19.2

The Electric Potential Difference

Since the electric force is , the work that it does as the charge moves from A to B in Figure 19-2 depends on the charge q0. It is useful, therefore, to express this work on a per-unit-charge basis, by dividing both sides of Equation 19.1 by the charge:

| (19.2)

Notice that the right-hand side of this equation is the difference between two terms, each of which is an electric potential energy divided by the test charge, EPE/q0. The quantity EPE/q0 is the electric potential energy per unit charge and is an important concept in electricity. It is called the electric potential or, simply, the potential and is referred to with the symbol V, as in Equation 19.3.

| Definition of Electric Potential
| The electric potential V at a given point is the electric potential energy EPE of a small test charge q0 situated at that point divided by the charge itself:

| (19.3)

SI Unit of Electric Potential: joule/coulomb = volt (V)

The SI unit of electric potential is a joule per coulomb, a quantity known as a volt. The name honors Alessandro Volta (1745–1827), who invented the voltaic pile, the forerunner of the battery. In spite of the similarity in names, the electric potential energy EPE and the electric potential V are not the same. The electric potential energy, as its name implies, is an energy and, therefore, is measured in joules. In contrast, the electric potential is an energy per unit charge and is measured in joules per coulomb, or volts.
We can now relate the work WAB done by the electric force when a charge q0 moves from A to B to the potential difference between the points. Combining Equations 19.2 and 19.3, we have:

| (19.4)

Often the “delta” notation is used to express the difference (final value minus initial value) in potentials and in potential energies: and . In terms of this notation, Equation 19.4 takes the following more compact form:

| (19.4)

Neither the potential V nor the potential energy EPE can be determined in an absolute sense, because only the differences ΔV and are measurable in terms of the work WAB. The gravitational potential energy has this same characteristic, since only the value at one height relative to that at some reference height has any significance. Example 1 emphasizes the relative nature of the electric potential.

| Example 1 | | | Work, Electric Potential Energy, and Electric Potential

In Figure 19-2, the work done by the electric force as the test charge moves from A to B is . (a) Find the value of the difference, , in the electric potential energies of the charge between these points. (b) Determine the potential difference, , between the points.
Reasoning The work done by the electric force when the charge travels from A to B is , according to Equation 19.1. Therefore, the difference in the electric potential energies (final value minus initial value) is . The potential difference, , is the difference in the electric potential energies divided by the charge q0, according to Equation 19.4.
Solution

a. | The difference in the electric potential energies of the charge between the points A and B is

| (19.1)

Thus, the charge has a higher electric potential energy at A than at B.
b. | The potential difference between A and B is

| (19.4)

This electric car stores energy from the charging station in its battery pack. The voltage of the battery pack and the magnitude of the charge sent through it by the charging station determine how much energy is stored. (© AP/Wide World Photos)

We can see, then, that the electric potential is higher at A than at B.

In Figure 19-1 the speed of the basketball increases as it falls from A to B. Since point A has a greater gravitational potential energy than point B, we see that an object of mass m accelerates when it moves from a region of higher potential energy toward a region of lower potential energy. Likewise, the positive charge in Figure 19-2 accelerates as it moves from A to B because of the electric repulsion from the upper plate and the attraction to the lower plate. Since point A has a higher electric potential than point B, we conclude that a positive charge accelerates from a region of higher electric potential toward a region of lower electric potential. On the other hand, a negative charge placed between the plates in Figure 19-2 behaves in the opposite fashion, since the electric force acting on the negative charge is directed opposite to that on the positive charge. A negative charge accelerates from a region of lower potential toward a region of higher potential. The next example illustrates the way positive and negative charges behave.

Problem solving insight

| Conceptual Example 2 | | | How Positive and Negative Charges Accelerate

Three points, A, B, and C, are located along a horizontal line, as Figure 19-3 illustrates. A positive test charge is released from rest at A and accelerates toward B. Upon reaching B, the test charge continues to accelerate toward C. Assuming that only motion along the line is possible, what will a negative test charge do when it is released from rest at B?

Figure 19-3 The electric potentials at points A, B, and C are different. Under the influence of these potentials, positive and negative charges accelerate in opposite directions.

