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Learning Objectives

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

Define electric potential and electric potential energy
Describe the relationship between potential difference and electrical potential energy
Explain electron volt and its usage in submicroscopic processes
Determine electric potential energy given potential difference and amount of charge

The information presented in this section supports the following AP® learning objectives and science practices:

2.C.1.1 The student is able to predict the direction and the magnitude of the force exerted on an object with an electric charge q placed in an electric field E using the mathematical model of the relation between an electric force and an electric field: F = qE; a vector relation. (S.P. 6.4, 7.2)
2.C.1.2 The student is able to calculate any one of the variables—electric force, electric charge, and electric field—at a point given the values and sign or direction of the other two quantities. (S.P. 2.2)
5.B.2.1 The student is able to calculate the expected behavior of a system using the object model (i.e., by ignoring changes in internal structure) to analyze a situation. Then, when the model fails, the student can justify the use of conservation of energy principles to calculate the change in internal energy due to changes in internal structure because the object is actually a system. (S.P. 1.4, 2.1)
5.B.3.1 The student is able to describe and make qualitative and/or quantitative predictions about everyday examples of systems with internal potential energy. (S.P. 2.2, 6.4, 7.2)
5.B.3.2 The student is able to make quantitative calculations of the internal potential energy of a system from a description or diagram of that system. (S.P. 1.4, 2.2)
5.B.3.3 The student is able to apply mathematical reasoning to create a description of the internal potential energy of a system from a description or diagram of the objects and interactions in that system. (S.P. 1.4, 2.2)
5.B.4.1 The student is able to describe and make predictions about the internal energy of systems. (S.P. 6.4, 7.2)
5.B.4.2 The student is able to calculate changes in kinetic energy and potential energy of a system, using information from representations of that system. (S.P. 1.4, 2.1, 2.2)

When a free positive charge $q$ is accelerated by an electric field, such as shown in Figure 2.2, it is given kinetic energy. The process is analogous to an object being accelerated by a gravitational field. It is as if the charge is going down an electrical hill where its electric potential energy is converted to kinetic energy. 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.

A charge plus q moves from a positive to a negative sheet of charge. The change in the electric potential energy equals the change in kinetic energy. This is similar to the change from gravitational potential energy to kinetic energy when an object of mass m rolls downhill.

Figure 2.2 A charge accelerated by an electric field is analogous to a mass going down a hill. In both cases potential energy is converted to another form. Work is done by a force, but since this force is conservative, we can write

$W = –Δ PE .$

The electrostatic or Coulomb force is conservative, which means that the work done on $q$ is independent of the path taken. This is exactly analogous to the gravitational force in the absence of dissipative forces such as friction. When a force is conservative, it is possible to define a potential energy associated with the force, and it is usually easier to deal with the potential energy—because it depends only on position—than to calculate the work directly.

We use the letters PE to denote electric potential energy, which has units of joules (J). The change in potential energy, $Δ PE,$ is crucial, since the work done by a conservative force is the negative of the change in potential energy; that is, $W = –Δ PE .$ For example, work $W$ done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative $Δ PE .$ There must be a minus sign in front of $Δ PE$ to make $W$ positive. PE 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.

Potential Energy

$W = –ΔPE .$ For example, work $W$ done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative $ΔPE .$ There must be a minus sign in front of $ΔPE$ to make $W$ positive. PE 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.

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 voltage—related to electric potential energy—than to deal with the Coulomb force directly.

Calculating the work directly is generally difficult, since $W = Fd cos θ$ and the direction and magnitude of $F$ can be complex for multiple charges, for odd-shaped objects, and along arbitrary paths. But we do know that, since $F = qE,$ the work, and hence $ΔPE,$ is proportional to the test charge $q.$ To have a physical quantity that is independent of test charge, we define electric potential $V$ (or simply potential, since electric is understood) to be the potential energy per unit charge

2.1

$V = \frac{PE}{q} .$

Electric Potential

This is the electric potential energy per unit charge

2.2

$V = \frac{PE}{q} .$

Since PE is proportional to $q$ , the dependence on $q$ cancels. Thus $V$ does not depend on $q .$ The change in potential energy $ΔPE$ is crucial, and so we are concerned with the difference in potential or potential difference $Δ V$ between two points, where

2.3

$Δ V = V_{B} - V_{A} = \frac{Δ PE}{q} .$

The potential difference between points A and B, $V_{B} - V_{A},$ is thus 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

2.4

$1 V = 1 \frac{J}{C}$

Potential Difference

The 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.

