Since the capacitors are connected in parallel, *they all have the same voltage $V$ across their plates*. However, each capacitor in the parallel network may store a different charge. To find the equivalent capacitance $C_{\mathrm{P}}$ of the parallel network, we note that the total charge $Q$ stored by the network is the sum of all the individual charges:

\[Q=Q_1+Q_2+Q_3.\]

On the left-hand side of this equation, we use the relation $Q=C_{\mathrm{P}}V$, which holds for the entire network. On the right-hand side of the equation, we use the relations $Q_1=C_1V$, $Q_2=C_2V$, and $Q_3=C_3V$ for the three capacitors in the network. In this way we obtain

\[C_{\mathrm{P}}V=C_1V+C_2V+C_3V.\]

This equation, when simplified, is the expression for the equivalent capacitance of the parallel network of three capacitors:

\[C_{\mathrm{P}}=C_1+C_2+C_3.\]

This expression is easily generalized to any number of capacitors connected in parallel in the network.

For capacitors connected in a **parallel combination**, the equivalent (net) capacitance is the sum of all individual capacitances in the network,

\begin{equation}C_{\mathrm{P}}=C_1+C_2+C_3+\ldots.\tag{4.2.2}\label{eq:4.2.2}\end{equation}

\begin{gather}.\tag{Figure 4.2.2}\label{fig:4.2.2}\end{gather}

\begin{eqnarray*}C_{\mathrm{P}}&=&C_1+C_2+C_3=1.0~\mu\mathrm{F}+5.0~\mu\mathrm{F}+8.0~\mu\mathrm{F}\\&=&14.0~\mu\mathrm{F}.\end{eqnarray*}

Capacitor networks are usually some combination of series and parallel connections, as shown in \ref{fig:4.2.3}. To find the net capacitance of such combinations, we identify parts that contain only series or only parallel connections, and find their equivalent capacitances. We repeat this process until we can determine the equivalent capacitance of the entire network. The following example illustrates this process.

\begin{gather}.\tag{Figure 4.2.3}\label{fig:4.2.3}\end{gather}
\[\frac{1}{C_{\mathrm{S}}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{1.000~\mu\mathrm{F}}+\frac{1}{5.000~\mu\mathrm{F}}=\frac{1.200}{\mu\mathrm{F}}\Rightarrow C_{\mathrm{S}}=0.833~\mu\mathrm{F}.\]

Capacitance $C_{\mathrm{S}}$ is connected in parallel with the third capacitance $C_3$, so we use Equation \ref{eq:4.2.2} to find the equivalent capacitance $C$ of the entire network:

\[C=C_{\mathrm{S}}+C_3=0.833~\mu\mathrm{F}+8.000~\mu\mathrm{F}=8.833~\mu\mathrm{F}.\]

\[C_{23}=C_2+C_3=2.0~\mu\mathrm{F}+4.0~\mu\mathrm{F}=6.0~\mu\mathrm{F}.\]

The entire three-capacitor combination is equivalent to two capacitors in series,

\[\frac{1}{C}=\frac{1}{12.0~\mu\mathrm{F}}+\frac{1}{6.0~\mu\mathrm{F}}=\frac{1}{4.0~\mu\mathrm{F}}\Rightarrow C=4.0~\mu\mathrm{F}.\]

Consider the equivalent two-capacitor combination in \ref{fig:4.2.4}(b). Since the capacitors are in series, they have the same charge, $Q_1=Q_{23}$. Also, the capacitors share the $12.0{\text -}\mathrm{V}$ potential difference, so

\[12.0~\mathrm{V}=V_1+V_{23}=\frac{Q_1}{C_1}+\frac{Q_{23}}{C_{23}}=\frac{Q_1}{12.0~\mu\mathrm{F}}}+\frac{Q_{1}}{6.0~\mu\mathrm{F}}}\Rightarrow Q_1=48.0~\mu\mathrm{C}.\]

Now the potential difference across capacitor $1$ is

\[V_1=\frac{Q_1}{C_1}=\frac{48.0~\mu\mathrm{C}}{12.0~\mu\mathrm{F}}=4.0~\mathrm{V}.\]

Because capacitors $2$ and $3$ are connected in parallel, they are at the same potential difference:

\[V_2=V_3=12.0~\mathrm{V}-4.0~\mathrm{V}=8.0~\mathrm{V}.\]

Hence, the charges on these two capacitors are, respectively,

\[Q_2=C_2V_2=(2.0~\mu\mathrm{F})(8.0~\mathrm{V})=16.0~\mu\mathrm{C},\]
\[Q_3=C_3V_3=(4.0~\mu\mathrm{F})(8.0~\mathrm{V})=32.0~\mu\mathrm{C}.\]

Determine the net capacitance $C$ of each network of capacitors shown below. Assume that $C_1=1.0~\mathrm{pF}$, $C_2=2.0~\mathrm{pF}$, $C_3=4.0~\mathrm{pF}$, $C_4=5.0~\mathrm{pF}$. Find the charge on each capacitor, assuming there is a potential difference of $12.0~\mathrm{V}$ across each network.

- Explain how energy is stored in a capacitor
- Use energy relations to determine the energy stored in a capacitor network

Most of us have seen dramatizations of medical personnel using a defibrillator to pass an electrical current through a patient’s heart to get it to beat normally. Often realistic in detail, the person applying the shock directs another person to “make it $400$ joules this time.” The energy delivered by the defibrillator is stored in a capacitor and can be adjusted to fit the situation. SI units of joules are often employed. Less dramatic is the use of capacitors in microelectronics to supply energy when batteries are charged (\ref{fig:4.3.1}). Capacitors are also used to supply energy for flash lamps on cameras.

\begin{gather}.\tag{Figure 4.3.1}\label{fig:4.3.1}\end{gather}The energy $U_C$ stored in a capacitor is electrostatic potential energy and is thus related to the charge $Q$ and voltage $V$ between the capacitor plates. A charged capacitor stores energy in the electrical field between its plates. As the capacitor is being charged, the electrical field builds up. When a charged capacitor is disconnected from a battery, its energy remains in the field in the space between its plates.

To gain insight into how this energy may be expressed (in terms of $Q$ and $V$), consider a charged, empty, parallel-plate capacitor; that is, a capacitor without a dielectric but with a vacuum between its plates. The space between its plates has a volume $Ad$, and it is filled with a uniform electrostatic field $E$. The total energy $U_C$ of the capacitor is contained within this space. The **energy density** $u_E$ in this space is simply $U_C$ divided by the volume $Ad$. If we know the energy density, the energy can be found as $U_C=u_E(Ad)$. We will learn in Electromagnetic Waves (after completing the study of Maxwell’s equations) that the energy density $u_E$ in a region of free space occupied by an electrical field $E$ depends only on the magnitude of the field and is

\begin{equation}u_E=\frac{1}{2}\epsilon_0E^2.\tag{4.3.1}\label{eq:4.3.1}\end{equation}

If we multiply the energy density by the volume between the plates, we obtain the amount of energy stored between the plates of a parallel-plate capacitor: $U_C=u_E(Ad)=\frac{1}{2}\epsilon_0E^2Ad=\frac{1}{2}\epsilon_0\frac{V^2}{d^2}Ad=\frac{1}{2}V^2\epsilon_0\frac{A}{d}=\frac{1}{2}V^2C$.

In this derivation, we used the fact that the electrical field between the plates is uniform so that $E=V/d$ and $C=\epsilon_0A/d$. Because $C=Q/V$, we can express this result in other equivalent forms:

\begin{equation}U_C=\frac{1}{2}V^2C=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}QV.\tag{4.3.2}\label{eq:4.3.2}\end{equation}

The expression in Equation \ref{eq:4.3.1} for the energy stored in a parallel-plate capacitor is generally valid for all types of capacitors. To see this, consider any uncharged capacitor (not necessarily a parallel-plate type). At some instant, we connect it across a battery, giving it a potential difference $V=q/C$ between its plates. Initially, the charge on the plates is $Q=0$. As the capacitor is being charged, the charge gradually builds up on its plates, and after some time, it reaches the value $Q$. To move an infinitesimal charge $dq$ from the negative plate to the positive plate (from a lower to a higher potential), the amount of work $dW$ that must be done on $dq$ is $dW=Vdq=\frac{q}{C}dq$.

This work becomes the energy stored in the electrical field of the capacitor. In order to charge the capacitor to a charge $Q$, the total work required is

\[W=\int_0^W(Q)dW=\int_0^Q\frac{q}{C}dq=\frac{1}{2}\frac{Q^2}{C}.\]Since the geometry of the capacitor has not been specified, this equation holds for any type of capacitor. The total work $W$ needed to charge a capacitor is the electrical potential energy $U_C$ stored in it, or $U_C=W$. When the charge is expressed in coulombs, potential is expressed in volts, and the capacitance is expressed in farads, this relation gives the energy in joules.

Knowing that the energy stored in a capacitor is $U_C=Q^2/(2C)$, we can now find the energy density $u_E$ stored in a vacuum between the plates of a charged parallel-plate capacitor. We just have to divide $U_C$ by the volume $Ad$ of space between its plates and take into account that for a parallel-plate capacitor, we have $E=\sigma/\epsilon_0$ and $C=\epsilon_0A/d$. Therefore, we obtain

\[u_E=\frac{U_C}{Ad}=\frac{1}{2}\frac{Q^2}{C}\frac{1}{Ad}=\frac{1}{2}=\frac{Q^2}{\epsilon_0A/d}\frac{1}{Ad}=\frac{1}{2}\frac{1}{\epsilon_0}\left(\frac{Q}{A}\right)^2=\frac{\sigma^2}{2\epsilon_0}=\frac{(E\epsilon_0)^2}{2\epsilon_0}=\frac{\epsilon_0}{2}E^2.\]We see that this expression for the density of energy stored in a parallel-plate capacitor is in accordance with the general relation expressed in Equation \ref{eq:4.3.1}. We could repeat this calculation for either a spherical capacitor or a cylindrical capacitor—or other capacitors—and in all cases, we would end up with the general relation given by Equation \ref{eq:4.3.1}.

\begin{eqnarray*}U_1&=&\frac{1}{2}C_1V_1^2=\frac{1}{2}(12.0~\mu\mathrm{F})(4.0~\mathrm{V})^2=96~\mu\mathrm{J},\\U_2&=&\frac{1}{2}C_2V_2^2=\frac{1}{2}(2.0~\mu\mathrm{F})(8.0~\mathrm{V})^2=64~\mu\mathrm{J},\\U_3&=&\frac{1}{2}C_3V_3^2=\frac{1}{2}(4.0~\mu\mathrm{F})(8.0~\mathrm{V})^2=130~\mu\mathrm{J}.\end{eqnarray*}

The total energy stored in this network is

\[U_C=U_1+U_2+U_3=96~\mu\mathrm{J}+64~\mu\mathrm{J}+130~\mu\mathrm{J}=0.29~\mathrm{mJ}.\]

The potential difference across a $5.0{\text -}\mathrm{pF}$ capacitor is $0.40~\mathrm{V}$. (a) What is the energy stored in this capacitor? (b) The potential difference is now increased to $1.20~\mathrm{V}$. By what factor is the stored energy increased?

In a cardiac emergency, a portable electronic device known as an automated external defibrillator (AED) can be a lifesaver. A **defibrillator** (\ref{fig:4.3.2}) delivers a large charge in a short burst, or a shock, to a person’s heart to correct abnormal heart rhythm (an arrhythmia). A heart attack can arise from the onset of fast, irregular beating of the heart—called cardiac or ventricular fibrillation. Applying a large shock of electrical energy can terminate the arrhythmia and allow the body’s natural pacemaker to resume its normal rhythm. Today, it is common for ambulances to carry AEDs. AEDs are also found in many public places. These are designed to be used by lay persons. The device automatically diagnoses the patient’s heart rhythm and then applies the shock with appropriate energy and waveform. CPR (cardiopulmonary resuscitation) is recommended in many cases before using a defibrillator.

- Describe the effects a dielectric in a capacitor has on capacitance and other properties
- Calculate the capacitance of a capacitor containing a dielectric

As we discussed earlier, an insulating material placed between the plates of a capacitor is called a dielectric. Inserting a dielectric between the plates of a capacitor affects its capacitance. To see why, let’s consider an experiment described in \ref{fig:4.4.1}. Initially, a capacitor with capacitance $C_0$ when there is air between its plates is charged by a battery to voltage $V_0$. When the capacitor is fully charged, the battery is disconnected. A charge $Q_0$ then resides on the plates, and the potential difference between the plates is measured to be $V_0$. Now, suppose we insert a dielectric that *totally* fills the gap between the plates. If we monitor the voltage, we find that the voltmeter reading has dropped to a *smaller* value $V$. We write this new voltage value as a fraction of the original voltage $V_0$, with a positive number $\kappa$ ($\kappa>1$):

\begin{equation}C=\frac{Q_0}{V}=\frac{Q_0}{V_0/\kappa}=\kappa\frac{Q_0}{V_0}=\kappa C_0.\tag{4.4.1}\label{eq:4.4.1}\end{equation}

This equation tells us that the *capacitance $C_0$* *of an empty (vacuum) capacitor can be increased by a factor of $\kappa$* *when we insert a dielectric material to completely fill the space between its plates*. Note that Equation \ref{eq:4.4.1} can also be used for an empty capacitor by setting $\kappa=1$. In other words, we can say that the dielectric constant of the vacuum is $1$, which is a reference value.

The principle expressed by Equation \ref{eq:4.4.1} is widely used in the construction industry (\ref{fig:4.4.2}). Metal plates in an electronic stud finder act effectively as a capacitor. You place a stud finder with its flat side on the wall and move it continually in the horizontal direction. When the finder moves over a wooden stud, the capacitance of its plates changes, because wood has a different dielectric constant than a gypsum wall. This change triggers a signal in a circuit, and thus the stud is detected.

\begin{gather}.\tag{Figure 4.4.2}\label{fig:4.4.2}\end{gather}The electrical energy stored by a capacitor is also affected by the presence of a dielectric. When the energy stored in an empty capacitor is $U_0$, the energy $U$ stored in a capacitor with a dielectric is smaller by a factor of $\kappa$,

\begin{equation}U=\frac{1}{2}\frac{Q^2}{C}\frac{1}{2}\frac{Q_0^2}{\kappa C_0}=\frac{1}{\kappa}U_0.\tag{4.4.2}\label{eq:4.4.2}\end{equation}

As a dielectric material sample is brought near an empty charged capacitor, the sample reacts to the electrical field of the charges on the capacitor plates. Just as we learned in Electric Charges and Fields on electrostatics, there will be the induced charges on the surface of the sample; however, they are not free charges like in a conductor, because a perfect insulator does not have freely moving charges. These induced charges on the dielectric surface are of an opposite sign to the free charges on the plates of the capacitor, and so they are attracted by the free charges on the plates. Consequently, the dielectric is “pulled” into the gap, and the work to polarize the dielectric material between the plates is done at the expense of the stored electrical energy, which is reduced, in accordance with Equation \ref{eq:4.4.2}.

Since the battery is disconnected before the dielectric is inserted, the plate charge is unaffected by the dielectric and remains at $0.800~\mathrm{nC}$.

c. With the dielectric, the potential difference becomes \[V=\frac{1}{\kappa}V_0=\frac{1}{2.1}40.0~\mathrm{V}=19.0~\mathrm{V}.\]d. The stored energy without the dielectric is

With the dielectric inserted, we use Equation \ref{eq:4.4.2} to find that the stored energy decreases to

\[U=\frac{1}{\kappa}U_0=\frac{1}{2.1}16.0~\mathrm{nJ}=7.6~\mathrm{nJ}.\]

When a dielectric is inserted into an isolated and charged capacitor, the stored energy decreases to $33\%$ of its original value. (a) What is the dielectric constant? (b) How does the capacitance change?

- Explain the polarization of a dielectric in a uniform electrical field
- Describe the effect of a polarized dielectric on the electrical field between capacitor plates
- Explain dielectric breakdown

We can understand the effect of a dielectric on capacitance by looking at its behavior at the molecular level. As we have seen in earlier chapters, in general, all molecules can be classified as either *polar* or *nonpolar*. There is a net separation of positive and negative charges in an isolated polar molecule, whereas there is no charge separation in an isolated nonpolar molecule (\ref{fig:4.5.1}). In other words, polar molecules have permanent *electric-dipole moments* and nonpolar molecules do not. For example, a molecule of water is polar, and a molecule of oxygen is nonpolar. Nonpolar molecules can become polar in the presence of an external electrical field, which is called *induced polarization*.

Let’s first consider a dielectric composed of polar molecules. In the absence of any external electrical field, the electric dipoles are oriented randomly, as illustrated in \ref{fig:4.5.2}(a). However, if the dielectric is placed in an external electrical field $\vec{\mathbf{E}}_0$, the polar molecules align with the external field, as shown in part (b) of the figure. Opposite charges on adjacent dipoles within the volume of dielectric neutralize each other, so there is no net charge within the dielectric (see the dashed circles in part (b)). However, this is not the case very close to the upper and lower surfaces that border the dielectric (the region enclosed by the dashed rectangles in part (b)), where the alignment does produce a net charge. Since the external electrical field merely aligns the dipoles, the dielectric as a whole is neutral, and the surface charges induced on its opposite faces are equal and opposite. These **induced surface charges** $+Q_i$ and $-Q_i$ produce an additional electrical field $\vec{\mathbf{E}}_i$ (an **induced electrical field**), which *opposes* the external field $\vec{\mathbf{E}}_0$, as illustrated in part (c).

The same effect is produced when the molecules of a dielectric are nonpolar. In this case, a nonpolar molecule acquires an **induced electric-dipole moment** because the external field $\vec{\mathbf{E}}_0$ causes a separation between its positive and negative charges. The induced dipoles of the nonpolar molecules align with $\vec{\mathbf{E}}_0$ in the same way as the permanent dipoles of the polar molecules are aligned (shown in part (b)). Hence, the electrical field within the dielectric is weakened regardless of whether its molecules are polar or nonpolar.

Therefore, when the region between the parallel plates of a charged capacitor, such as that shown in \ref{fig:4.5.3}(a), is filled with a dielectric, within the dielectric there is an electrical field $\vec{\mathbf{E}}_0$ due to the *free charge* $Q_0$ on the capacitor plates and an electrical field $\vec{\mathbf{E}}_i$ due to the induced charge $Q_i$ on the surfaces of the dielectric. Their vector sum gives the net electrical field $\vec{\mathbf{E}}$ within the dielectric between the capacitor plates (shown in part (b) of the figure):

\begin{equation}\vec{\mathbf{E}}=\vec{\mathbf{E}}_0+\vec{\mathbf{E}}_i.\tag{4.5.1}\label{eq:4.5.1}\end{equation}

This net field can be considered to be the field produced by an *effective charge* $Q_0-Q_i$ on the capacitor.

In most dielectrics, the net electrical field $\vec{\mathbf{E}}$ is proportional to the field $\vec{\mathbf{E}}_0$ produced by the free charge. In terms of these two electrical fields, the dielectric constant $\kappa$ of the material is defined as

\begin{equation}\kappa=\frac{E_0}{E}.\tag{4.5.2}\label{eq:4.5.2}\end{equation}

Since $\vec{\mathbf{E}}_0$ and $\vec{\mathbf{E}}_i$ point in opposite directions, the magnitude $E$ is smaller than the magnitude $E_0$ and therefore $\kappa>1$. Combining Equation \ref{eq:4.5.2} with Equation \ref{4.5.1}, and rearranging the terms, yields the following expression for the induced electrical field in a dielectric:

\begin{equation}\vec{\mathbf{E}}_i=\left(\frac{1}{\kappa}-1\right)\vec{\mathbf{E}}_0.\tag{4.5.3}\label{eq:4.5.3}\end{equation}

When the magnitude of an external electrical field becomes too large, the molecules of dielectric material start to become ionized. A molecule or an atom is ionized when one or more electrons are removed from it and become free electrons, no longer bound to the molecular or atomic structure. When this happens, the material can conduct, thereby allowing charge to move through the dielectric from one capacitor plate to the other. This phenomenon is called **dielectric breakdown**. (Figure 4.0.1 shows typical random-path patterns of electrical discharge during dielectric breakdown.) The critical value, $E_c$, of the electrical field at which the molecules of an insulator become ionized is called the **dielectric strength** of the material. The dielectric strength imposes a limit on the voltage that can be applied for a given plate separation in a capacitor. For example, the dielectric strength of air is $E_c=3.0~\mathrm{MV/m}$, so for an air-filled capacitor with a plate separation of $d=1.00~\mathrm{mm}$ the limit on the potential difference that can be safely applied across its plates without causing dielectric breakdown is $V=E_cd=(3.0\times10^6~\mathrm{V/m})(1.00\times10^{-3}~\mathrm{m})=3.0~\mathrm{kV}$.