Reasoning and Solution A negative charge will accelerate from a region of lower potential toward a region of higher potential. Therefore, before we can decide what the negative test charge will do, it is necessary to know how the electric potentials compare at points A, B, and C. This information can be deduced from the behavior of the positive test charge, which accelerates from a region of higher toward a region of lower potential. Since the positive test charge accelerates from A to B, the potential at A must exceed that at B. And since the positive test charge accelerates from B to C, the potential at B must exceed that at C. The potential at point B, then, must lie between that at points A and C, as Figure 19-3 illustrates. When the negative test charge is released from rest at B, it will accelerate toward the region of higher potential. In other words, it will begin moving toward A.

As a familiar application of electric potential energy and electric potential, Figure 19-4 shows a 12-V automobile battery with a headlight connected between its terminals. The positive terminal, point A, has a potential that is 12 V higher than the potential at the negative terminal, point B; in other words, . Positive charges would be repelled from the positive terminal and travel through the wires and headlight toward the negative terminal.* As the charges pass through the headlight, virtually all their potential energy is converted into heat, which causes the filament to glow “white hot” and emit light. When the charges reach the negative terminal, they no longer have any potential energy. The battery then gives the charges an additional “shot” of potential energy by moving them to the higher-potential positive terminal, and the cycle is repeated. In raising the potential energy of the charges, the battery does work on them and draws from its reserve of chemical energy to do so. Example 3 illustrates the concepts of electric potential energy and electric potential as applied to a battery.

Figure 19-4 A headlight connected to a 12-V battery.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 3 | | | Operating a Headlight

The wattage of the headlight in Figure 19-4 is 60.0 W. Determine the number of particles, each carrying a charge of (the magnitude of the charge on an electron), that pass between the terminals of the 12-V car battery when the headlight burns for one hour.
Reasoning The number of particles is the total charge that passes between the battery terminals in one hour divided by the magnitude of the charge on each particle. The total charge is that needed to convey the energy used by the headlight in one hour. This energy is related to the wattage of the headlight, which specifies the power or rate at which energy is used, and the time the light is on.
Knowns and Unknowns The following table summarizes the data provided:

Description | Symbol | Value | Comment
Wattage of headlight | P | 60.0 W |
Charge magnitude per particle | e |  |
Electric potential difference between battery terminals |  | 12 V | See Figure 19-4.
Time headlight is on | t | 3600 s | One hour.
Unknown Variable |  |  |
Number of particles | n | ? |

Modeling the Problem

| Step 1 The Number of Particles
| The number n of particles is the total charge q0 that passes between the battery terminals in one hour divided by the magnitude e of the charge on each particle, as expressed by Equation 1 at the right. The value of e is given. To evaluate q0, we proceed to Step 2.

| (1)

| Step 2 The Total Charge Provided by the Battery
| The battery must supply the total energy used by the headlight in one hour. The battery does this by supplying the charge q0 to convey this energy. The energy is the difference between the electric potential energy EPEA at terminal A and the electric potential energy EPEB at terminal B of the battery (see Figure 19-4). According to Equation 19.4, this total energy is , where is the electric potential difference between the battery terminals. Solving this expression for q0 gives

which can be substituted into Equation 1, as shown at the right. As the data table shows, a value is given for . In Step 3, we determine a value for .

| Step 3 The Energy Used by the Headlight
| The rate at which the headlight uses energy is the power P or wattage of the headlight. According to Equation 6.10b, the power is the total energy divided by the time t, so that . Solving for the total energy gives

Since P and t are given, we substitute this result into Equation 2, as indicated at the right.

Solution Combining the results of each step algebraically, we find that

The number of particles that pass between the battery terminals in one hour is

Related Homework: Problems 5, 52

Need more practice?

Interactive LearningWare19.1
The energy in a lightning bolt is enormous. Consider, for example, a lightning bolt in which 29 C of charge moves through a potential difference of . With the amount of energy in this bolt, how many days could a household function that uses 75 kWh of energy per day?

As used in connection with batteries, the volt is a familiar unit for measuring electric potential difference. The word “volt” also appears in another context, as part of a unit that is used to measure energy, particularly the energy of an atomic particle, such as an electron or a proton. This energy unit is called the electron volt (eV). Problem solving insight: One electron volt is the magnitude of the amount by which the potential energy of an electron changes when the electron moves through a potential difference of one volt. Since the magnitude of the change in potential energy is , it follows that

One million (10+6) electron volts of energy is referred to as one MeV, and one billion (10+9) electron volts of energy is one GeV, where the “G” stands for the prefix “giga” (pronounced “jig’a”).
CONCEPTS AT A GLANCE The Concepts-at-a-Glance chart in Figure 19-5 emphasizes that the total energy of an object, which is the sum of its kinetic and potential energies, is an important concept. Its significance lies in the fact that the total energy remains the same (is conserved) during the object’s motion, provided that nonconservative forces, such as friction, are either absent or do no net work. While the sum of the energies at each instant remains constant, energy may be converted from one form to another; for example, gravitational potential energy is converted into kinetic energy as a ball falls. As Figure 19-5 illustrates, we now include the electric potential energy EPE as part of the total energy that an object can have:

If the total energy is conserved as the object moves, then its final energy Ef is equal to its initial energy E0, or . Example 4 illustrates how the conservation of energy is applied to a charge moving in an electric field.