2.5

$1 V = 1 \frac{J}{C}$

The familiar term voltage is the common name for 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.

In summary, the relationship between potential difference—or voltage—and electrical potential energy is given by

2.6

$Δ V = \frac{ΔPE}{q} and ΔPE = q Δ V .$

Potential Difference and Electrical Potential Energy

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

2.7

$Δ V = \frac{ΔPE}{q} and ΔPE = q Δ V .$

The second equation is equivalent to the first.

Real World Connections: Electric Potential in Electronic Devices

You probably use devices with stored electric potential daily. Do you own or use any electronic devices that do not have to be attached to a wall socket? What happens if you use these items long enough? Do they cease functioning? What do you do in that case? Choose one of these types of electronic devices and determine how much electric potential, measured in volts, the item requires for proper functioning. Then estimate the amount of time between replenishments of potential. Describe how the time between replenishments of potential depends on use.

Answer

Ready examples include calculators and cell phones. The former will either be solar powered, or have replaceable batteries, probably four 1.5 V for a total of 6 V. The latter will need to be recharged with a specialized charger, which probably puts out 5 V. Times between replenishments will be highly dependent on which item is used, but should be less with more intense use.

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 since $ΔPE = q Δ V .$ The car battery can move more charge than the motorcycle battery, although both are 12-V batteries.

Example 2.1 Calculating Energy

Suppose you have a 12.0-V motorcycle battery that can move 5,000 C of charge, and a 12.0-V car battery that can move 60,000 C of charge. How much energy does each deliver? Assume that the numerical value of each charge is accurate to three significant figures.

Strategy

To say we have a 12.0-V battery means that its terminals have a 12.0-V potential difference. When such a battery moves charge, it puts the charge through a potential difference of 12.0 V, and the charge is given a change in potential energy equal to $ΔPE = q Δ V .$

So to find the energy output, we multiply the charge moved by the potential difference.

Solution

For the motorcycle battery, $q = 5,000 C$ and $Δ V = 12.0 V .$ The total energy delivered by the motorcycle battery is

2.8

$\begin{array}{l} {ΔPE}_{cycle} & = & (5,000 C) (12.0 V) \\ = & (5,000 C) (12.0 J/C) \\ = & 6.00 \times 10^{4} J. \end{array}$

Similarly, for the car battery, $q = 60, 000 C$ and

2.9

$\begin{array}{l} {ΔPE}_{car} & = & (60,000 C) (12.0 V) \\ = & 7.20 \times 10^{5} J. \end{array}$

Discussion

While voltage and energy are related, they are not the same thing. The voltages of the batteries are identical, but the energy supplied by each is quite different. Note also that as a battery is discharged, some of its energy is used internally and its terminal voltage drops, such as when headlights dim because of a low car battery. The energy supplied by the battery is still calculated as in this example, but not all of the energy is available for external use.

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 Figure 2.3. The change in potential is $Δ V = V_{B} {–V}_{A} = +12 V$ and the charge $q$ is negative, so that $ΔPE = q Δ V$ is negative, meaning the potential energy of the battery has decreased when $q$ has moved from A to B.

A headlight is connected to a 12 V battery. Negative charges move from the negative terminal of the battery to the positive terminal, resulting in a current flow and making the headlight glow. However, the positive terminal is at a greater potential than the negative terminal.

Figure 2.3 A battery moves negative charge from its negative terminal through a headlight to its positive terminal. Appropriate combinations of chemicals in the battery separate charges so that the negative terminal has an excess of negative charge, which is repelled by it and attracted to the excess positive charge on the other terminal. In terms of potential, the positive terminal is at a higher voltage than the negative. Inside the battery, both positive and negative charges move.