However, this limit becomes $60.0~\mathrm{kV}$ when the same capacitor is filled with Teflon™, whose dielectric strength is about $60.0~\mathrm{MV/m}$. Because of this limit imposed by the dielectric strength, the amount of charge that an air-filled capacitor can store is only $Q_0=\kappa_{\mathrm{air}}C_0(3.0~\mathrm{kV})$ and the charge stored on the same Teflon™-filled capacitor can be as much as

\begin{eqnarray*}Q&=&\kappa_{\mathrm{teflon}}C_0(60.0~\mathrm{kV})=\kappa_{\mathrm{teflon}}\frac{Q_0}{\kappa_{\mathrm{air}}(3.0~\mathrm{kV})}(60.0~\mathrm{kV})=20\frac{\kappa_{\mathrm{teflon}}}{\kappa_{\mathrm{air}}}Q_0\\&=&20\frac{2.1}{1.00059}Q_0\simeq42Q_0,\end{eqnarray*}which is about $42$ times greater than a charge stored on an air-filled capacitor. Typical values of dielectric constants and dielectric strengths for various materials are given in \ref{tab:4.5.1}. Notice that the dielectric constant $\kappa$ is exactly $1.0$ for a vacuum (the empty space serves as a reference condition) and very close to $1.0$ for air under normal conditions (normal pressure at room temperature). These two values are so close that, in fact, the properties of an air-filled capacitor are essentially the same as those of an empty capacitor.

\begin{gather}.\tag{Table 4.5.1}\label{tab:4.5.1}\end{gather}Material | Dielectric constant $\kappa$ | Dielectric strength $E_c[\times10^6~\mathrm{V/m}]$ |
---|---|---|

Vacuum | 1 | ∞ |

Dry air (1 atm) | 1.00059 | 3.0 |

Teflon™ | 2.1 | 60 to 173 |

Paraffin | 2.3 | 11 |

Silicon oil | 2.5 | 10 to 15 |

Polystyrene | 2.56 | 19.7 |

Nylon | 3.4 | 14 |

Paper | 3.7 | 16 |

Fused quartz | 3.78 | 8 |

Glass | 4 to 6 | 9.8 to 13.8 |

Concrete | 4.5 | – |

Bakelite | 4.9 | 24 |

Diamond | 5.5 | 2,000 |

Pyrex glass | 5.6 | 14 |

Mica | 6.0 | 118 |

Neoprene rubber | 6.7 | 15.7 to 26.7 |

Water | 80 | −− |

Sulfuric acid | 84 to 100 | −− |

Titanium dioxide | 86 to 173 | – |

Strontium titanate | 310 | 8 |

Barium titanate | 1,200 to 10,000 | – |

Calcium copper titanate | > 250,000 | – |

Not all substances listed in the table are good insulators, despite their high dielectric constants. Water, for example, consists of polar molecules and has a large dielectric constant of about $80$. In a water molecule, electrons are more likely found around the oxygen nucleus than around the hydrogen nuclei. This makes the oxygen end of the molecule slightly negative and leaves the hydrogens end slightly positive, which makes the molecule easy to align along an external electrical field, and thus water has a large dielectric constant. However, the polar nature of water molecules also makes water a good solvent for many substances, which produces undesirable effects, because any concentration of free ions in water conducts electricity.

The electrical field $E$ with the Teflon™ in place is

\[E=\frac{1}{\kappa}E_0=\frac{1}{2.1}2.0\times10^4~\mathrm{V/m}=9.5\times10^3~\mathrm{V/m}.\] b. The effective charge on the capacitor is the difference between the free charge $Q_0$ and the induced charge $Q_i$. The electrical field in the Teflon™ is caused by this effective charge. Thus \[E=\frac{1}{\epsilon_0}\sigma=\frac{1}{\epsilon_0}\frac{Q_0-Q_i}{A}.\]We invert this equation to obtain $Q_i$, which yields

\begin{eqnarray*}Q_i&\!\!\!=\!\!\!&Q_0-\epsilon_0AE\\&\!\!\!=\!\!\!&8.0\times10^{-10}~\mathrm{C}-\left(8.85\times10^{-12}~\frac{\mathrm{C}^2}{\mathrm{N}\cdot\mathrm{m}^2}\right)(4.5\times10^{-3}~\mathrm{m}^2)\left(9.5\times10^3~\frac{\mathrm{V}}{\mathrm{m}}\right)\\&\!\!\!=\!\!\!&4.2\times10^{-10}~\mathrm{C}=0.42~\mathrm{nC}.\end{eqnarray*}(b) For the filled capacitor, the free charge on the plates is

\[Q=CV=(\kappa C_0)V_0=\kappa(C_0V_0)=\kappa Q_0.\]The electrical field $E$ in the filled capacitor is due to the effective charge $Q-Q_i$ (\ref{fig:4.5.4}(b)). Since $E=E_0$, we have

\[\frac{Q-Q_i}{\epsilon_0A}=\frac{Q_0}{\epsilon_0A}.\] Solving this equation for $Q_i$, we obtain for the induced charge \[Q_i=Q-Q_0=\kappa Q_0-Q_0=(\kappa-1)Q_0.\]Continuing with Example 4.5.2, show that when the battery is connected across the plates the energy stored in dielectric-filled capacitor is $U=\kappa U_0$ (larger than the energy $U_0$ of an empty capacitor kept at the same voltage). Compare this result with the result $U=U_0/\kappa$ found previously for an isolated, charged capacitor.

Repeat the calculations of Example 4.4.1 for the case in which the battery remains connected while the dielectric is placed in the capacitor.

Capacitance | $C=\frac{Q}{V}$ |

Capacitance of a parallel-plate capacitor | $C=\epsilon_0\frac{A}{d}$ |

Capacitance of a vacuum spherical capacitor | $C=4\pi\epsilon_0\frac{R_1R_2}{R_2-R_1}$ |

Capacitance of a vacuum cylindrical capacitor | $C=\frac{2\pi\epsilon_0l}{\ln(r_2/R_1)}$ |

Capacitance of a series combination | $\frac{1}{C_{\mathrm{S}}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots$ |

Capacitance of a parallel combination | $C_{\mathrm{P}}=C_1+C_2+C_3+\ldots$ |

Energy density | $U_E=\frac{1}{2}\epsilon_0E^2$ |

Energy stored in a capacitor | $U_C=\frac{1}{2}V^2C=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}QV$ |

Capacitance of a capacitor with dielectric | $C=\kappa C_0$ |

Energy stored in an isolated capacitor with dielectric | $U=\frac{1}{\kappa}U_0$ |

Dielectric constant | $\kappa=\frac{E_0}{E}$ |

Induced electrical field in a dielectric | $\vec{\mathbf{E}}_\mathrm{i}=\left(\frac{1}{\kappa}-1\right)\vec{\mathbf{E}}_0$ |

- A capacitor is a device that stores an electrical charge and electrical energy. The amount of charge a vacuum capacitor can store depends on two major factors: the voltage applied and the capacitor’s physical characteristics, such as its size and geometry.
- The capacitance of a capacitor is a parameter that tells us how much charge can be stored in the capacitor per unit potential difference between its plates. Capacitance of a system of conductors depends only on the geometry of their arrangement and physical properties of the insulating material that fills the space between the conductors. The unit of capacitance is the farad, where $1~\mathrm{F}=1~\mathrm{C}/1~\mathrm{V}.$

- When several capacitors are connected in a series combination, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances.
- When several capacitors are connected in a parallel combination, the equivalent capacitance is the sum of the individual capacitances.
- When a network of capacitors contains a combination of series and parallel connections, we identify the series and parallel networks, and compute their equivalent capacitances step by step until the entire network becomes reduced to one equivalent capacitance.

- Capacitors are used to supply energy to a variety of devices, including defibrillators, microelectronics such as calculators, and flash lamps.
- The energy stored in a capacitor is the work required to charge the capacitor, beginning with no charge on its plates. The energy is stored in the electrical field in the space between the capacitor plates. It depends on the amount of electrical charge on the plates and on the potential difference between the plates.
- The energy stored in a capacitor network is the sum of the energies stored on individual capacitors in the network. It can be computed as the energy stored in the equivalent capacitor of the network.

- The capacitance of an empty capacitor is increased by a factor of $\kappa$ when the space between its plates is completely filled by a dielectric with dielectric constant $\kappa.$
- Each dielectric material has its specific dielectric constant.
- The energy stored in an empty isolated capacitor is decreased by a factor of $\kappa$ when the space between its plates is completely filled with a dielectric with dielectric constant $\kappa.$

- When a dielectric is inserted between the plates of a capacitor, equal and opposite surface charge is induced on the two faces of the dielectric. The induced surface charge produces an induced electrical field that opposes the field of the free charge on the capacitor plates.
- The dielectric constant of a material is the ratio of the electrical field in vacuum to the net electrical field in the material. A capacitor filled with dielectric has a larger capacitance than an empty capacitor.
- The dielectric strength of an insulator represents a critical value of electrical field at which the molecules in an insulating material start to become ionized. When this happens, the material can conduct and dielectric breakdown is observed.

4.1 $1.1\times10^{-3}~\mathrm{m}$

4.3 $3.59~\mathrm{cm},$ $17.98~\mathrm{cm}$

4.4 a. $25.0~\mathrm{pF};$ b. $9.2$
4.5 a. $C=0.86~\mathrm{pF},$ $Q_1=10~\mathrm{pC},$ $Q_2=3.4~\mathrm{pC},$ $Q_3=6.8~\mathrm{pC};$ b. $C=2.3~\mathrm{pF},$ $Q_1=12~\mathrm{pC},$ $Q_2=Q_3=16~\mathrm{pC};$

4.6 a. $4.0\times10^{-13}~\mathrm{J};$ b. $9$ times
4.7 a. $3.0;$ b. $C=3.0~\mathrm{C_0}$

4.9 a. $C_0=20~\mathrm{pF},$ $C=42~\mathrm{pF};$ b. $Q_0=0.8~\mathrm{nC},$ $Q=1.7~\mathrm{nC};$ c. $V_0=V=40~\mathrm{V};$ d. $U_0=16~\mathrm{nJ},$ $U=34~\mathrm{nJ}$

1. Does the capacitance of a device depend on the applied voltage? Does the capacitance of a device depend on the charge residing on it?

2. Would you place the plates of a parallel-plate capacitor closer together or farther apart to increase their capacitance?

3. The value of the capacitance is zero if the plates are not charged. True or false?

4. If the plates of a capacitor have different areas, will they acquire the same charge when the capacitor is connected across a battery?

5. Does the capacitance of a spherical capacitor depend on which sphere is charged positively or negatively?

6. If you wish to store a large amount of charge in a capacitor bank, would you connect capacitors in series or in parallel? Explain.

7. What is the maximum capacitance you can get by connecting three $1.0{\text -}\mu\mathrm{F}$ capacitors? What is the minimum capacitance?

8. If you wish to store a large amount of energy in a capacitor bank, would you connect capacitors in series or parallel? Explain.

9. Discuss what would happen if a conducting slab rather than a dielectric were inserted into the gap between the capacitor plates.

10. Discuss how the energy stored in an empty but charged capacitor changes when a dielectric is inserted if (a) the capacitor is isolated so that its charge does not change; (b) the capacitor remains connected to a battery so that the potential difference between its plates does not change.

11. Distinguish between dielectric strength and dielectric constant.

12. Water is a good solvent because it has a high dielectric constant. Explain.

13. Water has a high dielectric constant. Explain why it is then not used as a dielectric material in capacitors.

14. Elaborate on why molecules in a dielectric material experience net forces on them in a non-uniform electrical field but not in a uniform field.

15. Explain why the dielectric constant of a substance containing permanent molecular electric dipoles decreases with increasing temperature.

16. Give a reason why a dielectric material increases capacitance compared with what it would be with air between the plates of a capacitor. How does a dielectric material also allow a greater voltage to be applied to a capacitor? (The dielectric thus increases $C$ and permits a greater $V.$)

17. Elaborate on the way in which the polar character of water molecules helps to explain water’s relatively large dielectric constant.

18. Sparks will occur between the plates of an air-filled capacitor at a lower voltage when the air is humid than when it is dry. Discuss why, considering the polar character of water molecules.

19. What charge is stored in a $180.0{\text -}\mu\mathrm{F}$ capacitor when $120.0~\mathrm{V}$ is applied to it?

20. Find the charge stored when $5.50~\mathrm{V}$ is applied to an $8.00{\text -}\mathrm{pF}$ capacitor.

21. Calculate the voltage applied to a $2.00{\text -}\mu\mathrm{F}$ capacitor when it holds $3.10~\mu\mathrm{C}$ of charge.

22. What voltage must be applied to an $8.00{\text -}\mathrm{nF}$ capacitor to store $0.160~\mathrm{mC}$ of charge?

23. What capacitance is needed to store $3.00~\mu\mathrm{C}$ of charge at a voltage of $120~\mathrm{V}$?

24. What is the capacitance of a large Van de Graaff generator’s terminal, given that it stores $8.00~\mathrm{mC}$ of charge at a voltage of $12.0~\mathrm{MV}$?

25. The plates of an empty parallel-plate capacitor of capacitance $5.0~\mathrm{pF}$ are $2.0~\mathrm{mm}$ apart. What is the area of each plate?

26. A $60.0{\text -}\mathrm{pF}$ vacuum capacitor has a plate area of $0.010~\mathrm{m^2}.$ What is the separation between its plates?

27. A set of parallel plates has a capacitance of $5.0~\mu\mathrm{F}$ How much charge must be added to the plates to increase the potential difference between them by $100~\mathrm{V}$?

28. Consider Earth to be a spherical conductor of radius $6400~\mathrm{km}$ and calculate its capacitance.

29. If the capacitance per unit length of a cylindrical capacitor is $20~\mathrm{pF/m},$ what is the ratio of the radii of the two cylinders?

30. An empty parallel-plate capacitor has a capacitance of $20~\mu\mathrm{F}.$ How much charge must leak off its plates before the voltage across them is reduced by $100~\mathrm{V}$?

31. A $4.00{\text -}\mathrm{pF}$ is connected in series with an $8.00{\text -}\mathrm{pF}$ capacitor and a $400{\text -}\mathrm{V}$ potential difference is applied across the pair. (a) What is the charge on each capacitor? (b) What is the voltage across each capacitor?

32. Three capacitors, with capacitances of $C_1=2.0~\mu\mathrm{F},$ $C_2=3.0~\mu\mathrm{F},$ and $C_3=6.0~\mu\mathrm{F},$ respectively, are connected in parallel. A 500-V potential difference is applied across the combination. Determine the voltage across each capacitor and the charge on each capacitor.

33. Find the total capacitance of this combination of series and parallel capacitors shown below.

34. Suppose you need a capacitor bank with a total capacitance of $0.750~\mathrm{F}$ but you have only $1.50{\text -}\mathrm{mF}$ capacitors at your disposal. What is the smallest number of capacitors you could connect together to achieve your goal, and how would you connect them?

35. What total capacitances can you make by connecting a $5.00{\text -}\mu\mathrm{F}$ and a $8.00{\text -}\mu\mathrm{F}$ capacitor?

36. Find the equivalent capacitance of the combination of series and parallel capacitors shown below.

37. Find the net capacitance of the combination of series and parallel capacitors shown below.

38. A $40{\text -}\mathrm{pF}$ capacitor is charged to a potential difference of $500~\mathrm{V}.$ Its terminals are then connected to those of an uncharged $10{\text -}\mathrm{pF}$ capacitor. Calculate: (a) the original charge on the $40{\text -}\mathrm{pF}$ capacitor; (b) the charge on each capacitor after the connection is made; and (c) the potential difference across the plates of each capacitor after the connection.

39. A $2.0{\text -}\mu\mathrm{F}$ capacitor and a $4.0{\text -}\mu\mathrm{F}$ capacitor are connected in series across a $1.0{\text -}\mathrm{kV}$ potential. The charged capacitors are then disconnected from the source and connected to each other with terminals of like sign together. Find the charge on each capacitor and the voltage across each capacitor.

40. How much energy is stored in an $8.00{\text -}\mathrm{mF}$ capacitor whose plates are at a potential difference of $6.00~\mathrm{V}$?

41. A capacitor has a charge of $2.5~\mu\mathrm{C}$ when connected to a $6.0{\text -}\mathrm{V}$ battery. How much energy is stored in this capacitor?

42. How much energy is stored in the electrical field of a metal sphere of radius $2.0~\mathrm{m}$ that is kept at a $10.0{\text -}\mathrm{V}$ potential?

43. (a) What is the energy stored in the $10.0{\text -}\mu\mathrm{F}$ capacitor of a heart defibrillator charged to $9.00\times10^3~\mathrm{V}$? (b) Find the amount of the stored charge.

44. In open-heart surgery, a much smaller amount of energy will defibrillate the heart. (a) What voltage is applied to the $8.00{\text -}\mu\mathrm{F}$ capacitor of a heart defibrillator that stores $40.0~\mathrm{J}$ of energy? (b) Find the amount of the stored charge.

45. A $165{\text -}\mu\mathrm{F}$ capacitor is used in conjunction with a dc motor. How much energy is stored in it when $119~\mathrm{V}$ is applied?

46. Suppose you have a $9.00{\text -}\mathrm{V}$ battery, a $2.00{\text -}\mu\mathrm{F}$ capacitor, and a $7.40{\text -}\mu\mathrm{F}$ capacitor. (a) Find the charge and energy stored if the capacitors are connected to the battery in series. (b) Do the same for a parallel connection.

47. An anxious physicist worries that the two metal shelves of a wood frame bookcase might obtain a high voltage if charged by static electricity, perhaps produced by friction. (a) What is the capacitance of the empty shelves if they have area $1.00\times10^2~\mathrm{m}^2$ and are $0.200~\mathrm{m}$ apart? (b) What is the voltage between them if opposite charges of magnitude $2.00~\mathrm{nC}$ are placed on them? (c) To show that this voltage poses a small hazard, calculate the energy stored. (d) The actual shelves have an area $100$ times smaller than these hypothetical shelves. Are his fears justified?

48. A parallel-plate capacitor is made of two square plates $25~\mathrm{cm}$ on a side and $1.0~\mathrm{mm}$ apart. The capacitor is connected to a $50.0{\text -}\mathrm{V}$ battery. With the battery still connected, the plates are pulled apart to a separation of $2.00~\mathrm{mm}$. What are the energies stored in the capacitor before and after the plates are pulled farther apart? Why does the energy decrease even though work is done in separating the plates?

49. Suppose that the capacitance of a variable capacitor can be manually changed from $100~\mathrm{pF}$ to $800~\mathrm{pF}$ by turning a dial, connected to one set of plates by a shaft, from $0^{\circ}$ to $180^{\circ}.$ With the dial set at $180^{\circ}$ (corresponding to $C=800~\mathrm{pF}$), the capacitor is connected to a $500{\text -}\mathrm{V}$ source. After charging, the capacitor is disconnected from the source, and the dial is turned to $0^{\circ}.$ If friction is negligible, how much work is required to turn the dial from $180^{\circ}$ to $0^{\circ}$?

50. Show that for a given dielectric material, the maximum energy a parallel-plate capacitor can store is directly proportional to the volume of dielectric.

51. An air-filled capacitor is made from two flat parallel plates $1.0~\mathrm{mm}$ apart. The inside area of each plate is $8.0~\mathrm{cm}^2.$ (a) What is the capacitance of this set of plates? (b) If the region between the plates is filled with a material whose dielectric constant is $6.0,$ what is the new capacitance?

52. A capacitor is made from two concentric spheres, one with radius $5.00~\mathrm{cm},$ the other with radius $8.00~\mathrm{cm}.$ (a) What is the capacitance of this set of conductors? (b) If the region between the conductors is filled with a material whose dielectric constant is $6.00,$ what is the capacitance of the system?

53. A parallel-plate capacitor has charge of magnitude $9.00~\mu\mathrm{C}$ on each plate and capacitance $3.00~\mu\mathrm{C}$ when there is air between the plates. The plates are separated by $2.00~\mathrm{mm}$. With the charge on the plates kept constant, a dielectric with $\kappa=5$ is inserted between the plates, completely filling the volume between the plates. (a) What is the potential difference between the plates of the capacitor, before and after the dielectric has been inserted? (b) What is the electrical field at the point midway between the plates before and after the dielectric is inserted?

54. Some cell walls in the human body have a layer of negative charge on the inside surface. Suppose that the surface charge densities are $\pm0.50\times10^{-3}~\mathrm{C/m}^2,$ the cell wall is $5.0\times10^{-9}~\mathrm{m}$ thick, and the cell wall material has a dielectric constant of $\kappa=5.4.$ (a) Find the magnitude of the electric field in the wall between two charge layers. (b) Find the potential difference between the inside and the outside of the cell. Which is at higher potential? (c) A typical cell in the human body has volume $10^{-16}~\mathrm{m}^3.$ Estimate the total electrical field energy stored in the wall of a cell of this size when assuming that the cell is spherical. (*Hint*: Calculate the volume of the cell wall.)

55. A parallel-plate capacitor with only air between its plates is charged by connecting the capacitor to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates. (a) A voltmeter reads $45.0~\mathrm{V}$ when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads $11.5~\mathrm{V}$. What is the dielectric constant of the material? (b) What will the voltmeter read if the dielectric is now pulled away out so it fills only one-third of the space between the plates?

56. Two flat plates containing equal and opposite charges are separated by material $4.0~\mathrm{mm}$ thick with a dielectric constant of $5.0.$ If the electrical field in the dielectric is $1.5~\mathrm{MV/m},$ what are (a) the charge density on the capacitor plates, and (b) the induced charge density on the surfaces of the dielectric?