Figure 19-5 CONCEPTS AT A GLANCE Electric potential energy is another type of energy. The patient in the photograph is undergoing an X-ray examination. To produce X-rays, electrons are accelerated from rest and allowed to collide with a target material within the X-ray machine. As each electron is accelerated, electric potential energy is converted into kinetic energy, and the total energy is conserved during the process. (Jonathan Nourok/Stone/Getty)

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 4 | | | The Conservation of Energy

A particle has a mass of and a charge of . It is released from rest at point A and accelerates until it reaches point B, as Figure 19-6a shows. The particle moves on a horizontal straight line and does not rotate. The only forces acting on the particle are the gravitational force and an electrostatic force (neither is shown in the drawing). The electric potential at A is 25 V greater than that at B; in other words, . What is the translational speed of the particle at point B?

Figure 19-6 (a) A positive charge starts from rest at point A and accelerates toward point B. (b) A negative charge starts from rest at B and accelerates toward A.

Reasoning The translational speed of the particle is related to the particle’s translational kinetic energy, which forms one part of the total mechanical energy that the particle has. The total mechanical energy is conserved, because only the gravitational force and an electrostatic force, both of which are conservative forces, act on the particle (see Section 6.5). Thus, we will determine the speed at point B by utilizing the principle of conservation of mechanical energy.
Knowns and Unknowns We have the following data:

Description | Symbol | Value | Comment
Explicit Data |  |  |
Mass of particle | m |  |
Electric charge of particle | q0 |  |
Electric potential difference between points A and B |  | 25 V | See Figure 19-6a.
Implicit Data |  |  |
Speed at point A | vA | 0 m/s | Particle released from rest.
Vertical height above ground | h | Remains constant | Particle travels horizontally.
Angular speed |  | 0 rad/s | Particle does not rotate during motion.
Elastic force | Felastic | 0 N | No elastic force acts on particle.
Unknown Variable |  |  |
Speed at point B | vB | ? |

Modeling the Problem

| Step 1 Conservation of Total Mechanical Energy
| The particle’s total mechanical E is

Since the particle does not rotate, the angular speed is always zero (see the data table) and since there is no elastic force (see the data table), we may omit the terms and from this expression. With this in mind, we express the fact that (energy is conserved) as follows:

This equation can be simplified further, since the particle travels horizontally, so that (see the data table), with the result that

Solving for vB gives Equation 1 at the right. Values for vA and m are available, and we turn to Step 2 in order to evaluate .

| (1)

| Step 2 The Electric Potential Difference
| According to Equation 19.4, the difference in electric potential energies is related to the electric potential difference :

The terms q0 and are known, so we substitute this expression into Equation 1, as shown at the right.

Solution Combining the results of each step algebraically, we find that

The speed of the particle at point B is

Note that if the particle had a negative charge of and were released from rest at point B, it would arrive at point A with the same speed of 9.1 m/s (see Figure 19-6b). This result can be obtained by returning to Modeling Step 1 and solving for vA instead of vB.

Problem solving insight A positive charge accelerates from a region of higher potential toward a region of lower potential. In contrast, a negative charge accelerates from a region of lower potential toward a region of higher potential.

Related Homework: Problems 3, 6, 7

Check Your Understanding 1

An ion, starting from rest in a constant electric field, accelerates from point A to point B. Does the electric potential energy of the ion at point B depend on (a) the magnitude of its charge and (b) its mass? Does the speed of the ion at B depend on (c) the magnitude of its charge and (d) its mass? (The answers are given at the end of the book.)
Background: This question deals with how a charged particle moves between two locations that have different electric potential energies. The answers also require an understanding of how the kinetic energy of a particle depends on its mass and speed.
For similar questions (including calculational counterparts), consult Self-Assessment Test 19.1, which is described at the end of Section 19.3.

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