Making Connections: Potential Energy in a Battery

The previous example stated that the potential energy of a battery decreased with each electron it pushed out. However, shouldn’t this reduced internal energy reduce the potential, as well? Yes, it should. So why don’t we notice this?

Part of the answer is that the amount of energy taken by any one electron is extremely small, and therefore it doesn’t reduce the potential much. But the main reason is that the energy is stored in the battery as a chemical reaction waiting to happen, not as electric potential. This reaction only runs when a load is attached to both terminals of the battery. Any one set of chemical reactants has a certain maximum potential that it can provide; this is why larger batteries consist of cells attached in series, so that the overall potential increases additively. As these reactants get used up, each cell gives less potential to the electrons it is moving; eventually this potential falls below a useful threshold. Then the battery either needs to be charged, which reverses the chemical reaction and reconstitutes the original reactants, or changed.

Example 2.2 How Many Electrons Move through a Headlight Each Second?

When a 12.0-V car battery runs a single 30.0-W headlight, how many electrons pass through it each second?

Strategy

To find the number of electrons, we must first find the charge that moved in 1.00 s. The charge moved is related to voltage and energy through the equation $ΔPE = q Δ V .$ A 30.0 W lamp uses 30.0 joules per second. Since the battery loses energy, we have $ΔPE = –30.0 J$ and, since the electrons are going from the negative terminal to the positive, we see that $Δ V = +12.0 V .$

Solution

To find the charge $q$ moved, we solve the equation $ΔPE = q Δ V .$

2.10

$q = \frac{ΔPE}{Δ V}$

Entering the values for $Δ PE$ and $Δ V,$ we get

2.11

$q = \frac{–30.0 J}{+12.0 V} = \frac{–30.0 J}{+12.0 J/C} = –2.50 C.$

The number of electrons $n_{e}$ is the total charge divided by the charge per electron. That is

2.12

$n_{e} = \frac{–2.50 C}{–1.60 \times 10^{–19} {C/e}^{-}} = 1.56 \times 10^{19} electrons.$

Discussion

This is a very large number. It is no wonder that we do not ordinarily observe individual electrons with so many being present in ordinary systems. In fact, electricity had been in use for many decades before it was determined that the moving charges in many circumstances were negative. Positive charge moving in the opposite direction of negative charge often produces identical effects; this makes it difficult to determine which is moving or whether both are moving.

Applying the Science Practices: Work and Potential Energy in Point Charges

Consider a system consisting of two positive point charges, each 2.0 µC, placed 1.0 m away from each other. We can calculate the potential, that is, internal, energy of this configuration by computing the potential due to one of the charges, and then calculating the potential energy of the second charge in the potential of the first. Applying Equations (19.38) and (19.2) gives us a potential energy of 3.6 × 10^–2 J. If we move the charges closer to each other, say, to 0.50 m apart, the potential energy doubles. Note that, to create this second case, some outside force would have had to do work on this system to change the configuration, and hence it was not a closed system. However, because the electric force is conservative, we can use the work-energy theorem to state that, since there was no change in kinetic energy, all of the work done went into increasing the internal energy of the system. Also note that if the point charges had different signs they would be attracted to each other, so they would be capable of doing work on an outside system when the distance between them decreased. As work is done on the outside system, the internal energy in the two-charge system decreases.

A pair of two microcoulomb charges are initially one meter apart. Work is done to move them to a distance of half a meter apart.

Figure 2.4 Work is done by moving two charges with the same sign closer to each other, increasing the internal energy of the two-charge system.

The Electron Volt

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. Figure 2.5 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 is given kinetic energy that is later converted to another form—light in the television tube, for example. Note that downhill for the electron is uphill for a positive charge. Since energy is related to voltage by $ΔPE = q Δ V,$ we can think of the joule as a coulomb-volt.

In an electron gun the electrons move from the negatively charged plate to the positively charged plate. Their kinetic energy will be equal to the potential energy.

Figure 2.5 A typical electron gun accelerates electrons using a potential difference between two metal plates. The energy of the electron in electron volts is numerically the same as the voltage between the plates. For example, a 5,000-V potential difference produces 5,000 eV electrons.