57. For a Teflon™-filled, parallel-plate capacitor, the area of the plate is $50.0~\mathrm{cm}^2$ and the spacing between the plates is $0.50~\mathrm{mm}.$ If the capacitor is connected to a $200{\text -}\mathrm{V}$ battery, find (a) the free charge on the capacitor plates, (b) the electrical field in the dielectric, and (c) the induced charge on the dielectric surfaces.

58. Find the capacitance of a parallel-plate capacitor having plates with a surface area of $5.00~\mathrm{m}^2$ and separated by $0.100~\mathrm{mm}$ of Teflon™.

59. (a) What is the capacitance of a parallel-plate capacitor with plates of area $1.50~\mathrm{m}^2$ that are separated by $0.0200~\mathrm{mm}$ of neoprene rubber? (b) What charge does it hold when $9.00~\mathrm{V}$ is applied to it?

60. Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electrical field is $E=3.20\times10^5~\mathrm{V/m}.$ When the space is filled with dielectric, the electrical field is $E=2.50\times10^5~\mathrm{V/m}.$ (a) What is the surface charge density on each surface of the dielectric? (b) What is the dielectric constant?

61. The dielectric to be used in a parallel-plate capacitor has a dielectric constant of $3.60$ and a dielectric strength of $1.60\times10^7~\mathrm{V/m}.$ The capacitor has to have a capacitance of $1.25~\mathrm{nF}$ and must be able to withstand a maximum potential difference $5.5~\mathrm{kV}.$ What is the minimum area the plates of the capacitor may have?

62. When a $360{\text -}\mathrm{nF}$ air capacitor is connected to a power supply, the energy stored in the capacitor is $18.5~\mu\mathrm{J}.$ While the capacitor is connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by $23.2~\mu\mathrm{J}.$ (a) What is the potential difference between the capacitor plates? (b) What is the dielectric constant of the slab?

63. A parallel-plate capacitor has square plates that are $8.00~\mathrm{cm}$ on each side and $3.80~\mathrm{mm}$ apart. The space between the plates is completely filled with two square slabs of dielectric, each $8.00~\mathrm{cm}$ on a side and $1.90~\mathrm{mm}$ thick. One slab is Pyrex glass and the other slab is polystyrene. If the potential difference between the plates is $86.0~\mathrm{V},$ find how much electrical energy can be stored in this capacitor.

64. A capacitor is made from two flat parallel plates placed $0.40~\mathrm{mm}$ apart. When a charge of $0.020~\mu\mathrm{C}$ is placed on the plates the potential difference between them is $250~\mathrm{V}$. (a) What is the capacitance of the plates? (b) What is the area of each plate? (c) What is the charge on the plates when the potential difference between them is $500~\mathrm{V}$? (d) What maximum potential difference can be applied between the plates so that the magnitude of electrical fields between the plates does not exceed $3.0~\mathrm{MV/m}$?

65. An air-filled (empty) parallel-plate capacitor is made from two square plates that are $25~\mathrm{cm}$ on each side and $1.0~\mathrm{mm}$ apart. The capacitor is connected to a $50{\text -}\mathrm{V}$ battery and fully charged. It is then disconnected from the battery and its plates are pulled apart to a separation of $2.00~\mathrm{mm}.$ (a) What is the capacitance of this new capacitor? (b) What is the charge on each plate? (c) What is the electrical field between the plates?

66. Suppose that the capacitance of a variable capacitor can be manually changed from $100$ to $800~\mathrm{pF}$ by turning a dial connected to one set of plates by a shaft, from $0^{\circ}$ to $180^{\circ}.$ With the dial set at $180^{\circ}$ (corresponding to $C=800~\mathrm{pF}$), the capacitor is connected to a $500{\text -}\mathrm{V}$ source. After charging, the capacitor is disconnected from the source, and the dial is turned to $0^{\circ}.$ (a) What is the charge on the capacitor? (b) What is the voltage across the capacitor when the dial is set to $0^{\circ}$?

67. Earth can be considered as a spherical capacitor with two plates, where the negative plate is the surface of Earth and the positive plate is the bottom of the ionosphere, which is located at an altitude of approximately $70~\mathrm{km}.$ The potential difference between Earth’s surface and the ionosphere is about $350,000~\mathrm{V}.$ (a) Calculate the capacitance of this system. (b) Find the total charge on this capacitor. (c) Find the energy stored in this system.

68. A $4.00{\text -}\mu\mathrm{F}$ capacitor and a $6.00{\text -}\mu\mathrm{F}$ capacitor are connected in parallel across a $600{\text -}\mathrm{V}$ supply line. (a) Find the charge on each capacitor and voltage across each. (b) The charged capacitors are disconnected from the line and from each other. They are then reconnected to each other with terminals of unlike sign together. Find the final charge on each capacitor and the voltage across each.

69. Three capacitors having capacitances of $8.40,$ $8.40,$ and $4.20~\mu\mathrm{F},$ respectively, are connected in series across a $36.0{\text -}\mathrm{V}$ potential difference. (a) What is the charge on the $4.20{\text -}\mu\mathrm{F}$ capacitor? (b) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination?

70. A parallel-plate capacitor with capacitance $5.0~\mu\mathrm{F}$ is charged with a $12.0{\text -}\mathrm{V}$ battery, after which the battery is disconnected. Determine the minimum work required to increase the separation between the plates by a factor of $3.$

71. (a) How much energy is stored in the electrical fields in the capacitors (in total) shown below? (b) Is this energy equal to the work done by the $400{\text -}\mathrm{V}$ source in charging the capacitors?

72. Three capacitors having capacitances $8.4,$ $8.4,$ and $4.2~\mu\mathrm{F}$ are connected in series across a $36.0{\text -}\mathrm{V}$ potential difference. (a) What is the total energy stored in all three capacitors? (b) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other with the positively charged plates connected together. What is the total energy now stored in the capacitors?

73. (a) An $8.00{\text -}\mu\mathrm{F}$ capacitor is connected in parallel to another capacitor, producing a total capacitance of $5.00~\mu\mathrm{F}.$ What is the capacitance of the second capacitor? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

74. (a) On a particular day, it takes $9.60\times10^3~\mathrm{J}$ of electrical energy to start a truck’s engine. Calculate the capacitance of a capacitor that could store that amount of energy at $12.0~\mathrm{V}.$ (b) What is unreasonable about this result? (c) Which assumptions are responsible?

75. (a) A certain parallel-plate capacitor has plates of area $4.00~\mathrm{m}^2,$ separated by $0.0100~\mathrm{mm}$ of nylon, and stores $0.170~\mathrm{C}$ of charge. What is the applied voltage? (b) What is unreasonable about this result? (c) Which assumptions are responsible or inconsistent?

76. A prankster applies $450~\mathrm{V}$ to an $80.0{\text -}\mu\mathrm{F}$ capacitor and then tosses it to an unsuspecting victim. The victim’s finger is burned by the discharge of the capacitor through $0.200~\mathrm{g}$ of flesh. Estimate, what is the temperature increase of the flesh? Is it reasonable to assume that no thermodynamic phase change happened?

77. A spherical capacitor is formed from two concentric spherical conducting spheres separated by vacuum. The inner sphere has radius $12.5~\mathrm{cm}$ and the outer sphere has radius $14.8~\mathrm{cm}.$ A potential difference of $120~\mathrm{V}$ is applied to the capacitor. (a) What is the capacitance of the capacitor? (b) What is the magnitude of the electrical field at $r=12.6~\mathrm{cm},$ just outside the inner sphere? (c) What is the magnitude of the electrical field at $r=14.7~\mathrm{cm}$ just inside the outer sphere? (d) For a parallel-plate capacitor the electrical field is uniform in the region between the plates, except near the edges of the plates. Is this also true for a spherical capacitor?

78. The network of capacitors shown below are all uncharged when a $300{\text -}\mathrm{V}$ potential is applied between points $\mathrm{A}$ and $\mathrm{B}$ with the switch $\mathrm{S}$ open. (a) What is the potential difference $V_E-V_D$? (b) What is the potential at point $\mathrm{E}$ after the switch is closed? (c) How much charge flows through the switch after it is closed?

79. Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit the flash lasts for $1/675$ fraction of a second with an average light power output of $270~\mathrm{kW}.$ (a) If the conversion of electrical energy to light is $95\%$ efficient (because the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of $125~\mathrm{V}$ when the stored energy equals the value stored in part (a). What is the capacitance?

80. A spherical capacitor is formed from two concentric spherical conducting shells separated by a vacuum. The inner sphere has radius $12.5~\mathrm{cm}$ and the outer sphere has radius $14.8~\mathrm{cm}.$ A potential difference of $120~\mathrm{V}$ is applied to the capacitor. (a) What is the energy density at $r=12.6~\mathrm{cm},$ just outside the inner sphere? (b) What is the energy density at $r=14.7~\mathrm{cm},$ just inside the outer sphere? (c) For the parallel-plate capacitor the energy density is uniform in the region between the plates, except near the edges of the plates. Is this also true for the spherical capacitor?

81. A metal plate of thickness $t$ is held in place between two capacitor plates by plastic pegs, as shown below. The effect of the pegs on the capacitance is negligible. The area of each capacitor plate and the area of the top and bottom surfaces of the inserted plate are all $A.$ What is the capacitance of this system?

82. A parallel-plate capacitor is filled with two dielectrics, as shown below. When the plate area is $A$ and separation between plates is $d,$ show that the capacitance is given by $C=\epsilon_0\frac{A}{d}\frac{\kappa_1+\kappa_2}{2}.$

83. A parallel-plate capacitor is filled with two dielectrics, as shown below. Show that the capacitance is given by $C=2\epsilon_0\frac{A}{d}\frac{\kappa_1\kappa_2}{\kappa_1+\kappa_2}.$

84. A capacitor has parallel plates of area $12~\mathrm{cm}^2$ separated by $2.0~\mathrm{mm}.$ The space between the plates is filled with polystyrene. (a) Find the maximum permissible voltage across the capacitor to avoid dielectric breakdown. (b) When the voltage equals the value found in part (a), find the surface charge density on the surface of the dielectric.

- Describe an electrical current
- Define the unit of electrical current
- Explain the direction of current flow

Up to now, we have considered primarily static charges. When charges did move, they were accelerated in response to an electrical field created by a voltage difference. The charges lost potential energy and gained kinetic energy as they traveled through a potential difference where the electrical field did work on the charge.

Although charges do not require a material to flow through, the majority of this chapter deals with understanding the movement of charges through a material. The rate at which the charges flow past a location—that is, the amount of charge per unit time—is known as the *electrical current*. When charges flow through a medium, the current depends on the voltage applied, the material through which the charges flow, and the state of the material. Of particular interest is the motion of charges in a conducting wire. In previous chapters, charges were accelerated due to the force provided by an electrical field, losing potential energy and gaining kinetic energy. In this chapter, we discuss the situation of the force provided by an electrical field in a conductor, where charges lose kinetic energy to the material reaching a constant velocity, known as the “*drift velocity*.” This is analogous to an object falling through the atmosphere and losing kinetic energy to the air, reaching a constant terminal velocity.

If you have ever taken a course in first aid or safety, you may have heard that in the event of electric shock, it is the current, not the voltage, which is the important factor on the severity of the shock and the amount of damage to the human body. Current is measured in units called amperes; you may have noticed that circuit breakers in your home and fuses in your car are rated in amps (or amperes). But what is the ampere and what does it measure?

Electrical current is defined to be the rate at which charge flows. When there is a large current present, such as that used to run a refrigerator, a large amount of charge moves through the wire in a small amount of time. If the current is small, such as that used to operate a handheld calculator, a small amount of charge moves through the circuit over a long period of time.

The average electrical current $I$ is the rate at which charge flows,

\begin{equation}I_{\mathrm{ave}}=\frac{\Delta Q}{\Delta t},\tag{5.1.1}\label{eq:5.1.1}\end{equation} where $\Delta Q$ is the amount of charge passing through a given area in time $\Delta t$ (\ref{fig:5.1.1}). The SI unit for current is theMost electrical appliances are rated in amperes (or amps) required for proper operation, as are fuses and circuit breakers.

\begin{gather}.\tag{Figure 5.1.1}\label{fig:5.1.1}\end{gather}(a) What is the average current involved when a truck battery sets in motion $720~\mathrm{C}$ of charge in $4.00~\mathrm{s}$ while starting an engine? (b) How long does it take $1.00~\mathrm{C}$ of charge to flow from the battery?

b. Solving the relationship $I=\frac{\Delta Q}{\Delta t}$ for time $\Delta t$ and entering the known values for charge and current gives

\[\Delta t=\frac{\Delta Q}{I}=\frac{1.00~\mathrm{C}}{180~\mathrm{C/s}}=5.56\times10^{-3}~\mathrm{s}=5.56~\mathrm{ms}.\]

Handheld calculators often use small solar cells to supply the energy required to complete the calculations needed to complete your next physics exam. The current needed to run your calculator can be as small as $0.30~\mathrm{mA}$. How long would it take for $1.00~\mathrm{C}$ of charge to flow from the solar cells? Can solar cells be used, instead of batteries, to start traditional internal combustion engines presently used in most cars and trucks?

Circuit breakers in a home are rated in amperes, normally in a range from $10~\mathrm{A}$ to $30~\mathrm{A}$, and are used to protect the residents from harm and their appliances from damage due to large currents. A single $15{\text -}\mathrm{A}$ circuit breaker may be used to protect several outlets in the living room, whereas a single $20{\text -}\mathrm{A}$ circuit breaker may be used to protect the refrigerator in the kitchen. What can you deduce from this about current used by the various appliances?

In the previous paragraphs, we defined the current as the charge that flows through a cross-sectional area per unit time. In order for charge to flow through an appliance, such as the headlight shown in \ref{fig:5.1.4}, there must be a complete path (or circuit) from the positive terminal to the negative terminal. Consider a simple circuit of a car battery, a switch, a headlight lamp, and wires that provide a current path between the components. In order for the lamp to light, there must be a complete path for current flow. In other words, a charge must be able to leave the positive terminal of the battery, travel through the component, and back to the negative terminal of the battery. The switch is there to control the circuit. Part (a) of the figure shows the simple circuit of a car battery, a switch, a conducting path, and a headlight lamp. Also shown is the **schematic** of the circuit [part (b)]. A schematic is a graphical representation of a circuit and is very useful in visualizing the main features of a circuit. Schematics use standardized symbols to represent the components in a circuits and solid lines to represent the wires connecting the components. The battery is shown as a series of long and short lines, representing the historic voltaic pile. The lamp is shown as a circle with a loop inside, representing the filament of an incandescent bulb. The switch is shown as two points with a conducting bar to connect the two points and the wires connecting the components are shown as solid lines. The schematic in part (c) shows the direction of current flow when the switch is closed.

When the switch is closed in \ref{fig:5.1.4}(c), there is a complete path for charges to flow, from the positive terminal of the battery, through the switch, then through the headlight and back to the negative terminal of the battery. Note that the direction of current flow is from positive to negative. The direction of **conventional current** is always represented in the direction that positive charge would flow, from the positive terminal to the negative terminal.

The conventional current flows from the positive terminal to the negative terminal, but depending on the actual situation, positive charges, negative charges, or both may move. In metal wires, for example, current is carried by electrons—that is, negative charges move. In ionic solutions, such as salt water, both positive and negative charges move. This is also true in nerve cells. A Van de Graaff generator, used for nuclear research, can produce a current of pure positive charges, such as protons. In the Tevatron Accelerator at Fermilab, before it was shut down in 2011, beams of protons and antiprotons traveling in opposite directions were collided. The protons are positive and therefore their current is in the same direction as they travel. The antiprotons are negativity charged and thus their current is in the opposite direction that the actual particles travel.

A closer look at the current flowing through a wire is shown in \ref{fig:5.1.5}. The figure illustrates the movement of charged particles that compose a current. The fact that conventional current is taken to be in the direction that positive charge would flow can be traced back to American scientist and statesman Benjamin **Franklin** in the 1700s. Having no knowledge of the particles that make up the atom (namely the proton, electron, and neutron), Franklin believed that electrical current flowed from a material that had more of an “electrical fluid” and to a material that had less of this “electrical fluid.” He coined the term *positive* for the material that had more of this electrical fluid and *negative* for the material that lacked the electrical fluid. He surmised that current would flow from the material with more electrical fluid—the positive material—to the negative material, which has less electrical fluid. Franklin called this direction of current a positive current flow. This was pretty advanced thinking for a man who knew nothing about the atom.

We now know that a material is positive if it has a greater number of protons than electrons, and it is negative if it has a greater number of electrons than protons. In a conducting metal, the current flow is due primarily to electrons flowing from the negative material to the positive material, but for historical reasons, we consider the positive current flow and the current is shown to flow from the positive terminal of the battery to the negative terminal.

It is important to realize that an electrical field is present in conductors and is responsible for producing the current (\ref{fig:5.1.5}). In previous chapters, we considered the static electrical case, where charges in a conductor quickly redistribute themselves on the surface of the conductor in order to cancel out the external electrical field and restore equilibrium. In the case of an electrical circuit, the charges are prevented from ever reaching equilibrium by an external source of electric potential, such as a battery. The energy needed to move the charge is supplied by the electric potential from the battery.

Although the electrical field is responsible for the motion of the charges in the conductor, the work done on the charges by the electrical field does not increase the kinetic energy of the charges. We will show that the electrical field is responsible for keeping the electric charges moving at a “drift velocity.”

- Define the drift velocity of charges moving through a metal
- Define the vector current density
- Describe the operation of an incandescent lamp

When electrons move through a conducting wire, they do not move at a constant velocity, that is, the electrons do not move in a straight line at a constant speed. Rather, they interact with and collide with atoms and other free electrons in the conductor. Thus, the electrons move in a zig-zag fashion and drift through the wire. We should also note that even though it is convenient to discuss the direction of current, current is a scalar quantity. When discussing the velocity of charges in a current, it is more appropriate to discuss the current density. We will come back to this idea at the end of this section.

Electrical signals move very rapidly. Telephone conversations carried by currents in wires cover large distances without noticeable delays. Lights come on as soon as a light switch is moved to the ‘on’ position. Most electrical signals carried by currents travel at speeds on the order of $10^8~\mathrm{m/s}$, a significant fraction of the speed of light. Interestingly, the individual charges that make up the current move much slower on average, typically drifting at speeds on the order of $10^{-4}~\mathrm{m/s}$. How do we reconcile these two speeds, and what does it tell us about standard conductors?

The high speed of electrical signals results from the fact that the force between charges acts rapidly at a distance. Thus, when a free charge is forced into a wire, as in \ref{fig:5.2.1}, the incoming charge pushes other charges ahead of it due to the repulsive force between like charges. These moving charges push on charges farther down the line. The density of charge in a system cannot easily be increased, so the signal is passed on rapidly. The resulting electrical shock wave moves through the system at nearly the speed of light. To be precise, this fast-moving signal, or shock wave, is a rapidly propagating change in the electrical field.

\begin{gather}.\tag{Figure 5.2.1}\label{fig:5.2.1}\end{gather}Good conductors have large numbers of free charges. In metals, the free charges are free electrons. (In fact, good electrical conductors are often good heat conductors too, because large numbers of free electrons can transport thermal energy as well as carry electrical current.) \ref{fig:5.2.2} shows how free electrons move through an ordinary conductor. The distance that an individual electron can move between collisions with atoms or other electrons is quite small. The electron paths thus appear nearly random, like the motion of atoms in a gas. But there is an electrical field in the conductor that causes the electrons to drift in the direction shown (opposite to the field, since they are negative). The **drift velocity** $\vec{\mathbf{v}}_{\mathrm{d}}$ is the average velocity of the free charges. Drift velocity is quite small, since there are so many free charges. If we have an estimate of the density of free electrons in a conductor, we can calculate the drift velocity for a given current. The larger the density, the lower the velocity required for a given current.

Free-electron collisions transfer energy to the atoms of the conductor. The electrical field does work in moving the electrons through a distance, but that work does not increase the kinetic energy (nor speed) of the electrons. The work is transferred to the conductor’s atoms, often increasing temperature. Thus, a continuous power input is required to keep a current flowing. (An exception is superconductors, for reasons we shall explore in a later chapter. Superconductors can have a steady current without a continual supply of energy—a great energy savings.) For a conductor that is not a superconductor, the supply of energy can be useful, as in an incandescent light bulb filament (\ref{fig:5.2.3}). The supply of energy is necessary to increase the temperature of the tungsten filament, so that the filament glows.

\begin{gather}.\tag{Figure 5.2.3}\label{fig:5.2.3}\end{gather}We can obtain an expression for the relationship between current and drift velocity by considering the number of free charges in a segment of wire, as illustrated in \ref{fig:5.2.4} The number of free charges per unit volume, or the number density of free charges, is given the symbol $n$ where $n=\frac{\mathrm{number~of~charges}}{\mathrm{volume}}$. The value of $n$ depends on the material. The shaded segment has a volume $Av_{\mathrm{d}}dt$, so that the number of free charges in the volume is $nAv_{\mathrm{d}}dt$. The charge $dQ$ in this segment is thus $qnAv_{\mathrm{d}}dt$ where $q$ is the amount of charge on each carrier. (The magnitude of the charge of electrons is $q=1.60\times10^{-19}~\mathrm{C}$.) Current is charge moved per unit time; thus, if all the original charges move out of this segment in time $dt$, the current is

\[I=\frac{dQ}{dt}=qnAv_{\mathrm{d}}.\]

Rearranging terms gives

\begin{equation}v_{\mathrm{d}}=\frac{I}{nqA}\tag{5.2.1}\label{eq:5.2.1}\end{equation}

where $v_{\mathrm{d}}$ is the drift velocity, $n$ is the free charge density, $A$ is the cross-sectional area of the wire, and $I$ is the current through the wire. The carriers of the current each have charge $q$ and move with a drift velocity of magnitude $v_{\mathrm{d}}$.

\begin{gather}.\tag{Figure 5.2.4}\label{fig:5.2.4}\end{gather}Note that simple drift velocity is not the entire story. The speed of an electron is sometimes much greater than its drift velocity. In addition, not all of the electrons in a conductor can move freely, and those that do move might move somewhat faster or slower than the drift velocity. So what do we mean by free electrons?

Atoms in a metallic conductor are packed in the form of a lattice structure. Some electrons are far enough away from the atomic nuclei that they do not experience the attraction of the nuclei as strongly as the inner electrons do. These are the free electrons. They are not bound to a single atom but can instead move freely among the atoms in a “sea” of electrons. When an electrical field is applied, these free electrons respond by accelerating. As they move, they collide with the atoms in the lattice and with other electrons, generating thermal energy, and the conductor gets warmer. In an insulator, the organization of the atoms and the structure do not allow for such free electrons.

As you know, electric power is usually supplied to equipment and appliances through round wires made of a conducting material (copper, aluminum, silver, or gold) that are stranded or solid. The diameter of the wire determines the current-carrying capacity—the larger the diameter, the greater the current-carrying capacity. Even though the current-carrying capacity is determined by the diameter, wire is not normally characterized by the diameter directly. Instead, wire is commonly sold in a unit known as “gauge.” Wires are manufactured by passing the material through circular forms called “drawing dies.” In order to make thinner wires, manufacturers draw the wires through multiple dies of successively thinner diameter. Historically, the gauge of the wire was related to the number of drawing processes required to manufacture the wire. For this reason, the larger the gauge, the smaller the diameter. In the United States, the American Wire Gauge (AWG) was developed to standardize the system. Household wiring commonly consists of $10{\text -}\mathrm{gauge}$ ($2.588{\text -}\mathrm{mm}$ diameter) to $14{\text -}\mathrm{gauge}$ ($1.626{\text -}\mathrm{mm}$ diameter) wire. A device used to measure the gauge of wire is shown in \ref{fig:5.2.5}.

\begin{gather}.\tag{Figure 5.2.5}\label{fig:5.2.5}\end{gather}
\begin{eqnarray*}n&=&\frac{1 e^{-}}{\mathrm{atom}}\times\frac{6.02\times10^{23}~\mathrm{atoms}}{\mathrm{mol}}\times\frac{1~\mathrm{mol}}{63.54~\mathrm{g}}\times\frac{1000~\mathrm{g}}{\mathrm{kg}}\times\frac{8.80\times10^3~\mathrm{kg}}{1~\mathrm{m}^3}\\&=&8.34\times10^{28}e^{-}/\mathrm{m}^3.\end{eqnarray*}

The cross-sectional area of the wire is

\[A=\pi r^2=\pi\left(\frac{2.05\times10^{-3}~\mathrm{m}}{2}\right)^2=3.30\times10^{-6}~\mathrm{m}^2.\]

Rearranging $I=nqAv_{\mathrm{d}}$ to isolate drift velocity gives

\begin{eqnarray*}v_{\mathrm{d}}&=&\frac{I}{nqA}=\frac{20.0~\mathrm{A}}{(8.34\times10^{28}e^-/\mathrm{m}^3)(-1.60\times10^{-19}~\mathrm{C}/e^-)(3.30\times10^{-6}~\mathrm{m}^2)}\\&=&-4.54\times10^{-4}~\mathrm{m/s}.\end{eqnarray*}

In Example 5.2.2, the drift velocity was calculated for a $2.053{\text -}\mathrm{mm}$ diameter ($12{\text -}\mathrm{gauge}$) copper wire carrying a $20{\text -}\mathrm{A}$ current. Would the drift velocity change for a $1.628{\text -}\mathrm{mm}$ diameter ($14{\text -}\mathrm{gauge}$) wire carrying the same $20{\text -}\mathrm{A}$ current?

Although it is often convenient to attach a negative or positive sign to indicate the overall direction of motion of the charges, current is a scalar quantity, $I=\frac{dQ}{dt}$. It is often necessary to discuss the details of the motion of the charge, instead of discussing the overall motion of the charges. In such cases, it is necessary to discuss the current density, $\vec{\mathbf{J}}$, a vector quantity. The **current density** is the flow of charge through an infinitesimal area, divided by the area. The current density must take into account the local magnitude and direction of the charge flow, which varies from point to point. The unit of current density is ampere per meter squared, and the direction is defined as the direction of net flow of positive charges through the area.

The relationship between the current and the current density can be seen in \ref{fig:5.2.6}. The differential current flow through the area $d\vec{\mathbf{A}}$ is found as

\[dI=\vec{\mathbf{J}}\cdot d\vec{\mathbf{A}}=JdA\cos\theta,\]

where $\theta$ is the angle between the area and the current density. The total current passing through area $d\vec{\mathbf{A}}$ can be found by integrating over the area,

\begin{equation}I=\iint_{\mathrm{area}}\vec{\mathbf{J}}\cdot d\vec{\mathbf{A}}.\tag{5.2.2}\label{eq:5.2.2}\end{equation}

Consider the magnitude of the current density, which is the current divided by the area:

\[J=\frac{I}{A}=\frac{n|q|Av_{\mathrm{d}}}{A}=n|q|v_{\mathrm{d}}.\]

Thus, the current density is $\vec{\mathbf{J}}=nq\vec{\mathbf{v}}_{\mathrm{d}}$. If $q$ is positive, $\vec{\mathbf{v}}_{\mathrm{d}}$ is in the same direction as the electrical field $\vec{\mathbf{E}}$. If $q$ is negative, $\vec{\mathbf{v}}_{\mathrm{d}}$ is in the opposite direction of $\vec{\mathbf{E}}$. Either way, the direction of the current density $\vec{\mathbf{J}}$ is in the direction of the electrical field $\vec{\mathbf{E}}$.

\begin{gather}.\tag{Figure 5.2.6}\label{fig:5.2.6}\end{gather}
\[J=\frac{I}{A}=\frac{0.87~\mathrm{A}}{5.26\times10^{-6}~\mathrm{m}^2}=1.65\times10^5~\frac{\mathrm{A}}{\mathrm{m}^2}.\]

The current density is proportional to the current and inversely proportional to the area. If the current density in a conducting wire increases, what would happen to the drift velocity of the charges in the wire?

What is the significance of the current density? The current density is proportional to the current, and the current is the number of charges that pass through a cross-sectional area per second. The charges move through the conductor, accelerated by the electric force provided by the electrical field. The electrical field is created when a voltage is applied across the conductor. In Ohm’s Law, we will use this relationship between the current density and the electrical field to examine the relationship between the current through a conductor and the voltage applied.

- Differentiate between resistance and resistivity
- Define the term conductivity
- Describe the electrical component known as a resistor
- State the relationship between resistance of a resistor and its length, cross-sectional area, and resistivity
- State the relationship between resistivity and temperature

What drives current? We can think of various devices—such as batteries, generators, wall outlets, and so on—that are necessary to maintain a current. All such devices create a potential difference and are referred to as voltage sources. When a voltage source is connected to a conductor, it applies a potential difference $V$ that creates an electrical field. The electrical field, in turn, exerts force on free charges, causing current. The amount of current depends not only on the magnitude of the voltage, but also on the characteristics of the material that the current is flowing through. The material can resist the flow of the charges, and the measure of how much a material resists the flow of charges is known as the *resistivity*. This resistivity is crudely analogous to the friction between two materials that resists motion.

When a voltage is applied to a conductor, an electrical field $\vec{\mathbf{E}}$ is created, and charges in the conductor feel a force due to the electrical field. The current density $\vec{\mathbf{J}}$ that results depends on the electrical field and the properties of the material. This dependence can be very complex. In some materials, including metals at a given temperature, the current density is approximately proportional to the electrical field. In these cases, the current density can be modeled as

\[\vec{\mathbf{J}}=\sigma\vec{\mathbf{E}},\]where $\sigma$ is the **electrical conductivity**. The electrical conductivity is analogous to thermal conductivity and is a measure of a material’s ability to conduct or transmit electricity. Conductors have a higher electrical conductivity than insulators. Since the electrical conductivity is $\sigma=J/E$, the units are

\[\sigma=\frac{[J]}{[E]}=\frac{\mathrm{A/m}^2}{\mathrm{V/m}}=\frac{\mathrm{A}}{\mathrm{V}\cdot\mathrm{m}}.\]

Here, we define a unit named the **ohm** with the Greek symbol uppercase omega, $\Omega$. The unit is named after Georg Simon Ohm, whom we will discuss later in this chapter. The $\Omega$ is used to avoid confusion with the number $0$. One ohm equals one volt per amp: $1~\Omega=1~\mathrm{V/A}$. The units of electrical conductivity are therefore $(\Omega\cdot\mathrm{m})^{-1}$.

Conductivity is an intrinsic property of a material. Another intrinsic property of a material is the **resistivity**, or electrical resistivity. The resistivity of a material is a measure of how strongly a material opposes the flow of electrical current. The symbol for resistivity is the lowercase Greek letter rho, $\rho$, and resistivity is the reciprocal of electrical conductivity:

\[\rho=\frac{1}{\sigma}.\]

The unit of resistivity in SI units is the ohm-meter $(\Omega\cdot\mathrm{m})$. We can define the resistivity in terms of the electrical field and the current density,

\begin{equation}\rho=\frac{E}{J}.\tag{5.3.1}\label{eq:5.3.1}\end{equation}

The greater the resistivity, the larger the field needed to produce a given current density. The lower the resistivity, the larger the current density produced by a given electrical field. Good conductors have a high conductivity and low resistivity. Good insulators have a low conductivity and a high resistivity. \ref{tab:5.3.1} lists resistivity and conductivity values for various materials.

\begin{gather}.\tag{Table 5.3.1}\label{tab:5.3.1}\end{gather}Material | Conductivity, $\sigma$ $(\Omega\cdot\mathrm{m})^{-1}$ | Resistivity, $\rho$ $(\Omega\cdot\mathrm{m})$ | Temperature Coefficient, $\alpha$ $(^{\circ}\mathrm{C})^{-1}$ |
---|---|---|---|

Conductors |
|||

Silver | $6.29\times10^7$ | $1.59\times10^{-8}$ | $0.0038$ |

Copper | $5.95\times10^7$ | $1.68\times10^{-8}$ | $0.0039$ |

Gold | $4.10\times10^7$ | $2.44\times10^{-8}$ | $0.0034$ |

Aluminum | $3.77\times10^7$ | $2.65\times10^{-8}$ | $0.0039$ |

Tungsten | $1.79\times10^7$ | $5.60\times10^{-8}$ | $0.0045$ |

Iron | $1.03\times10^7$ | $9.71\times10^{-8}$ | $0.0065$ |

Platinum | $0.94\times10^7$ | $10.60\times10^{-8}$ | $0.0039$ |

Steel | $0.50\times10^7$ | $20.00\times10^{-8}$ | |

Lead | $0.45\times10^7$ | $22.00\times10^{-8}$ | |

Manganin (Cu, Mn, Ni alloy) | $0.21\times10^7$ | $48.20\times10^{-8}$ | $0.000002$ |

Constantan (Cu, Ni alloy) | $0.20\times10^7$ | $49.00\times10^{-8}$ | $0.00003$ |

Mercury | $0.10\times10^7$ | $98.00\times10^{-8}$ | $0.0009$ |

Nichrome (Ni, Fe, Cr alloy) | $0.10\times10^7$ | $100.00\times10^{-8}$ | $0.0004$ |

Semiconductors[1] |
|||

Carbon (pure) | $2.86\times10^{-6}$ | $3.50\times10^{-5}$ | $−0.0005$ |

Carbon | $(2.86-1.67)\times10^{-6}$ | $(3.5-60)\times10^{-5}$ | $−0.0005$ |

Germanium (pure) | $600\times10^{-3}$ | $−0.048$ | |

Germanium | $(1-600)\times10^{-3}$ | $−0.050$ | |

Silicon (pure) | $2300$ | $−0.075$ | |

Silicon | $0.1-2300$ | $−0.07$ | |

Insulators |
|||

Amber | $2.00\times10^{-15}$ | $5\times10^{14}$ | |

Glass | $10^{-9}-10^{-14}$ | $10^9-10^{14}$ | |

Lucite | $<10^{-13}$ | $>10^{13}$ | |

Mica | $10^{-11}-10^{-15}$ | $10^{11}-10^{15}$ | |

Quartz (fused) | $2.00\times10^{-15}$ | $75\times10^{16}$ | |

Rubber (hard) | $10^{-13}-10^{-16}$ | $10^{13}-10^{16}$ | |

Sulfur | $10^{-15}$ | $10^{15}$ | |

Teflon^{TM} |
$<10^{-13}$ | $>10^{13}$ | |

Wood | $10^{-8}-10^{-11}$ | $10^8-10^{11}$ |

The materials listed in the table are separated into categories of conductors, semiconductors, and insulators, based on broad groupings of resistivity. Conductors have the smallest resistivity, and insulators have the largest; semiconductors have intermediate resistivity. Conductors have varying but large, free charge densities, whereas most charges in insulators are bound to atoms and are not free to move. Semiconductors are intermediate, having far fewer free charges than conductors, but having properties that make the number of free charges depend strongly on the type and amount of impurities in the semiconductor. These unique properties of semiconductors are put to use in modern electronics, as we will explore in later chapters.

The resistance of the wire is

\[R=\rho\frac{L}{A}=(1.68\times10^{-8}~\Omega\cdot\mathrm{m})\frac{5.00~\mathrm{m}}{3.31\times10^{-6}~\mathrm{m}^2}=0.025~\Omega.\] Finally, we can find the electrical field: \[E=\rho J=(1.68\times10^{-8}~\Omega\cdot\mathrm{m})\left(3.02\times10^3~\mathrm{A/m}^2\right)=5.07\times10^{-5}~\mathrm{V/m}.\]Copper wires use routinely used for extension cords and house wiring for several reasons. Copper has the highest electrical conductivity rating, and therefore the lowest resistivity rating, of all nonprecious metals. Also important is the tensile strength, where the tensile strength is a measure of the force required to pull an object to the point where it breaks. The tensile strength of a material is the maximum amount of tensile stress it can take before breaking. Copper has a high tensile strength, $2\times10^8~\mathrm{N/m}^2$. A third important characteristic is ductility. Ductility is a measure of a material’s ability to be drawn into wires and a measure of the flexibility of the material, and copper has a high ductility. Summarizing, for a conductor to be a suitable candidate for making wire, there are at least three important characteristics: low resistivity, high tensile strength, and high ductility. What other materials are used for wiring and what are the advantages and disadvantages?

View this interactive simulation to see what the effects of the cross-sectional area, the length, and the resistivity of a wire are on the resistance of a conductor. Adjust the variables using slide bars and see if the resistance becomes smaller or larger.

Looking back at \ref{tab:5.3.1}, you will see a column labeled “Temperature Coefficient.” The resistivity of some materials has a strong temperature dependence. In some materials, such as copper, the resistivity increases with increasing temperature. In fact, in most conducting metals, the resistivity increases with increasing temperature. The increasing temperature causes increased vibrations of the atoms in the lattice structure of the metals, which impede the motion of the electrons. In other materials, such as carbon, the resistivity decreases with increasing temperature. In many materials, the dependence is approximately linear and can be modeled using a linear equation:

\begin{equation}\rho\approx\rho_0[1+\alpha(T-T_0)],\tag{5.3.2}\label{eq:5.3.2}\end{equation}

where $\rho$ is the resistivity of the material at temperature $T$, $\alpha$ is the temperature coefficient of the material, and $\rho_0$ is the resistivity at $T_0$, usually taken as $T_0=20.00~^{\circ}\mathrm{C}$.

Note also that the temperature coefficient $\alpha$ is negative for the semiconductors listed in \ref{tab:5.3.1}, meaning that their resistivity decreases with increasing temperature. They become better conductors at higher temperature, because increased thermal agitation increases the number of free charges available to carry current. This property of decreasing $\rho$ with temperature is also related to the type and amount of impurities present in the semiconductors.

We now consider the resistance of a wire or component. The resistance is a measure of how difficult it is to pass current through a wire or component. Resistance depends on the resistivity. The resistivity is a characteristic of the material used to fabricate a wire or other electrical component, whereas the resistance is a characteristic of the wire or component.

To calculate the resistance, consider a section of conducting wire with cross-sectional area $A$, length $L$, and resistivity $\rho$. A battery is connected across the conductor, providing a potential difference $\Delta V$ across it (\ref{fig:5.3.1}). The potential difference produces an electrical field that is proportional to the current density, according to $\vec{\mathbf{E}}=\rho\vec{\mathbf{J}}$.

\begin{gather}.\tag{Figure 5.3.1}\label{fig:5.3.1}\end{gather}The magnitude of the electrical field across the segment of the conductor is equal to the voltage divided by the length, $E=V/L$, and the magnitude of the current density is equal to the current divided by the cross-sectional area, $J=I/A$. Using this information and recalling that the electrical field is proportional to the resistivity and the current density, we can see that the voltage is proportional to the current:

\begin{eqnarray*}E&=&\rho J\\\Rightarrow\frac{V}{L}&=&\rho\frac{I}{A}\\\Rightarrow V&=&\left(\rho\frac{L}{A}\right)I.\end{eqnarray*}The ratio of the voltage to the current is defined as the **resistance** $R$:

\begin{equation}R\equiv\frac{V}{I}.\tag{5.3.3}\label{eq:5.3.3}\end{equation}

The resistance of a cylindrical segment of a conductor is equal to the resistivity of the material times the length divided by the area:

\begin{equation}R\equiv\frac{V}{I}=\rho\frac{L}{A}.\tag{5.3.4}\label{eq:5.3.4}\end{equation}

The unit of resistance is the ohm, $\Omega$. For a given voltage, the higher the resistance, the lower the current.

A common component in electronic circuits is the resistor. The resistor can be used to reduce current flow or provide a voltage drop. \ref{fig:5.3.2} shows the symbols used for a resistor in schematic diagrams of a circuit. Two commonly used standards for circuit diagrams are provided by the American National Standard Institute (ANSI, pronounced “AN-see”) and the International Electrotechnical Commission (IEC). Both systems are commonly used. We use the ANSI standard in this text for its visual recognition, but we note that for larger, more complex circuits, the IEC standard may have a cleaner presentation, making it easier to read.

\begin{gather}.\tag{Figure 5.3.2}\label{fig:5.3.2}\end{gather}A resistor can be modeled as a cylinder with a cross-sectional area $A$ and a length $L$, made of a material with a resistivity $\rho$ (\ref{fig:5.3.3}). The resistance of the resistor is $R=\rho\frac{L}{A}$.

\begin{gather}.\tag{Figure 5.3.3}\label{fig:5.3.3}\end{gather}The most common material used to make a resistor is carbon. A carbon track is wrapped around a ceramic core, and two copper leads are attached. A second type of resistor is the metal film resistor, which also has a ceramic core. The track is made from a metal oxide material, which has semiconductive properties similar to carbon. Again, copper leads are inserted into the ends of the resistor. The resistor is then painted and marked for identification. A resistor has four colored bands, as shown in \ref{fig:5.3.4}.

\begin{gather}.\tag{Figure 5.3.4}\label{fig:5.3.4}\end{gather}Resistances range over many orders of magnitude. Some ceramic insulators, such as those used to support power lines, have resistances of $10^{12}~\Ohm$ or more. A dry person may have a hand-to-foot resistance of $10^{5}~\Ohm$, whereas the resistance of the human heart is about $10^{3}~\Ohm$. A meter-long piece of large-diameter copper wire may have a resistance of $10^{-5}~\Ohm$, and superconductors have no resistance at all at low temperatures. As we have seen, resistance is related to the shape of an object and the material of which it is composed.

The resistance of an object also depends on temperature, since $R_0$ is directly proportional to $\rho$. For a cylinder, we know $R=\rho\frac{L}{A}$, so if $L$ and $A$ do not change greatly with temperature, $R$ has the same temperature dependence as $\rho$. (Examination of the coefficients of linear expansion shows them to be about two orders of magnitude less than typical temperature coefficients of resistivity, so the effect of temperature on $L$ and $A$ is about two orders of magnitude less than on $\rho$.) Thus,

\begin{equation}R=R_0(1+\alpha\Delta T)\tag{5.3.5}\label{eq:5.3.5}\end{equation}

is the temperature dependence of the resistance of an object, where $R_0$ is the original resistance (usually taken to be $20.00~^{\circ}\mathrm{C}$) and $R$ is the resistance after a temperature change $\Delta T$. The color code gives the resistance of the resistor at a temperature of $T=20.00~^{\circ}\mathrm{C}$.

Numerous thermometers are based on the effect of temperature on resistance (\ref{fig:5.3.5}). One of the most common thermometers is based on the thermistor, a semiconductor crystal with a strong temperature dependence, the resistance of which is measured to obtain its temperature. The device is small, so that it quickly comes into thermal equilibrium with the part of a person it touches.

\begin{gather}.\tag{Figure 5.3.5}\label{fig:5.3.5}\end{gather}
\[R=R_0(1+\alpha\Delta T)=(0.350~\Omega)\left[1+\left(4.5\times10^{-3}~(^{\circ}\mathrm{C})^{-1}\right)(2850~^{\circ}\mathrm{C})\right]=4.8~\Omega.\]

A strain gauge is an electrical device to measure strain, as shown below. It consists of a flexible, insulating backing that supports a conduction foil pattern. The resistance of the foil changes as the backing is stretched. How does the strain gauge resistance change? Is the strain gauge affected by temperature changes?

\begin{eqnarray*}dR&=&\frac{\rho}{A}dr=\frac{\rho}{2\pi rL}dr,\\R&=&\int_{r_i}^{r_0}dR=\int_{r_i}^{r_0}\frac{\rho}{2\pi rL}dr=\frac{\rho}{2\pi L}\int_{r_i}^{r_0}\frac{1}{r}dr=\frac{\rho}{2\pi L}\ln\frac{r_0}{r_i}.\end{eqnarray*}

The resistance between the two conductors of a coaxial cable depends on the resistivity of the material separating the two conductors, the length of the cable and the inner and outer radius of the two conductor. If you are designing a coaxial cable, how does the resistance between the two conductors depend on these variables?

View this simulation to see how the voltage applied and the resistance of the material the current flows through affects the current through the material. You can visualize the collisions of the electrons and the atoms of the material effect the temperature of the material.

- Describe Ohm’s law
- Recognize when Ohm’s law applies and when it does not

We have been discussing three electrical properties so far in this chapter: current, voltage, and resistance. It turns out that many materials exhibit a simple relationship among the values for these properties, known as Ohm’s law. Many other materials do not show this relationship, so despite being called Ohm’s law, it is not considered a law of nature, like Newton’s laws or the laws of thermodynamics. But it is very useful for calculations involving materials that do obey Ohm’s law.

The current that flows through most substances is directly proportional to the voltage $V$ applied to it. The German physicist Georg Simon **Ohm** (1787–1854) was the first to demonstrate experimentally that the current in a metal wire is *directly proportional to the voltage applied*:

This important relationship is the basis for **Ohm’s law**. It can be viewed as a cause-and-effect relationship, with voltage the cause and current the effect. This is an empirical law, which is to say that it is an experimentally observed phenomenon, like friction. Such a linear relationship doesn’t always occur. Any material, component, or device that obeys Ohm’s law, where the current through the device is proportional to the voltage applied, is known as an **ohmic** material or ohmic component. Any material or component that does not obey Ohm’s law is known as a **nonohmic** material or nonohmic component.

In a paper published in 1827, Georg Ohm described an experiment in which he measured voltage across and current through various simple electrical circuits containing various lengths of wire. A similar experiment is shown in \ref{fig:5.4.1}. This experiment is used to observe the current through a resistor that results from an applied voltage. In this simple circuit, a resistor is connected in series with a battery. The voltage is measured with a voltmeter, which must be placed across the resistor (in parallel with the resistor). The current is measured with an ammeter, which must be in line with the resistor (in series with the resistor).

\begin{gather}.\tag{Figure 5.4.1}\label{fig:5.4.1}\end{gather}In this updated version of Ohm’s original experiment, several measurements of the current were made for several different voltages. When the battery was hooked up as in \ref{fig:5.4.1}(a), the current flowed in the clockwise direction and the readings of the voltmeter and ammeter were positive. Does the behavior of the current change if the current flowed in the opposite direction? To get the current to flow in the opposite direction, the leads of the battery can be switched. When the leads of the battery were switched, the readings of the voltmeter and ammeter readings were negative because the current flowed in the opposite direction, in this case, counterclockwise. Results of a similar experiment are shown in \ref{fig:5.4.2}.

\begin{gather}.\tag{Figure 5.4.2}\label{fig:5.4.2}\end{gather}In this experiment, the voltage applied across the resistor varies from $−10.00$ to $+10.00~\mathrm{V}$, by increments of $1.00~\mathrm{V}$. The current through the resistor and the voltage across the resistor are measured. A plot is made of the voltage versus the current, and the result is approximately linear. The slope of the line is the resistance, or the voltage divided by the current. This result is known as **Ohm’s law**:

\begin{equation}V=IR,\tag{5.4.1}\label{eq:5.4.1}\end{equation}

where $V$ is the voltage measured in volts across the object in question, $I$ is the current measured through the object in amps, and $R$ is the resistance in units of ohms. As stated previously, any device that shows a linear relationship between the voltage and the current is known as an ohmic device. A resistor is therefore an ohmic device.

(b) First, the resistance is temperature dependent so the new resistance after the resistor has been heated can be found using $R=R_0(1+\alpha\Delta T)$. The current can be found using Ohm’s law in the form $I=V/R$.

The current through the heated resistor is

\[I=\frac{V}{R}=\frac{9.00~\mathrm{V}}{2.94\times10^{3}~\Omega}=3.06\times10^{-3}~\mathrm{A}=3.06~\mathrm{mA}.\]

The voltage supplied to your house varies as $V(t)=V_{\mathrm{max}}\sin\,(2\pi ft)$. If a resistor is connected across this voltage, will Ohm’s law $V=IR$ still be valid?

See how the equation form of Ohm’s law relates to a simple circuit. Adjust the voltage and resistance, and see the current change according to Ohm’s law. The sizes of the symbols in the equation change to match the circuit diagram.

Nonohmic devices do not exhibit a linear relationship between the voltage and the current. One such device is the semiconducting circuit element known as a diode. A **diode** is a circuit device that allows current flow in only one direction. A diagram of a simple circuit consisting of a battery, a diode, and a resistor is shown in \ref{fig:5.4.3}. Although we do not cover the theory of the diode in this section, the diode can be tested to see if it is an ohmic or a nonohmic device.

A plot of current versus voltage is shown in \ref{fig:5.4.4}. Note that the behavior of the diode is shown as current versus voltage, whereas the resistor operation was shown as voltage versus current. A diode consists of an anode and a cathode. When the anode is at a negative potential and the cathode is at a positive potential, as shown in part (a), the diode is said to have reverse bias. With reverse bias, the diode has an extremely large resistance and there is very little current flow—essentially zero current—through the diode and the resistor. As the voltage applied to the circuit increases, the current remains essentially zero, until the voltage reaches the breakdown voltage and the diode conducts current, as shown in \ref{fig:5.4.4}. When the battery and the potential across the diode are reversed, making the anode positive and the cathode negative, the diode conducts and current flows through the diode if the voltage is greater than $0.7~\mathrm{V}$. The resistance of the diode is close to zero. (This is the reason for the resistor in the circuit; if it were not there, the current would become very large.) You can see from the graph in \ref{fig:5.4.4} that the voltage and the current do not have a linear relationship. Thus, the diode is an example of a nonohmic device.

\begin{gather}.\tag{Figure 5.4.4}\label{fig:5.4.4}\end{gather}Ohm’s law is commonly stated as $V=IR$, but originally it was stated as a microscopic view, in terms of the current density, the conductivity, and the electrical field. This microscopic view suggests the proportionality $V\propto I$ comes from the drift velocity of the free electrons in the metal that results from an applied electrical field. As stated earlier, the current density is proportional to the applied electrical field. The reformulation of Ohm’s law is credited to Gustav Kirchhoff, whose name we will see again in the next chapter.

- Express electrical power in terms of the voltage and the current
- Describe the power dissipated by a resistor in an electric circuit
- Calculate the energy efficiency and cost effectiveness of appliances and equipment

In an electric circuit, electrical energy is continuously converted into other forms of energy. For example, when a current flows in a conductor, electrical energy is converted into thermal energy within the conductor. The electrical field, supplied by the voltage source, accelerates the free electrons, increasing their kinetic energy for a short time. This increased kinetic energy is converted into thermal energy through collisions with the ions of the lattice structure of the conductor. Power is defined as the rate at which work is done by a force measured in watts. Power can also be defined as the rate at which energy is transferred. In this section, we discuss the time rate of energy transfer, or power, in an electric circuit.

Power is associated by many people with electricity. Power transmission lines might come to mind. We also think of light bulbs in terms of their power ratings in watts. What is the expression for **electric power**?

Let us compare a $25{\text -}\mathrm{W}$ bulb with a $60{\text -}\mathrm{W}$ bulb (\ref{fig:5.5.1}(a)). The $60{\text -}\mathrm{W}$ bulb glows brighter than the $25{\text -}\mathrm{W}$ bulb. Although it is not shown, a $60{\text -}\mathrm{W}$ light bulb is also warmer than the $25{\text -}\mathrm{W}$ bulb. The heat and light is produced by from the conversion of electrical energy. The kinetic energy lost by the electrons in collisions is converted into the internal energy of the conductor and radiation. How are voltage, current, and resistance related to electric power?

\begin{gather}.\tag{Figure 5.5.1}\label{fig:5.5.1}\end{gather}To calculate electric power, consider a voltage difference existing across a material (\ref{fig:5.5.2}). The electric potential $V_1$ is higher than the electric potential at $V_2$, and the voltage difference is negative $V=V_2-V_1$. As discussed in Electric Potential, an electrical field exists between the two potentials, which points from the higher potential to the lower potential. Recall that the electrical potential is defined as the potential energy per charge, $V=\Delta U/q$, and the charge $\Delta Q$ loses potential energy moving through the potential difference.

\begin{gather}.\tag{Figure 5.5.2}\label{fig:5.5.2}\end{gather}If the charge is positive, the charge experiences a force due to the electrical field $\vec{\mathbf{F}}=m\vec{\mathbf{a}}=\Delta Q\vec{\mathbf{E}}$. This force is necessary to keep the charge moving. This force does not act to accelerate the charge through the entire distance $\Delta L$ because of the interactions of the charge with atoms and free electrons in the material. The speed, and therefore the kinetic energy, of the charge do not increase during the entire trip across $\Delta L$, and charge passing through area $A_2$ has the same drift velocity $v_{\mathrm{d}$ as the charge that passes through area $A_1$. However, work is done on the charge, by the electrical field, which changes the potential energy. Since the change in the electrical potential difference is negative, the electrical field is found to be

\[E=-\frac{(V_2-V_1)}{\Delta L}=\frac{V}{\Delta L}.\]

The work done on the charge is equal to the electric force times the length at which the force is applied,

\[W=F\Delta L=(\Delta QE)\Delta L=\left(\Delta Q\frac{V}{\Delta L}\Delta L=\Delta QV=\Delta U.\right)\]

The charge moves at a drift velocity $v_{\mathrm{d}}$ so the work done on the charge results in a loss of potential energy, but the average kinetic energy remains constant. The lost electrical potential energy appears as thermal energy in the material. On a microscopic scale, the energy transfer is due to collisions between the charge and the molecules of the material, which leads to an increase in temperature in the material. The loss of potential energy results in an increase in the temperature of the material, which is dissipated as radiation. In a resistor, it is dissipated as heat, and in a light bulb, it is dissipated as heat and light.

The power dissipated by the material as heat and light is equal to the time rate of change of the work:

\[P=\frac{\Delta U}{\Delta t}=-\frac{\Delta AV}{\Delta t}=IV.\]

With a resistor, the voltage drop across the resistor is dissipated as heat. Ohm’s law states that the voltage across the resistor is equal to the current times the resistance, $V=IR$. The power dissipated by the resistor is therefore

\[P=IV=I(IR)=I^2R~~~~~\mathrm{or}~~~~~P=IV=\left(\frac{V}{R}\right)V=\frac{V^2}{R}.\]

If a resistor is connected to a battery, the power dissipated as radiant energy by the wires and the resistor is equal to $P=IV=I^2R=\frac{V^2}{R}$ The power supplied from the battery is equal to current times the voltage, $P=IV$.

The electric power gained or lost by any device has the form

\begin{equation}P=IV\tag{5.5.1}\label{eq:5.5.1}\end{equation} The power dissipated by a resistor has the form \begin{equation}P=I^2R=\frac{V^2}{R}\tag{5.5.2}\label{eq:5.5.2}\end{equation}Different insights can be gained from the three different expressions for electric power. For example, $P=V^2/R$ implies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in $P=V^2/R$, the effect of applying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a $25{\text -}\mathrm{W}$ bulb, its power nearly quadruples to about $100~\mathrm{W}$, burning it out. If the bulb’s resistance remained constant, its power would be exactly $100~\mathrm{W}$, but at the higher temperature, its resistance is higher, too.

Electric motors have a reasonably high efficiency. A $100{\text -}\mathrm{hp}$ motor can have an efficiency of $90\%$ and a $1{\text -}\mathrm{hp}$ motor can have an efficiency of $80\%$. Why is it important to use high-performance motors?

A fuse (\ref{fig:5.5.3}) is a device that protects a circuit from currents that are too high. A fuse is basically a short piece of wire between two contacts. As we have seen, when a current is running through a conductor, the kinetic energy of the charge carriers is converted into thermal energy in the conductor. The piece of wire in the fuse is under tension and has a low melting point. The wire is designed to heat up and break at the rated current. The fuse is destroyed and must be replaced, but it protects the rest of the circuit. Fuses act quickly, but there is a small time delay while the wire heats up and breaks.

\begin{gather}.\tag{Figure 5.5.3}\label{fig:5.5.3}\end{gather}Circuit breakers are also rated for a maximum current, and open to protect the circuit, but can be reset. Circuit breakers react much faster. The operation of circuit breakers is not within the scope of this chapter and will be discussed in later chapters. Another method of protecting equipment and people is the ground fault circuit interrupter (GFCI), which is common in bathrooms and kitchens. The GFCI outlets respond very quickly to changes in current. These outlets open when there is a change in magnetic field produced by current-carrying conductors, which is also beyond the scope of this chapter and is covered in a later chapter.

The more electric appliances you use and the longer they are left on, the higher your electric bill. This familiar fact is based on the relationship between energy and power. You pay for the energy used. Since $P=\frac{dE}{dt}$, we see that

\[E=\int Pdt\]is the energy used by a device using power $P$ for a time interval $t$. If power is delivered at a constant rate, then then the energy can be found by $E=Pt$. For example, the more light bulbs burning, the greater $P$ used; the longer they are on, the greater $t$ is.

The energy unit on electric bills is the kilowatt-hour $(\mathrm{kW}\cdot\mathrm{h})$, consistent with the relationship $E=Pt$. It is easy to estimate the cost of operating electrical appliances if you have some idea of their power consumption rate in watts or kilowatts, the time they are on in hours, and the cost per kilowatt-hour for your electric utility. Kilowatt-hours, like all other specialized energy units such as food calories, can be converted into joules. You can prove to yourself that $1~\mathrm{kW}\cdot\mathrm{h}=3.6\times10^{6}~\mathrm{J}$.

The electrical energy ($E$) used can be reduced either by reducing the time of use or by reducing the power consumption of that appliance or fixture. This not only reduces the cost but also results in a reduced impact on the environment. Improvements to lighting are some of the fastest ways to reduce the electrical energy used in a home or business. About $20\%$ of a home’s use of energy goes to lighting, and the number for commercial establishments is closer to $40\%$. Fluorescent lights are about four times more efficient than incandescent lights—this is true for both the long tubes and the compact fluorescent lights (CFLs). (See \ref{fig:5.5.1}(b).) Thus, a $60{\text -}\mathrm{W}$ incandescent bulb can be replaced by a $15{\text -}\mathrm{W}$ CFL, which has the same brightness and colour. CFLs have a bent tube inside a globe or a spiral-shaped tube, all connected to a standard screw-in base that fits standard incandescent light sockets. (Original problems with colour, flicker, shape, and high initial investment for CFLs have been addressed in recent years.)

The heat transfer from these CFLs is less, and they last up to $10$ times longer than incandescent bulbs. The significance of an investment in such bulbs is addressed in the next example. New white LED lights (which are clusters of small LED bulbs) are even more efficient (twice that of CFLs) and last five times longer than CFLs.

(b) Multiply the energy by the cost.

Is the efficiency of the various light bulbs the only consideration when comparing the various light bulbs?

Changing light bulbs from incandescent bulbs to CFL or LED bulbs is a simple way to reduce energy consumption in homes and commercial sites. CFL bulbs operate with a much different mechanism than do incandescent lights. The mechanism is complex and beyond the scope of this chapter, but here is a very general description of the mechanism. CFL bulbs contain argon and mercury vapor housed within a spiral-shaped tube. The CFL bulbs use a “ballast” that increases the voltage used by the CFL bulb. The ballast produce an electrical current, which passes through the gas mixture and excites the gas molecules. The excited gas molecules produce ultraviolet (UV) light, which in turn stimulates the fluorescent coating on the inside of the tube. This coating fluoresces in the visible spectrum, emitting visible light. Traditional fluorescent tubes and CFL bulbs had a short time delay of up to a few seconds while the mixture was being “warmed up” and the molecules reached an excited state. It should be noted that these bulbs do contain mercury, which is poisonous, but if the bulb is broken, the mercury is never released. Even if the bulb is broken, the mercury tends to remain in the fluorescent coating. The amount is also quite small and the advantage of the energy saving may outweigh the disadvantage of using mercury.

The CFL light bulbs are being replaced with LED light bulbs, where LED stands for “light-emitting diode.” The diode was briefly discussed as a nonohmic device, made of semiconducting material, which essentially permits current flow in one direction. LEDs are a special type of diode made of semiconducting materials infused with impurities in combinations and concentrations that enable the extra energy from the movement of the electrons during electrical excitation to be converted into visible light.

Commercial LEDs are quickly becoming the standard for commercial and residential lighting, replacing incandescent and CFL bulbs. They are designed for the visible spectrum and are constructed from gallium doped with arsenic and phosphorous atoms. The color emitted from an LED depends on the materials used in the semiconductor and the current. In the early years of LED development, small LEDs found on circuit boards were red, green, and yellow, but LED light bulbs can now be programmed to produce millions of colors of light as well as many different hues of white light.

The energy savings can be significant when replacing an incandescent light bulb or a CFL light bulb with an LED light. Light bulbs are rated by the amount of power that the bulb consumes, and the amount of light output is measured in lumens. The lumen ($\mathrm{lm}$) is the SI-derived unit of luminous flux and is a measure of the total quantity of visible light emitted by a source. A $60{\text -}\mathrm{W}$ incandescent light bulb can be replaced with a $13{\text -}$ to $15{\text -}\mathrm{W}$ CFL bulb or a $6{\text -}$ to $8{\text -}\mathrm{W}$ LED bulb, all three of which have a light output of approximately $800~\mathrm{lm}$. A table of light output for some commonly used light bulbs appears in \ref{tab:5.5.1}.

The life spans of the three types of bulbs are significantly different. An LED bulb has a life span of $50,000$ hours, whereas the CFL has a lifespan of $8000$ hours and the incandescent lasts a mere $1200$ hours. The LED bulb is the most durable, easily withstanding rough treatment such as jarring and bumping. The incandescent light bulb has little tolerance to the same treatment since the filament and glass can easily break. The CFL bulb is also less durable than the LED bulb because of its glass construction. The amount of heat emitted is $3.4~\mathrm{btu/h}$ for the $8{\text -}\mathrm{W}$ LED bulb, $85~\mathrm{btu/h}$ for the $60{\text -}\mathrm{W}$ incandescent bulb, and $30~\mathrm{btu/h}$ for the CFL bulb. As mentioned earlier, a major drawback of the CFL bulb is that it contains mercury, a neurotoxin, and must be disposed of as hazardous waste. From these data, it is easy to understand why the LED light bulb is quickly becoming the standard in lighting.

\begin{gather}.\tag{Table 5.5.1}\label{tab:5.5.1}\end{gather}Light Output (lumens) | LED Light Bulbs (watts) | Incandescent Light Bulbs (watts) | CFL Light Bulbs (watts) |
---|---|---|---|

$450$ | $4−5$ | $40$ | $9−13$ |

$800$ | $6−8$ | $60$ | $13−15$ |

$1100$ | $9−13$ | $75$ | $18−25$ |

$1600$ | $16−20$ | $100$ | $23−30$ |

$2600$ | $25−28$ | $150$ | $30−55$ |

In this chapter, we have discussed relationships between voltages, current, resistance, and power. \ref{fig:5.5.4} shows a summary of the relationships between these measurable quantities for ohmic devices. (Recall that ohmic devices follow Ohm’s law $V=IR$.) For example, if you need to calculate the power, use the pink section, which shows that $P=VI$, $P=\frac{V^2}{R}$, and $P=I^2R$.

\begin{gather}.\tag{Figure 5.5.4}\label{fig:5.5.4}\end{gather}Which equation you use depends on what values you are given, or you measure. For example if you are given the current and the resistance, use $P=I^2R$. Although all the possible combinations may seem overwhelming, don’t forget that they all are combinations of just two equations, Ohm’s law ($V=IR$) and power ($P=IV$).

- Describe the phenomenon of superconductivity
- List applications of superconductivity

Touch the power supply of your laptop computer or some other device. It probably feels slightly warm. That heat is an unwanted byproduct of the process of converting household electric power into a current that can be used by your device. Although electric power is reasonably efficient, other losses are associated with it. As discussed in the section on power and energy, transmission of electric power produces $I^2R$ line losses. These line losses exist whether the power is generated from conventional power plants (using coal, oil, or gas), nuclear plants, solar plants, hydroelectric plants, or wind farms. These losses can be reduced, but not eliminated, by transmitting using a higher voltage. It would be wonderful if these line losses could be eliminated, but that would require transmission lines that have zero resistance. In a world that has a global interest in not wasting energy, the reduction or elimination of this unwanted thermal energy would be a significant achievement. Is this possible?

In 1911, Heike **Kamerlingh Onnes** of Leiden University, a Dutch physicist, was looking at the temperature dependence of the resistance of the element mercury. He cooled the sample of mercury and noticed the familiar behavior of a linear dependence of resistance on temperature; as the temperature decreased, the resistance decreased. Kamerlingh Onnes continued to cool the sample of mercury, using liquid helium. As the temperature approached $4.2~\mathrm{K}$ ($-269.2~^{\circ}\mathrm{C}$), the resistance abruptly went to zero (\ref{fig:5.6.1}). This temperature is known as the **critical temperature** $T_c$ for mercury. The sample of mercury entered into a phase where the resistance was absolutely zero. This phenomenon is known as **superconductivity**. (*Note:* If you connect the leads of a three-digit ohmmeter across a conductor, the reading commonly shows up as $0.00~\Omega$. The resistance of the conductor is not actually zero, it is less than $0.01~\Omega$.) There are various methods to measure very small resistances, such as the four-point method, but an ohmmeter is not an acceptable method to use for testing resistance in superconductivity.

As research continued, several other materials were found to enter a superconducting phase, when the temperature reached near absolute zero. In 1941, an alloy of niobium-nitride was found that could become superconducting at $T_c=16~\mathrm{K}$ ($-257~^{\circ}\mathrm{C}$) and in 1953, vanadium-silicon was found to become superconductive at $T_c=17.5~\mathrm{K}$ ($-255.7~^{\circ}\mathrm{C}$). The temperatures for the transition into superconductivity were slowly creeping higher. Strangely, many materials that make good conductors, such as copper, silver, and gold, do not exhibit superconductivity. Imagine the energy savings if transmission lines for electric power-generating stations could be made to be superconducting at temperatures near room temperature! A resistance of zero ohms means no $I^2R$ losses and a great boost to reducing energy consumption. The problem is that $T_c=17.5~\mathrm{K}$ is still very cold and in the range of liquid helium temperatures. At this temperature, it is not cost effective to transmit electrical energy because of the cooling requirements.

A large jump was seen in 1986, when a team of researchers, headed by Dr. Ching Wu Chu of Houston University, fabricated a brittle, ceramic compound with a transition temperature of $T_c=92~\mathrm{K}$ ($-181~^{\circ}\mathrm{C}$) The ceramic material, composed of yttrium barium copper oxide (YBCO), was an insulator at room temperature. Although this temperature still seems quite cold, it is near the boiling point of liquid nitrogen, a liquid commonly used in refrigeration. You may have noticed refrigerated trucks traveling down the highway labeled as “Liquid Nitrogen Cooled.”

YBCO ceramic is a material that could be useful for transmitting electrical energy because the cost saving of reducing the $I^2R$ losses are larger than the cost of cooling the superconducting cable, making it financially feasible. There were and are many engineering problems to overcome. For example, unlike traditional electrical cables, which are flexible and have a decent tensile strength, ceramics are brittle and would break rather than stretch under pressure. Processes that are rather simple with traditional cables, such as making connections, become difficult when working with ceramics. The problems are difficult and complex, and material scientists and engineers are coming up with innovative solutions.

An interesting consequence of the resistance going to zero is that once a current is established in a superconductor, it persists without an applied voltage source. Current loops in a superconductor have been set up and the current loops have been observed to persist for years without decaying.

Zero resistance is not the only interesting phenomenon that occurs as the materials reach their transition temperatures. A second effect is the exclusion of magnetic fields. This is known as the **Meissner effect** (\ref{fig:5.6.2}). A light, permanent magnet placed over a superconducting sample will levitate in a stable position above the superconductor. High-speed trains have been developed that levitate on strong superconducting magnets, eliminating the friction normally experienced between the train and the tracks. In Japan, the Yamanashi Maglev test line opened on April 3, 1997. In April 2015, the MLX01 test vehicle attained a speed of $374~\mathrm{mph}$ ($603~\mathrm{km/h}$).

\ref{tab:5.6.1} shows a select list of elements, compounds, and high-temperature superconductors, along with the critical temperatures for which they become superconducting. Each section is sorted from the highest critical temperature to the lowest. Also listed is the critical magnetic field for some of the materials. This is the strength of the magnetic field that destroys superconductivity. Finally, the type of the superconductor is listed.

There are two types of superconductors. There are 30 pure metals that exhibit zero resistivity below their critical temperature and exhibit the Meissner effect, the property of excluding magnetic fields from the interior of the superconductor while the superconductor is at a temperature below the critical temperature. These metals are called Type I superconductors. The superconductivity exists only below their critical temperatures and below a critical magnetic field strength. Type I superconductors are well described by the BCS theory (described next). Type I superconductors have limited practical applications because the strength of the critical magnetic field needed to destroy the superconductivity is quite low.

Type II superconductors are found to have much higher critical magnetic fields and therefore can carry much higher current densities while remaining in the superconducting state. A collection of various ceramics containing barium-copper-oxide have much higher critical temperatures for the transition into a superconducting state. Superconducting materials that belong to this subcategory of the Type II superconductors are often categorized as high-temperature superconductors.

Type I superconductors, along with some Type II superconductors can be modeled using the **BCS theory**, proposed by John **Bardeen**, Leon **Cooper**, and Robert **Schrieffer**. Although the theory is beyond the scope of this textbook, a short summary of the theory is provided here. The theory considers pairs of electrons and how they are coupled together through lattice-vibration interactions. Through the interactions with the crystalline lattice, electrons near the Fermi energy level feel a small attractive force and form pairs (**Cooper pairs**), and the coupling is known as a phonon interaction. Single electrons are fermions, which are particles that obey the Pauli exclusion principle. The Pauli exclusion principle in quantum mechanics states that two identical fermions (particles with half-integer spin) cannot occupy the same quantum state simultaneously. Each electron has four quantum numbers ($n,l,m_l,m_s$) The principal quantum number ($n$) describes the energy of the electron, the orbital angular momentum quantum number ($l$) indicates the most probable distance from the nucleus, the magnetic quantum number ($m_l$) describes the energy levels in the subshell, and the electron spin quantum number ($m_s$) describes the orientation of the spin of the electron, either up or down. As the material enters a superconducting state, pairs of electrons act more like bosons, which can condense into the same energy level and need not obey the Pauli exclusion principle. The electron pairs have a slightly lower energy and leave an energy gap above them on the order of $0.001~\mathrm{eV}$. This energy gap inhibits collision interactions that lead to ordinary resistivity. When the material is below the critical temperature, the thermal energy is less than the band gap and the material exhibits zero resistivity.

Material | Symbol or Formula | Critical Temperature
T_{c} ($\mathrm{K}$) |
Critical
Magnetic Field
H_{c} ($\mathrm{T}$) |
Type |
---|---|---|---|---|

Elements |
||||

Lead | Pb | 7.19 | 0.08 | I |

Lanthanum | La | ($\alpha$) 4.90 − ($\beta$) 6.30 | I | |

Tantalum | Ta | 4.48 | 0.09 | I |

Mercury | Hg | ($\alpha$) 4.15 − ($\beta$) 3.95 | 0.04 | I |

Tin | Sn | 3.72 | 0.03 | I |

Indium | In | 3.40 | 0.03 | I |

Thallium | Tl | 2.39 | 0.03 | I |

Rhenium | Re | 2.40 | 0.03 | I |

Thorium | Th | 1.37 | 0.013 | I |

Protactinium | Pa | 1.40 | I | |

Aluminum | Al | 1.20 | 0.01 | I |

Gallium | Ga | 1.10 | 0.005 | I |

Zinc | Zn | 0.86 | 0.014 | I |

Titanium | Ti | 0.39 | 0.01 | I |

Uranium | U | ($\alpha$) 0.68 − ($\beta$) 1.80 | I | |

Cadmium | Cd | 11.4 | 4.00 | I |

Compounds |
||||

Niobium-germanium | Nb_{3}Ge |
23.20 | 37.00 | II |

Niobium-tin | Nb_{3}Sn |
18.30 | 30.00 | II |

Niobium-nitrite | NbN | 16.00 | II | |

Niobium-titanium | NbTi | 10.00 | 15.00 | II |

High-Temperature Oxides |
||||

HgBa_{2}CaCu_{2}O_{8} |
134.00 | II | ||

Tl_{2}Ba_{2}Ca_{2}Cu_{3}O_{10} |
125.00 | II | ||

YBa_{2}Cu_{3}O_{7} |
92.00 | 120.00 | II |

Superconductors can be used to make superconducting magnets. These magnets are 10 times stronger than the strongest electromagnets. These magnets are currently in use in magnetic resonance imaging (MRI), which produces high-quality images of the body interior without dangerous radiation.

Another interesting application of superconductivity is the **SQUID** (superconducting quantum interference device). A SQUID is a very sensitive magnetometer used to measure extremely subtle magnetic fields. The operation of the SQUID is based on superconducting loops containing Josephson junctions. A **Josephson junction** is the result of a theoretical prediction made by B. D. Josephson in an article published in 1962. In the article, Josephson described how a supercurrent can flow between two pieces of superconductor separated by a thin layer of insulator. This phenomenon is now called the Josephson effect. The SQUID consists of a superconducting current loop containing two Josephson junctions, as shown in \ref{fig:5.6.3}. When the loop is placed in even a very weak magnetic field, there is an interference effect that depends on the strength of the magnetic field.

Superconductivity is a fascinating and useful phenomenon. At critical temperatures near the boiling point of liquid nitrogen, superconductivity has special applications in MRIs, particle accelerators, and high-speed trains. Will we reach a state where we can have materials enter the superconducting phase at near room temperatures? It seems a long way off, but if scientists in 1911 were asked if we would reach liquid-nitrogen temperatures with a ceramic, they might have thought it implausible.

Average electrical current | $I_{\mathrm{ave}}=\frac{\Delta Q}{\Delta t}$ |

Definition of an ampere | $1~\mathrm{A}=1~\mathrm{C/s}$ |

Electrical current | $I=\frac{dQ}{dt}$ |

Drift velocity | $v_d=\frac{I}{nqA}$ |

Current density | $I=\iint_{\mathrm{area}}\vec{\mathbf{J}}\cdot d\vec{\mathbf{A}}$ |

Resistivity | $\rho=\frac{E}{J}$ |

Common expression of Ohm’s law | $V=IR$ |

Resistivity as a function of temperature | $\rho=\rho_0[1+\alpha(T-T_0)]$ |

Definition of resistance | $R\equiv\frac{V}{I}$ |

Resistance of a cylinder of material | $R=\rho\frac{L}{A}$ |

Temperature dependence of resistance | $R=R_0(1+\alpha\Delta T)$ |

Electric power | $P=IV$ |

Power dissipated by a resistor | $P=I^2R=\frac{V^2}{R}$ |

- The average electrical current $I_{\mathrm{ave}}$ is the rate at which charge flows, given by $I_{\mathrm{ave}}=\frac{\Delta Q}{\Delta t}$ where $\Delta Q$ is the amount of charge passing through an area in time $\Delta t.$
- The instantaneous electrical current, or simply the current $I,$ is the rate at which charge flows. Taking the limit as the change in time approaches zero, we have $I=\frac{dQ}{dt},$ where $\frac{dQ}{dt}$ is the time derivative of the charge.
- The direction of conventional current is taken as the direction in which positive charge moves. In a simple direct-current (DC) circuit, this will be from the positive terminal of the battery to the negative terminal.
- The SI unit for current is the ampere, or simply the amp ($\mathrm{A}$), where $1~\mathrm{A}=1~\mathrm{C/s}.$
- Current consists of the flow of free charges, such as electrons, protons, and ions.

- The current through a conductor depends mainly on the motion of free electrons.
- When an electrical field is applied to a conductor, the free electrons in a conductor do not move through a conductor at a constant speed and direction; instead, the motion is almost random due to collisions with atoms and other free electrons.
- Even though the electrons move in a nearly random fashion, when an electrical field is applied to the conductor, the overall velocity of the electrons can be defined in terms of a drift velocity.
- The current density is a vector quantity defined as the current through an infinitesimal area divided by the area.
- The current can be found from the current density, $I=\iint_{\mathrm{area}}\vec{\mathbf{J}}\cdot d\vec{\mathbf{A}}.$
- An incandescent light bulb is a filament of wire enclosed in a glass bulb that is partially evacuated. Current runs through the filament, where the electrical energy is converted to light and heat.

- Resistance has units of ohms ($\Omega$), related to volts and amperes by $1~\mathrm{\Omega}=1~\mathrm{V/A}.$
- The resistance $R$ of a cylinder of length $L$ and cross-sectional area $A$ is $R=\frac{\rho L}{A},$ where $\rho$
- Values of $\rho$ in Table 5.3.1 show that materials fall into three groups—conductors, semiconductors, and insulators.
- Temperature affects resistivity; for relatively small temperature changes $\Delta T,$ resistivity is $\rho=\rho_0(1+\alpha\Delta T),$ where $\rho_0$ is the original resistivity and $\alpha$ is the temperature coefficient of resistivity.
- The resistance $R$ of an object also varies with temperature: $R=R_0(1+\alpha\Delta T),$ where $R_0$ is the original resistance, and $R$ is the resistance after the temperature change.

- Ohm’s law is an empirical relationship for current, voltage, and resistance for some common types of circuit elements, including resistors. It does not apply to other devices, such as diodes.
- One statement of Ohm’s law gives the relationship among current $I,$ voltage $V,$ and resistance $R$ in a simple circuit as $V=IR.$
- Another statement of Ohm’s law, on a microscopic level, is $J=\sigma E.$

- Electric power is the rate at which electric energy is supplied to a circuit or consumed by a load.
- Power dissipated by a resistor depends on the square of the current through the resistor and is equal to $P=I^2R=\frac{V^2}{R}.$
- The SI unit for electric power is the watt and the SI unit for electric energy is the joule. Another common unit for electric energy, used by power companies, is the kilowatt-hour ($\mathrm{kW}\cdot\mathrm{h}$).
- The total energy used over a time interval can be found by $E=\int Pdt.$

- Superconductivity is a phenomenon that occurs in some materials when cooled to very low critical temperatures, resulting in a resistance of exactly zero and the expulsion of all magnetic fields.
- Materials that are normally good conductors (such as copper, gold, and silver) do not experience superconductivity.
- Superconductivity was first observed in mercury by Heike Kamerlingh Onnes in 1911. In 1986, Dr. Ching Wu Chu of Houston University fabricated a brittle, ceramic compound with a critical temperature close to the temperature of liquid nitrogen.
- Superconductivity can be used in the manufacture of superconducting magnets for use in MRIs and high-speed, levitated trains.

5.1 The time for $1.00~\mathrm{C}$ of charge to flow would be $\Delta t=\frac{\Delta Q}{I},$ slightly less than an hour. This is quite different from the $5.55~\mathrm{ms}$ for the truck battery. The calculator takes a very small amount of energy to operate, unlike the truck’s starter motor. There are several reasons that vehicles use batteries and not solar cells. Aside from the obvious fact that a light source to run the solar cells for a car or truck is not always available, the large amount of current needed to start the engine cannot easily be supplied by present-day solar cells. Solar cells can possibly be used to charge the batteries. Charging the battery requires a small amount of energy when compared to the energy required to run the engine and the other accessories such as the heater and air conditioner. Present day solar-powered cars are powered by solar panels, which may power an electric motor, instead of an internal combustion engine.

5.2 The total current needed by all the appliances in the living room (a few lamps, a television, and your laptop) draw less current and require less power than the refrigerator.

5.3 The diameter of the $14$-gauge wire is smaller than the diameter of the $12$-gauge wire. Since the drift velocity is inversely proportional to the cross-sectional area, the drift velocity in the $14$-gauge wire is larger than the drift velocity in the $12$-gauge wire carrying the same current. The number of electrons per cubic meter will remain constant.

5.4 The current density in a conducting wire increases due to an increase in current. The drift velocity is inversely proportional to the current $\left(v_d=\frac{I}{nqA}\right),$ so the drift velocity would decrease.

5.5 Silver, gold, and aluminum are all used for making wires. All four materials have a high conductivity, silver having the highest. All four can easily be drawn into wires and have a high tensile strength, though not as high as copper. The obvious disadvantage of gold and silver is the cost, but silver and gold wires are used for special applications, such as speaker wires. Gold does not oxidize, making better connections between components. Aluminum wires do have their drawbacks. Aluminum has a higher resistivity than copper, so a larger diameter is needed to match the resistance per length of copper wires, but aluminum is cheaper than copper, so this is not a major drawback. Aluminum wires do not have as high of a ductility and tensile strength as copper, but the ductility and tensile strength is within acceptable levels. There are a few concerns that must be addressed in using aluminum and care must be used when making connections. Aluminum has a higher rate of thermal expansion than copper, which can lead to loose connections and a possible fire hazard. The oxidation of aluminum does not conduct and can cause problems. Special techniques must be used when using aluminum wires and components, such as electrical outlets, must be designed to accept aluminum wires.

5.6 The foil pattern stretches as the backing stretches, and the foil tracks become longer and thinner. Since the resistance is calculated as $R=\rho\frac{L}{A},$ the resistance increases as the foil tracks are stretched. When the temperature changes, so does the resistivity of the foil tracks, changing the resistance. One way to combat this is to use two strain gauges, one used as a reference and the other used to measure the strain. The two strain gauges are kept at a constant temperature

5.7 The longer the length, the smaller the resistance. The greater the resistivity, the higher the resistance. The larger the difference between the outer radius and the inner radius, that is, the greater the ratio between the two, the greater the resistance. If you are attempting to maximize the resistance, the choice of the values for these variables will depend on the application. For example, if the cable must be flexible, the choice of materials may be limited.

5.8 Yes, Ohm’s law is still valid. At every point in time the current is equal to $I(t)=V(t)/R,$ so the current is also a function of time, $I(t)=\frac{V_{\mathrm{max}}}{R}\sin(2\pi ft).$

5.9 Even though electric motors are highly efficient $10-20\%$ of the power consumed is wasted, not being used for doing useful work. Most of the $10-20\%$ of the power lost is transferred into heat dissipated by the copper wires used to make the coils of the motor. This heat adds to the heat of the environment and adds to the demand on power plants providing the power. The demand on the power plant can lead to increased greenhouse gases, particularly if the power plant uses coal or gas as fuel.

5.10 No, the efficiency is a very important consideration of the light bulbs, but there are many other considerations. As mentioned above, the cost of the bulbs and the life span of the bulbs are important considerations. For example, CFL bulbs contain mercury, a neurotoxin, and must be disposed of as hazardous waste. When replacing incandescent bulbs that are being controlled by a dimmer switch with LED, the dimmer switch may need to be replaced. The dimmer switches for LED lights are comparably priced to the incandescent light switches, but this is an initial cost which should be considered. The spectrum of light should also be considered, but there is a broad range of color temperatures available, so you should be able to find one that fits your needs. None of these considerations mentioned are meant to discourage the use of LED or CFL light bulbs, but they are considerations.

1. Can a wire carry a current and still be neutral—that is, have a total charge of zero? Explain.

2. Car batteries are rated in ampere-hours ($\mathrm{kW}\cdot\mathrm{h}$). To what physical quantity do ampere-hours correspond (voltage, current, charge, energy, power,…)?

3. When working with high-power electric circuits, it is advised that whenever possible, you work “one-handed” or “keep one hand in your pocket.” Why is this a sensible suggestion?

4. Incandescent light bulbs are being replaced with more efficient LED and CFL light bulbs. Is there any obvious evidence that incandescent light bulbs might not be that energy efficient? Is energy converted into anything but visible light?

5. It was stated that the motion of an electron appears nearly random when an electrical field is applied to the conductor. What makes the motion nearly random and differentiates it from the random motion of molecules in a gas?

6. Electric circuits are sometimes explained using a conceptual model of water flowing through a pipe. In this conceptual model, the voltage source is represented as a pump that pumps water through pipes and the pipes connect components in the circuit. Is a conceptual model of water flowing through a pipe an adequate representation of the circuit? How are electrons and wires similar to water molecules and pipes? How are they different?

7. An incandescent light bulb is partially evacuated. Why do you suppose that is?

8. The $IR$ drop across a resistor means that there is a change in potential or voltage across the resistor. Is there any change in current as it passes through a resistor? Explain.

9. Do impurities in semiconducting materials listed in Table 5.3.1 supply free charges? (*Hint*: Examine the range of resistivity for each and determine whether the pure semiconductor has the higher or lower conductivity.)

10. Does the resistance of an object depend on the path current takes through it? Consider, for example, a rectangular bar—is its resistance the same along its length as across its width?

11. If aluminum and copper wires of the same length have the same resistance, which has the larger diameter? Why?

12. In Determining Field from Potential, resistance was defined $R\equiv\frac{V}{I}.$ In this section, we presented Ohm’s law, which is commonly expressed as $V=IR.$ The equations look exactly alike. What is the difference between Ohm’s law and the definition of resistance?

13. Shown below are the results of an experiment where four devices were connected across a variable voltage source. The voltage is increased and the current is measured. Which device, if any, is an ohmic device?

14. The current $I$ is measured through a sample of an ohmic material as a voltage $V$ is applied. (a) What is the current when the voltage is doubled to $2V$ (assume the change in temperature of the material is negligible)? (b) What is the voltage applied is the current measured is $0.2I$ (assume the change in temperature of the material is negligible)? What will happen to the current if the material if the voltage remains constant, but the temperature of the material increases significantly?

15. Common household appliances are rated at $110~\mathrm{V},$ but power companies deliver voltage in the kilovolt range and then step the voltage down using transformers to $110~\mathrm{V}$ to be used in homes. You will learn in later chapters that transformers consist of many turns of wire, which warm up as current flows through them, wasting some of the energy that is given off as heat. This sounds inefficient. Why do the power companies transport electric power using this method?

16. Your electric bill gives your consumption in units of kilowatt-hour ($\mathrm{kW}\cdot\mathrm{h}$). Does this unit represent the amount of charge, current, voltage, power, or energy you buy?

17. Resistors are commonly rated at $\frac{1}{8}~\mathrm{W},$ $\frac{1}{4}~\mathrm{W},$ $\frac{1}{2}~\mathrm{W},$ $1~\mathrm{W}$ and $2~\mathrm{W}$ for use in electrical circuits. If a current of $I=2.00~\mathrm{A}$ is accidentally passed through a $R=1.00~\Omega$ resistor rated at $1~\mathrm{W},$ what would be the most probable outcome? Is there anything that can be done to prevent such an accident?

18. An immersion heater is a small appliance used to heat a cup of water for tea by passing current through a resistor. If the voltage applied to the appliance is doubled, will the time required to heat the water change? By how much? Is this a good idea?

19. What requirement for superconductivity makes current superconducting devices expensive to operate?

20. Name two applications for superconductivity listed in this section and explain how superconductivity is used in the application. Can you think of a use for superconductivity that is not listed?

21. A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like protons or electrons. You may have seen it used to make human hair stand on end or produce large sparks. One application of the Van de Graaff generator is to create X-rays by bombarding a hard metal target with the beam. Consider a beam of protons at $1.00~\mathrm{keV}$ and a current of $5.00~\mathrm{mA}$ produced by the generator. (a) What is the speed of the protons? (b) How many protons are produced each second?

22. A cathode ray tube (CRT) is a device that produces a focused beam of electrons in a vacuum. The electrons strike a phosphor-coated glass screen at the end of the tube, which produces a bright spot of light. The position of the bright spot of light on the screen can be adjusted by deflecting the electrons with electrical fields, magnetic fields, or both. Although the CRT tube was once commonly found in televisions, computer displays, and oscilloscopes, newer appliances use a liquid crystal display (LCD) or plasma screen. You still may come across a CRT in your study of science. Consider a CRT with an electron beam average current of $25.00~\mu\mathrm{A}.$ How many electrons strike the screen every minute?

23. How many electrons flow through a point in a wire in $3.00~\mathrm{s}$ if there is a constant current of $I=4.00~\mathrm{A}$?

24. A conductor carries a current that is decreasing exponentially with time. The current is modeled as $I=I_0e^{-t/\tau},$ where $I_0=3.00~\mathrm{A}$ is the current at time $t=0.00~\mathrm{s}$ and $\tau=0.50~\mathrm{s}$ is the time constant. How much charge flows through the conductor between $t=0.00~\mathrm{s}$ and $t=3\tau$?

25. The quantity of charge through a conductor is modelled as $Q=(4.00~\mathrm{C/s}^4)t^4-(1.00~\mathrm{C/s})t+6.00~\mathrm{mC}.$ What is the current at time $t=3.00~\mathrm{s}$?

26. The current through a conductor is modelled as $I(t)=I_m\sin[2\pi(60~\mathrm{Hz})t].$ Write an equation for the charge as a function of time.

27. The charge on a capacitor in a circuit is modelled as $Q(t)=Q_{\mathrm{max}}\cos(\omega t+\phi).$ What is the current through the circuit as a function of time?

28. An aluminum wire $1.628~\mathrm{mm}$ in diameter ($14$-gauge) carries a current of $3.00~\mathrm{A}.$ (a) What is the absolute value of the charge density in the wire? (b) What is the drift velocity of the electrons? (c) What would be the drift velocity if the same gauge copper were used instead of aluminum? The density of copper is $8.96~\mathrm{g/cm}^3$ and the density of aluminum is $2.70~\mathrm{g/cm}^3$ The molar mass of aluminum is $26.98~\mathrm{g/mol}$ and the molar mass of copper is $63.5~\mathrm{g/mol}.$ Assume each atom of metal contributes one free electron.

29. The current of an electron beam has a measured current of $I=50.00~\mu\mathrm{A}$ with a radius of $1.00~\mathrm{mm}^2.$ What is the magnitude of the current density of the beam?

30. A high-energy proton accelerator produces a proton beam with a radius of $r=0.90~\mathrm{mm}.$ The beam current is $i=9.00~\mu\mathrm{A}$ and is constant. The charge density of the beam is $n=6.00\times10^{11}$ protons per cubic meter. (a) What is the current density of the beam? (b) What is the drift velocity of the beam? (c) How much time does it take for $1.00\times10^{10}$ protons to be emitted by the accelerator?

31. Consider a wire of a circular cross-section with a radius of $R=3.00~\mathrm{mm}.$ The magnitude of the current density is modelled as $J=cr^2=\left(5.00\times10^6~\mathrm{A}{m^4}\right)r^2.$ What is the current through the inner section of the wire from the centre to $r=0.5R$?

32. The current of an electron beam has a measured current of $I=50.00~\mu\mathrm{A}$ with a radius of $1.00~\mathrm{mm}^2.$ What is the magnitude of the current density of the beam?

33. The current supplied to an air conditioner unit is $4.00$ amps. The air conditioner is wired using a $10$-gauge (diameter $2.588~\mathrm{mm}$) wire. The charge density is $n=8.48\times10^{28}~\mathrm{electrons}{m^3}.$ Find the magnitude of (a) current density and (b) the drift velocity.

34. What current flows through the bulb of a $3.00{\text -}\mathrm{V}$ flashlight when its hot resistance is $3.60~\Omega$?

35. Calculate the effective resistance of a pocket calculator that has a $1.35{\text -}\mathrm{V}$ battery and through which $0.200~\mathrm{mA}$ flows.

36. How many volts are supplied to operate an indicator light on a DVD player that has a resistance of $140~\Omega,$ given that $25.0~\mathrm{mA}$ passes through it?

37. What is the resistance of a $20.0{\text -}\mathrm{m}$-long piece of $12$-gauge copper wire having a $2.053{\text -}\mathrm{mm}$ diameter?

38. The diameter of $0$-gauge copper wire is $8.252~\mathrm{mm}.$ Find the resistance of a $1.00{\text -}\mathrm{km}$ length of such wire used for power transmission.

39. If the $0.100{\text -}\mathrm{mm}$-diameter tungsten filament in a light bulb is to have a resistance of $0.200~\Omega$ at $20.0^{\circ}\mathrm{C},$ how long should it be?

40. A lead rod has a length of $30.00~\mathrm{cm}$ and a resistance of $5.00~\mu\Omega.$ What is the radius of the rod?

41. Find the ratio of the diameter of aluminum to copper wire, if they have the same resistance per unit length (as they might in household wiring).

42. What current flows through a $2.54{\text -}\mathrm{cm}$-diameter rod of pure silicon that is $20.0~\mathrm{cm}$ long, when $1.00\times10^3~\mathrm{V}$ is applied to it? (Such a rod may be used to make nuclear-particle detectors, for example.)

43. (a) To what temperature must you raise a copper wire, originally at $20.0~^{\circ}\mathrm{C},$ to double its resistance, neglecting any changes in dimensions? (b) Does this happen in household wiring under ordinary circumstances?

44. A resistor made of nichrome wire is used in an application where its resistance cannot change more than $1.00\%$ from its value at $20.0~^{\circ}\mathrm{C}.$ Over what temperature range can it be used?

45. Of what material is a resistor made if its resistance is $40.0\%$ greater at $100.0~^{\circ}\mathrm{C}$ than at $20.0~^{\circ}\mathrm{C}$?

46. An electronic device designed to operate at any temperature in the range from $-10.0~^{\circ}\mathrm{C}$ to $55.0~^{\circ}\mathrm{C}$ contains pure carbon resistors. By what factor does their resistance increase over this range?

47. (a) Of what material is a wire made, if it is $25.0~\mathrm{m}$ long with a diameter of $0.100~\mathrm{mm}$ and has a resistance of $77.7~\Omega$ at $20.0~^{\circ}\mathrm{C}$? (b) What is its resistance at $150.0~^{\circ}\mathrm{C}$?

48. Assuming a constant temperature coefficient of resistivity, what is the maximum percent decrease in the resistance of a constantan wire starting at $20.0~^{\circ}\mathrm{C}$?

49. A copper wire has a resistance of $0.500~\Omega$ at $20.0~^{\circ}\mathrm{C},$ and an iron wire has a resistance of $0.525~\Omega$ at the same temperature. At what temperature are their resistances equal?

50. A $2.2{\text -}\mathrm{k}\Omega$ resistor is connected across a D cell battery ($1.5~\mathrm{V}$). What is the current through the resistor?

51. A resistor rated at $250~\mathrm{k}\Omega$ is connected across two D cell batteries (each $1.50~\mathrm{V}$) in series, with a total voltage of $3.00~\mathrm{V}.$ The manufacturer advertises that their resistors are within $5\%$ of the rated value. What are the possible minimum current and maximum current through the resistor?

52. A resistor is connected in series with a power supply of $20.00~\mathrm{V}.$ The current measure is $0.50~\mathrm{A}.$ What is the resistance of the resistor?

53. A resistor is placed in a circuit with an adjustable voltage source. The voltage across and the current through the resistor and the measurements are shown below. Estimate the resistance of the resistor.

54. The following table show the measurements of a current through and the voltage across a sample of material. Plot the data, and assuming the object is an ohmic device, estimate the resistance.

$I~(\mathrm{A})$ | $V~(\mathrm{V})$ |
---|---|

$0$ | $3$ |

$2$ | $23$ |

$4$ | $39$ |

$6$ | $58$ |

$8$ | $77$ |

$10$ | $100$ |

$12$ | $119$ |

$14$ | $142$ |

$16$ | $162$ |

55. A $20.0{\text -}\mathrm{V}$ battery is used to supply current to a $10{\text -}\mathrm{k}\Omega$ resistor. Assume the voltage drop across any wires used for connections is negligible. (a) What is the current through the resistor? (b) What is the power dissipated by the resistor? (c) What is the power input from the battery, assuming all the electrical power is dissipated by the resistor? (d) What happens to the energy dissipated by the resistor?

56. What is the maximum voltage that can be applied to a $10{\text -}\mathrm{k}\Omega$ resistor rated at $14~\mathrm{W}$?

57. A heater is being designed that uses a coil of $14$-gauge nichrome wire to generate $300~\mathrm{W}$ using a voltage of $V=110~\mathrm{V}.$ How long should the engineer make the wire?

58. An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A $100{\text -}\mathrm{W}$ incandescent bulb can be replaced by a $16{\text -}\mathrm{W}$ LED bulb. Both produce $1600$ lumens of light. Assuming the cost of electricity is $\$0.10$ per kilowatt-hour, how much does it cost to run the bulb for one year if it runs for four hours a day?

59. The power dissipated by a resistor with a resistance of $R=100~\Omega$ is $P=2.0~\mathrm{W}.$ What are the current through and the voltage drop across the resistor?

60. Running late to catch a plane, a driver accidentally leaves the headlights on after parking the car in the airport parking lot. During takeoff, the driver realizes the mistake. Having just replaced the battery, the driver knows that the battery is a $12{\text -}\mathrm{V}$ automobile battery, rated at $100~\mathrm{A}\cdot\mathrm{h}.$ The driver, knowing there is nothing that can be done, estimates how long the lights will shine, assuming there are two $12{\text -}\mathrm{V}$ headlights, each rated at $40~\mathrm{W}.$ What did the driver conclude?

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of $3.00~\mathrm{A}$ and a voltage of $110~\mathrm{V},$ a lamp that contains a $100{\text -}\mathrm{W}$ bulb, an overhead light with a $60{\text -}\mathrm{W}$ bulb, and various other small devices adding up to $3.00~\mathrm{W}.$ (a) Assuming the power plant that supplies $110~\mathrm{V}$ electricity to the dorm is $10~\mathrm{km}$ away and the two aluminum transmission cables use $0$-gauge wire with a diameter of $8.252~\mathrm{mm},$ estimate the percentage of the total power supplied by the power company that is lost in the transmission. (b) What would be the result is the power company delivered the electric power at $110~\mathrm{kV}$?

62. A $0.50{\text -}\mathrm{W},$ $220{\text -}\Omega$ resistor carries the maximum current possible without damaging the resistor. If the current were reduced to half the value, what would be the power consumed?

63. Consider a power plant is located $60~\mathrm{km}$ away from a residential area uses $0$-gauge ($A=42.40~\mathrm{mm}^2$) wire of copper to transmit power at a current of $I=100.00~\mathrm{A}.$ How much more power is dissipated in the copper wires than it would be in superconducting wires?

64. A wire is drawn through a die, stretching it to four times its original length. By what factor does its resistance increase?

65. Digital medical thermometers determine temperature by measuring the resistance of a semiconductor device called a thermistor (which has $\alpha=-0.06/^{\circ}\mathrm{C}$) when it is at the same temperature as the patient. What is a patient’s temperature if the thermistor’s resistance at that temperature is $82.0\%$ of its value at $37.0~^{\circ}\mathrm{C}$ (normal body temperature)?

66. Electrical power generators are sometimes “load tested” by passing current through a large vat of water. A similar method can be used to test the heat output of a resistor. A $R=30~\Omega$ resistor is connected to a $9.0{\text -}\mathrm{V}$ battery and the resistor leads are waterproofed and the resistor is placed in $1.0~\mathrm{kg}$ of room temperature water $T=20~^{\circ}\mathrm{C}.$ Current runs through the resistor for $20$ minutes. Assuming all the electrical energy dissipated by the resistor is converted to heat, what is the final temperature of the water?

67. A $12$-gauge gold wire has a length of $1$ meter. (a) What would be the length of a silver $12$-gauge wire with the same resistance? (b) What are their respective resistances at the temperature of boiling water?

68. What is the change in temperature required to decrease the resistance for a carbon resistor by $10\%$?

69. A coaxial cable consists of an inner conductor with radius $r_i=0.25~\mathrm{cm}$ and an outer radius of $r_o=0.5~\mathrm{cm}$ and has a length of $10$ metres. Plastic, with a resistivity of $\rho=2.00\times10^{13}~\Omega\cdot\mathrm{m},$ separates the two conductors. What is the resistance of the cable?

70. A $10.00$-metre long wire cable that is made of copper has a resistance of $0.051$ ohms. (a) What is the weight if the wire was made of copper? (b) What is the weight of a $10.00$-metre-long wire of the same gauge made of aluminum? (c)What is the resistance of the aluminum wire? The density of copper is $8960~\mathrm{kg/m}^3$ and the density of aluminum is $2760~\mathrm{kg/m}^3.$

71. A nichrome rod that is $3.00~\mathrm{mm}$ long with a cross-sectional area of $1.00~\mathrm{mm}^2$ is used for a digital thermometer. (a) What is the resistance at room temperature? (b) What is the resistance at body temperature?

72. The temperature in Philadelphia, PA can vary between $68.00~^{\circ}\mathrm{F}$ and $100.00~^{\circ}\mathrm{F}$ in one summer day. By what percentage will an aluminum wire’s resistance change during the day?

73. When $100.0~\mathrm{V}$ is applied across a $5$-gauge (diameter $4.621~\mathrm{mm}$) wire that is $10~\mathrm{m}$ long, the magnitude of the current density is $2.0\times10^8~\mathrm{A/m}^2.$ What is the resistivity of the wire?

74. A wire with a resistance of $5.0~\Omega$ is drawn out through a die so that its new length is twice times its original length. Find the resistance of the longer wire. You may assume that the resistivity and density of the material are unchanged.

75. What is the resistivity of a wire of $5$-gauge wire ($A=16.8\times10^{-6}~\mathrm{m}^2$), $5.00~\mathrm{m}$ length, and $5.10~\mathrm{m}\Omega$ resistance?

76. Coils are often used in electrical and electronic circuits. Consider a coil which is formed by winding $1000$ turns of insulated $20$-gauge copper wire (area $0.52~\mathrm{mm}^2$) in a single layer on a cylindrical non-conducting core of radius $2.0~\mathrm{mm}.$ What is the resistance of the coil? Neglect the thickness of the insulation.

77. Currents of approximately $0.06~\mathrm{A}$ can be potentially fatal. Currents in that range can make the heart fibrillate (beat in an uncontrolled manner). The resistance of a dry human body can be approximately $100~\mathrm{k}\Omega.$ (a) What voltage can cause $0.2~\mathrm{A}$ through a dry human body? (b) When a human body is wet, the resistance can fall to $100~\Omega.$ What voltage can cause harm to a wet body?

78. A $20.00$-ohm, $5.00$-watt resistor is placed in series with a power supply. (a) What is the maximum voltage that can be applied to the resistor without harming the resistor? (b) What would be the current through the resistor?

79. A battery with an emf of $24.00~\mathrm{V}$ delivers a constant current of $2.00~\mathrm{mA}$ to an appliance. How much work does the battery do in three minutes?

80. A $12.00{\text -}\mathrm{V}$ battery has an internal resistance of a tenth of an ohm. (a) What is the current if the battery terminals are momentarily shorted together? (b) What is the terminal voltage if the battery delivers $0.25$ amps to a circuit?

81. A $10$-gauge copper wire has a cross-sectional area $A=5.26~\mathrm{mm}^2$ and carries a current of $I=5.00~\mathrm{A}.$ The density of copper is $89.50~\mathrm{g/cm^}3.$ One mole of copper atoms ($6.02\times10^{23}$ atoms) has a mass of approximately $63.50~\mathrm{g}.$ What is the magnitude of the drift velocity of the electrons, assuming that each copper atom contributes one free electron to the current?

82. The current through a $12$-gauge wire is given as $I(t)=(5.00~\mathrm{A})\sin[2\pi(60~\mathrm{Hz})t].$ What is the current density at time $15.00~\mathrm{ms}$?

83. A particle accelerator produces a beam with a radius of $1.25~\mathrm{mm}$ with a current of $2.00~\mathrm{mA}.$ Each proton has a kinetic energy of $10.00~\mathrm{MeV}.$ (a) What is the velocity of the protons? (b) What is the number ($n$) of protons per unit volume? (b) How many electrons pass a cross sectional area each second?

84. In this chapter, most examples and problems involved direct current (DC). DC circuits have the current flowing in one direction, from positive to negative. When the current was changing, it was changed linearly from $I=-I_{\mathrm{max}}$ to $I=+I_{\mathrm{max}}$ and the voltage changed linearly from $V=-V_{\mathrm{max}}$ to $V=+V_{\mathrm{max}},$ where $V_{\mathrm{max}}=I_{\mathrm{max}}R.$ Suppose a voltage source is placed in series with a resistor of $R=10~\Omega$ that supplied a current that alternated as a sine wave, for example, $I(t)=(3.00~\mathrm{A})\sin\left(\frac{2\pi}{4.00~\mathrm{s}}t\right).$ (a) What would a graph of the voltage drop across the resistor $V(t)$ versus time look like? (b) What would a plot of $V(t)$ versus $I(t)$ for one period look like? (*Hint*: If you are not sure, try plotting $V(t)$ versus $I(t)$ using a spreadsheet.)

85. A current of $I=25~\mathrm{A}$ is drawn from a $100{\text -}\mathrm{V}$ battery for $30$ seconds. By how much is the chemical energy reduced?

86. Consider a square rod of material with sides of length $L=3.00~\mathrm{cm}$ with a current density of $\vec{\mathbf{J}}=J_0e^{\alpha x}\hat{\mathbf{k}}=(0.35~\mathrm{A/m}^2)e^{(2.1\times10^{-3}~\mathrm{m}^{-1})x}\hat{\mathbf{k}}$ as shown below. Find the current that passes through the face of the rod.

87. A resistor of an unknown resistance is placed in an insulated container filled with $0.75~\mathrm{kg}$ of water. A voltage source is connected in series with the resistor and a current of $1.2$ amps flows through the resistor for $10$ minutes. During this time, the temperature of the water is measured and the temperature change during this time is $\Delta T=10.00~^{\circ}\mathrm{C}.$ (a) What is the resistance of the resistor? (b) What is the voltage supplied by the power supply?

88. The charge that flows through a point in a wire as a function of time is modelled as $q(t)=q_0e^{-t/T}=(10.0~\mathrm{C})e^{-t/(5~\mathrm{s})}.$ (a) What is the initial current through the wire at time $t=0.00~\mathrm{s}$? (b) Find the current at time $t=\frac{1}{2}T.$ (c) At what time $t$ will the current be reduced by one-half $I=\frac{1}{2}I_0$?

89. Consider a resistor made from a hollow cylinder of carbon as shown below. The inner radius of the cylinder is $R_i=0.20~\mathrm{mm}$ and the outer radius is $R_o=0.30~\mathrm{mm}.$ The length of the resistor is $L=0.90~\mathrm{mm}.$ The resistivity of the carbon is $\rho=3.5\times10^{-5}~\Omega\cdot\mathrm{m}.$ (a) Prove that the resistance perpendicular from the axis is $R=\frac{\rho}{2\pi L}\ln\left(\frac{R_o}{R_i}\right).$ (b) What is the resistance?

90. What is the current through a cylindrical wire of radius $R=0.1~\mathrm{mm}$ if the current density is $J=\frac{J_0}{R}r,$ where $J_0=32000~\mathrm{A/m}^2$?

91. A student uses a $100.00{\text -}\mathrm{W},$ $115.00{\text -}\mathrm{V}$ radiant heater to heat the student’s dorm room, during the hours between sunset and sunrise, $\mathrm{6:00~p.m.}$ to $\mathrm{7:00~a.m.}$ (a) What current does the heater operate at? (b) How many electrons move through the heater? (c) What is the resistance of the heater? (d) How much heat was added to the dorm room?

92. A $12{\text -}\mathrm{V}$ car battery is used to power a $20.00{\text -}\mathrm{W},$ $12.00{\text -}\mathrm{V}$ lamp during the physics club camping trip/star party. The cable to the lamp is $2.00$ metres long, $14$-gauge copper wire with a charge density of $n=9.50\times10^{28}~\mathrm{m}^{-3}.$ (a) What is the current draw by the lamp? (b) How long would it take an electron to get from the battery to the lamp?

93. A physics student uses a $115.00{\text -}\mathrm{V}$ immersion heater to heat $400.00$ grams (almost two cups) of water for herbal tea. During the two minutes it takes the water to heat, the physics student becomes bored and decides to figure out the resistance of the heater. The student starts with the assumption that the water is initially at the temperature of the room $T_i=25.00~^{\circ}\mathrm{C}$ and reaches $T_f=100.00~^{\circ}\mathrm{C}.$ The specific heat of the water is $c=4180~\mathrm{J/kg}.$ What is the resistance of the heater?

- Describe the electromotive force (emf) and the internal resistance of a battery
- Explain the basic operation of a battery

If you forget to turn off your car lights, they slowly dim as the battery runs down. Why don’t they suddenly blink off when the battery’s energy is gone? Their gradual dimming implies that the battery output voltage decreases as the battery is depleted. The reason for the decrease in output voltage for depleted batteries is that all voltage sources have two fundamental parts—a source of electrical energy and an internal resistance. In this section, we examine the energy source and the internal resistance.

Voltage has many sources, a few of which are shown in \ref{fig:6.1.1}. All such devices create a **potential difference** and can supply current if connected to a circuit. A special type of potential difference is known as **electromotive force** (emf). The emf is not a force at all, but the term ‘electromotive force’ is used for historical reasons. It was coined by Alessandro Volta in the 1800s, when he invented the first battery, also known as the **voltaic pile**. Because the electromotive force is not a force, it is common to refer to these sources simply as sources of emf (pronounced as the letters “ee-em-eff”), instead of sources of electromotive force.

If the electromotive force is not a force at all, then what is the emf and what is a source of emf? To answer these questions, consider a simple circuit of a $12{\text -}\mathrm{V}$ lamp attached to a $12{\text -}\mathrm{V}$ battery, as shown in \ref{fig:6.1.2}. The **battery** can be modeled as a two-terminal device that keeps one terminal at a higher electric potential than the second terminal. The higher electric potential is sometimes called the positive terminal and is labeled with a plus sign. The lower-potential terminal is sometimes called the negative terminal and labeled with a minus sign. This is the source of the emf.

When the emf source is not connected to the lamp, there is no net flow of charge within the emf source. Once the battery is connected to the lamp, charges flow from one terminal of the battery, through the lamp (causing the lamp to light), and back to the other terminal of the battery. If we consider positive (conventional) current flow, positive charges leave the positive terminal, travel through the lamp, and enter the negative terminal.

Positive current flow is useful for most of the circuit analysis in this chapter, but in metallic wires and resistors, electrons contribute the most to current, flowing in the opposite direction of positive current flow. Therefore, it is more realistic to consider the movement of electrons for the analysis of the circuit in \ref{fig:6.1.2}. The electrons leave the negative terminal, travel through the lamp, and return to the positive terminal. In order for the emf source to maintain the potential difference between the two terminals, negative charges (electrons) must be moved from the positive terminal to the negative terminal. The emf source acts as a charge pump, moving negative charges from the positive terminal to the negative terminal to maintain the potential difference. This increases the potential energy of the charges and, therefore, the electric potential of the charges.

The force on the negative charge from the electric field is in the opposite direction of the electric field, as shown in \ref{fig:6.1.2}. In order for the negative charges to be moved to the negative terminal, work must be done on the negative charges. This requires energy, which comes from chemical reactions in the battery. The potential is kept high on the positive terminal and low on the negative terminal to maintain the potential difference between the two terminals. The emf is equal to the work done on the charge per unit charge ($\mathcal{E}=dW/dq$) when there is no current flowing. Since the unit for work is the joule and the unit for charge is the coulomb, the unit for emf is the volt ($1~\mathrm{V}=1~\mathrm{J/C}$).

The **terminal voltage** $V_{\mathrm{terminal}}$ of a battery is voltage measured across the terminals of the battery when there is no load connected to the terminal. An ideal battery is an emf source that maintains a constant terminal voltage, independent of the current between the two terminals. An ideal battery has no internal resistance, and the terminal voltage is equal to the emf of the battery. In the next section, we will show that a real battery does have internal resistance and the terminal voltage is always less than the emf of the battery.

The combination of chemicals and the makeup of the terminals in a battery determine its emf. The **lead acid battery** used in cars and other vehicles is one of the most common combinations of chemicals. \ref{fig:6.1.3} shows a single cell (one of six) of this battery. The cathode (positive) terminal of the cell is connected to a lead oxide plate, whereas the anode (negative) terminal is connected to a lead plate. Both plates are immersed in sulfuric acid, the electrolyte for the system.

Knowing a little about how the chemicals in a lead-acid battery interact helps in understanding the potential created by the battery. \ref{fig:6.1.4} shows the result of a single chemical reaction. Two electrons are placed on the **anode**, making it negative, provided that the cathode supplies two electrons. This leaves the **cathode** positively charged, because it has lost two electrons. In short, a separation of charge has been driven by a chemical reaction.

Note that the reaction does not take place unless there is a complete circuit to allow two electrons to be supplied to the cathode. Under many circumstances, these electrons come from the anode, flow through a resistance, and return to the cathode. Note also that since the chemical reactions involve substances with resistance, it is not possible to create the emf without an internal resistance.

\begin{gather}.\tag{Figure 6.1.4}\label{fig:6.1.4}\end{gather}The amount of resistance to the flow of current within the voltage source is called the **internal resistance**. The internal resistance $r$ of a battery can behave in complex ways. It generally increases as a battery is depleted, due to the oxidation of the plates or the reduction of the acidity of the electrolyte. However, internal resistance may also depend on the magnitude and direction of the current through a voltage source, its temperature, and even its history. The internal resistance of rechargeable nickel-cadmium cells, for example, depends on how many times and how deeply they have been depleted. A simple model for a battery consists of an idealized emf source $\mathcal{E}$ and an internal resistance $r$ (\ref{fig:6.1.5}).

Suppose an external resistor, known as the load resistance $R$, is connected to a voltage source such as a battery, as in \ref{fig:6.1.6}. The figure shows a model of a battery with an emf $\mathcal{E}$, an internal resistance $r$, and a load resistor $R$ connected across its terminals. Using conventional current flow, positive charges leave the positive terminal of the battery, travel through the resistor, and return to the negative terminal of the battery. The terminal voltage of the battery depends on the emf, the internal resistance, and the current, and is equal to

\begin{equation}V_{\mathrm{terminal}}=\mathcal{E}-Ir\tag{6.1.1}\label{eq:6.1.1}\end{equation}

For a given emf and internal resistance, the terminal voltage decreases as the current increases due to the potential drop $Ir$ of the internal resistance.

\begin{gather}.\tag{Figure 6.1.6}\label{fig:6.1.6}\end{gather}A graph of the potential difference across each element the circuit is shown in \ref{fig:6.1.7}. A current $I$ runs through the circuit, and the potential drop across the internal resistor is equal to $Ir$. The terminal voltage is equal to $\mathcal{E}-Ir$, which is equal to the **potential drop **across the load resistor $IR=\mathcal{E}-Ir$. As with potential energy, it is the change in voltage that is important. When the term “voltage” is used, we assume that it is actually the change in the potential, or $\Delta V$. However, $\Delta$ is often omitted for convenience.

The current through the load resistor is $I=\frac{\mathcal{E}}{r+R}$. We see from this expression that the smaller the internal resistance $r$, the greater the current the voltage source supplies to its load $R$. As batteries are depleted, $r$ increases. If $r$ becomes a significant fraction of the load resistance, then the current is significantly reduced, as the following example illustrates.

Enter the known values into the equation $V_{\mathrm{terminal}}=\mathcal{E}-Ir$ to get the terminal voltage:

\[V_{\mathrm{terminal}}=\mathcal{E}-Ir=12.00~\mathrm{V}-(1.188~\mathrm{A})(0.100~\Omega)=11.90~\mathrm{V}.\]The terminal voltage here is only slightly lower than the emf, implying that the current drawn by this light load is not significant.

b. Similarly, with $R_{\mathrm{load}}=0.500~\Omega$, the current is \[I=\frac{\mathcal{E}}{R+r}=\frac{12.00~\mathrm{V}}{0.600~\Omega}=20.00~\mathrm{A}.\]The terminal voltage is now

\[V_{\mathrm{terminal}}=\mathcal{E}-Ir=12.00~\mathrm{V}-(20.00~\mathrm{A})(0.100~\Omega)=10.00~\mathrm{V}.\]The terminal voltage exhibits a more significant reduction compared with emf, implying $0.500~\Omega$ is a heavy load for this battery. A “heavy load” signifies a larger draw of current from the source but not a larger resistance.

c. The power dissipated by the $0.500{\text -}\Omega$ load can be found using the formula $P=I^2R$. Entering the known values gives \[P=I^2R=(20.0~\mathrm{A})^2(0.500~\Omega)=2.00\times10^2~\mathrm{W}.\]Note that this power can also be obtained using the expression $\frac{V^2}{R}$ or $IV$, where $V$ is the terminal voltage ($10.0~\mathrm{V}$ in this case).

d. Here, the internal resistance has increased, perhaps due to the depletion of the battery, to the point where it is as great as the load resistance. As before, we first find the current by entering the known values into the expression, yielding \[I=\frac{\mathcal{E}}{R+r}=\frac{12.00~\mathrm{V}}{1.00~\Omega}=12.0~\mathrm{A}.\]Now the terminal voltage is

\[V_{\mathrm{terminal}}=\mathcal{E}-Ir=12.00~\mathrm{V}-(12.00~\mathrm{A})(0.500~\Omega)=6.00~\mathrm{V},\]and the power dissipated by the load is

\[P=I^2R=(12.00~\mathrm{A})^2(0.500~\Omega)=72.0\times10^2~\mathrm{W}.\]We see that the increased internal resistance has significantly decreased the terminal voltage, current, and power delivered to a load.

If you place a wire directly across the two terminal of a battery, effectively shorting out the terminals, the battery will begin to get hot. Why do you suppose this happens?

**Battery testers**, such as those in \ref{fig:6.1.8}, use small load resistors to intentionally draw current to determine whether the terminal potential drops below an acceptable level. Although it is difficult to measure the internal resistance of a battery, battery testers can provide a measurement of the internal resistance of the battery. If internal resistance is high, the battery is weak, as evidenced by its low terminal voltage.

Some batteries can be recharged by passing a current through them in the direction opposite to the current they supply to an appliance. This is done routinely in cars and in batteries for small electrical appliances and electronic devices (\ref{fig:6.1.9}). The voltage output of the battery charger must be greater than the emf of the battery to reverse the current through it. This causes the terminal voltage of the battery to be greater than the emf, since $V=\mathcal{E}-Ir$ and $I$ is now negative.

\begin{gather}.\tag{Figure 6.1.9}\label{fig:6.1.9}\end{gather}It is important to understand the consequences of the internal resistance of emf sources, such as batteries and solar cells, but often, the analysis of circuits is done with the terminal voltage of the battery, as we have done in the previous sections. The terminal voltage is referred to as simply as $V$, dropping the subscript “terminal.” This is because the internal resistance of the battery is difficult to measure directly and can change over time.

- Define the term equivalent resistance
- Calculate the equivalent resistance of resistors connected in series
- Calculate the equivalent resistance of resistors connected in parallel

In Current and Resistance, we described the term ‘resistance’ and explained the basic design of a resistor. Basically, a resistor limits the flow of charge in a circuit and is an ohmic device where $V=IR$. Most circuits have more than one resistor. If several resistors are connected together and connected to a battery, the current supplied by the battery depends on the **equivalent resistance** of the circuit.

The equivalent resistance of a combination of resistors depends on both their individual values and how they are connected. The simplest combinations of resistors are series and parallel connections (\ref{fig:6.2.1}). In a **series circuit**, the output current of the first resistor flows into the input of the second resistor; therefore, the current is the same in each resistor. In a **parallel circuit**, all of the resistor leads on one side of the resistors are connected together and all the leads on the other side are connected together. In the case of a parallel configuration, each resistor has the same potential drop across it, and the currents through each resistor may be different, depending on the resistor. The sum of the individual currents equals the current that flows into the parallel connections.

Resistors are said to be in series whenever the current flows through the resistors sequentially. Consider \ref{fig:6.2.2}, which shows three resistors in series with an applied voltage equal to $V_{ab}$. Since there is only one path for the charges to flow through, the current is the same through each resistor. The equivalent resistance of a set of resistors in a series connection is equal to the algebraic sum of the individual resistances.

\begin{gather}.\tag{Figure 6.2.2}\label{fig:6.2.2}\end{gather}In \ref{fig:6.2.2}, the current coming from the voltage source flows through each resistor, so the current through each resistor is the same. The current through the circuit depends on the voltage supplied by the voltage source and the resistance of the resistors. For each resistor, a potential drop occurs that is equal to the loss of electric potential energy as a current travels through each resistor. According to Ohm’s law, the potential drop $V$ across a resistor when a current flows through it is calculated using the equation $V=IR$, where $I$ is the current in amps ($\mathrm{A}$) and $R$ is the resistance in ohms ($\Omega$). Since energy is conserved, and the voltage is equal to the potential energy per charge, the sum of the voltage applied to the circuit by the source and the potential drops across the individual resistors around a loop should be equal to zero:

\[\sum_{i=1}^{N}V_i=0.\] This equation is often referred to as Kirchhoff’s loop law, which we will look at in more detail later in this chapter. For \ref{fig:6.2.2}, the sum of the potential drop of each resistor and the voltage supplied by the voltage source should equal zero: \begin{eqnarray*}V-V_1-V_2-V_3&=&0,\\\Rightarrow V&=&V_1+V_2+V_3\\&=&IR_1+IR_2+IR_3,\\\Rightarrow I&=&\frac{V}{R_1+R_2+R_3}=\frac{V}{R_{\mathrm{eq}}}.\end{eqnarray*} Since the current through each component is the same, the equality can be simplified to an equivalent resistance, which is just the sum of the resistances of the individual resistors.Any number of resistors can be connected in series. If $N$ resistors are connected in series, the equivalent resistance is

\begin{equation}R_{\mathrm{eq}}=R_1+R_2+R_3+\ldots+R_{N-1}+R_N=\sum_{i=1}^NR_i.\tag{6.2.1}\label{eq:6.2.1}\end{equation}

One result of components connected in a series circuit is that if something happens to one component, it affects all the other components. For example, if several lamps are connected in series and one bulb burns out, all the other lamps go dark.

Note that the sum of the potential drops across each resistor is equal to the voltage supplied by the battery.

d. The power dissipated by a resistor is equal to $P=I^2R$, and the power supplied by the battery is equal to $P=I\mathcal{E}$: \[P_1=P_2=P_3=P_4=(0.1~\mathrm{A})^2(20~\Omega)=0.2~\mathrm{W},\] \[P_5=(0.1~\mathrm{A})^2(10~\Omega)=0.1~\mathrm{W},\] \[P_{\mathrm{dissipated}}=0.2~\mathrm{W}+0.2~\mathrm{W}+0.2~\mathrm{W}+0.2~\mathrm{W}+0.1~\mathrm{W}=0.9~\mathrm{W},\] \[P_{\mathrm{source}}=I\mathcal{E}=(0.1~\mathrm{A})(9~\mathrm{V})=0.9~\mathrm{W}.\]Some strings of miniature holiday lights are made to short out when a bulb burns out. The device that causes the short is called a shunt, which allows current to flow around the open circuit. A “short” is like putting a piece of wire across the component. The bulbs are usually grouped in series of nine bulbs. If too many bulbs burn out, the shunts eventually open. What causes this?

Let’s briefly summarize the major features of resistors in series:

- Series resistances add together to get the equivalent resistance: \[R_{\mathrm{eq}}=R_1+R_2+R_3+\ldots+R_{N-1}+R_N=\sum_{i=1}^NR_i.\]
- The same current flows through each resistor in series.
- Individual resistors in series do not get the total source voltage, but divide it. The total potential drop across a series configuration of resistors is equal to the sum of the potential drops across each resistor.

\ref{fig:6.2.4}shows resistors in parallel, wired to a voltage source. Resistors are in parallel when one end of all the resistors are connected by a continuous wire of negligible resistance and the other end of all the resistors are also connected to one another through a continuous wire of negligible resistance. The potential drop across each resistor is the same. Current through each resistor can be found using Ohm’s law $I=V/R$, where the voltage is constant across each resistor. For example, an automobile’s headlights, radio, and other systems are wired in parallel, so that each subsystem utilizes the full voltage of the source and can operate completely independently. The same is true of the wiring in your house or any building.

\begin{gather}.\tag{Figure 6.2.4}\label{fig:6.2.4}\end{gather}The current flowing from the voltage source in \ref{fig:6.2.4} depends on the voltage supplied by the voltage source and the equivalent resistance of the circuit. In this case, the current flows from the voltage source and enters a junction, or node, where the circuit splits flowing through resistors $R_1$ and $R_2$. As the charges flow from the battery, some go through resistor $R_1$ and some flow through resistor $R_2$. The sum of the currents flowing into a junction must be equal to the sum of the currents flowing out of the junction:

\[\sum I_{\mathrm{in}}=\sum I_{\mathrm{out}}.\]This equation is referred to as Kirchhoff’s junction rule and will be discussed in detail in the next section. In \ref{fig:6.2.4}, the junction rule gives $I=I_1+I_2$. There are two loops in this circuit, which leads to the equations $V=I_1R_1$ and $I_1R_1=I_2R_2$ Note the voltage across the resistors in parallel are the same ($V=V_1=V_2$) and the current is additive:

\begin{eqnarray*}I&=&I_1+I_2\\&=&\frac{V_1}{R_1}+\frac{V_2}{R_2}\\&=&\frac{V}{R_1}+\frac{V}{R_2}\\&=&V\left(\frac{1}{R_1}+\frac{1}{R_2}\right)=\frac{V}{R_{\mathrm{eq}}}\\\Rightarrow R_{\mathrm{eq}}&=&\left(\frac{1}{R_1}+\frac{1}{R_2}\right)^{-1}.\end{eqnarray*}

Generalizing to any number of $N$ resistors, the equivalent resistance $R_{\mathrm{eq}}$ of a parallel connection is related to the individual resistances by

\begin{equation}R_{\mathrm{eq}}&=&\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\ldots+\frac{1}{R_{N-1}}+\frac{1}{R_N}\right)^{-1}=\left(\sum_{i=1}^N\frac{1}{R_i}\right)^{-1}.\tag{6.2.2}\label{eq:6.2.2}\end{equation}

This relationship results in an equivalent resistance $R_{\mathrm{eq}}$ that is less than the smallest of the individual resistances. When resistors are connected in parallel, more current flows from the source than would flow for any of them individually, so the total resistance is lower.