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 V. In equation form,

2.13

$\begin{array}{l} 1 eV & = & (1.60 \times 10^{–19} C) (1 V) = (1.60 \times 10^{–19} C) (1 J/C) \\ = & 1.60 \times 10^{–19} J. \end{array}$

Electron Volt

2.14

$\begin{array}{l} 1 eV & = & (1.60 \times 10^{–19} C) (1 V) = (1.60 \times 10^{–19} C) (1 J/C) \\ = & 1.60 \times 10^{–19} J. \end{array}$

An electron accelerated through a potential difference of 1 V is given an energy of 1 eV. It follows that an electron accelerated through 50 V is given 50 eV. A potential difference of 100,000 V (100 kV) will give an electron an energy of 100,000 eV (100 keV), and so on. Similarly, an ion with a double positive charge accelerated through 100 V will be given 200 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.

Connections: Energy Units

The electron volt (eV) is the most common energy unit for submicroscopic processes. This will be particularly noticeable in the chapters on modern physics. Energy is so important to so many subjects that there is a tendency to define a special energy unit for each major topic. There are, for example, calories for food energy, kilowatt-hours for electrical energy, and therms for natural gas energy.

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 eV of energy is required to break up certain organic molecules. If a proton is accelerated from rest through a potential difference of 30 kV, it is given an energy of 30 keV (30,000 eV) and it can break up as many as 6,000 of these molecules $(30,000 eV \div 5 eV per molecule = 6,000 molecules).$ Nuclear decay energies are on the order of 1 MeV (1,000,000 eV) per event and can, thus, produce significant biological damage.

Conservation of Energy

The total energy of a system is conserved if there is no net addition—or subtraction—of 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, $KE + PE = constant .$ A loss of PE of a charged particle becomes an increase in its KE. Here PE is the electric potential energy. Conservation of energy is stated in equation form as

2.15

$KE + PE = constant$

2.16

${KE}_{i} + {PE}_{i} {= KE}_{f} + {PE}_{f},$

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.

Example 2.3 Electrical Potential Energy Converted to Kinetic Energy

Calculate the final speed of a free electron accelerated from rest through a potential difference of 100 V. Assume that this numerical value is accurate to three significant figures.

Strategy

We have a system with only conservative forces. Assuming the electron is accelerated in a vacuum, and neglecting the gravitational force—we will check on this assumption later— all of the electrical potential energy is converted into kinetic energy. We can identify the initial and final forms of energy to be ${KE}_{i} = 0, {KE}_{f} = ½ {mv}^{2}, {PE}_{i} = qV {, and PE}_{f} = 0.$

Solution

Conservation of energy states that

2.17

${KE}_{i} + {PE}_{i} {= KE}_{f} + {PE}_{f} .$

Entering the forms identified above, we obtain

2.18

$qV = \frac{{mv}^{2}}{2} .$

We solve this for $v .$

2.19

$v = \sqrt{\frac{2 qV}{m}} .$

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

2.20

$\begin{array}{l} v & = & \sqrt{\frac{2 (–1.60 \times 10^{–19} C) (–100 J/C)}{9.11 \times 10^{–31} kg}} \\ = & 5.93 \times 10^{6} m/s. \end{array}$

Discussion

Note that both the charge and the initial voltage are negative, as in Figure 2.5. From the discussions in Electric Charge and Electric Field, we know that electrostatic forces on small particles are generally very large compared with the gravitational force. The large final speed confirms that the gravitational force is indeed negligible here. The large speed also indicates how easy it is to accelerate electrons with small voltages because of their very small mass. Voltages much higher than the 100 V in this problem are typically used in electron guns or electron emitters. Those higher voltages produce electron speeds so great that relativistic effects must be taken into account. That is why a low voltage is considered accurately in this example.

Making Connections: Kinetic and Potential Energy in Point Charges

Now consider another system of two point charges. One has a mass of 1,000 kg and a charge of 50.0 µC, and is initially stationary. The other has a mass of 1.00 kg, a charge of 10.0 µC, and is initially traveling directly at the first point charge at 10.0 m/s from very far away. What will be the closest approach of these two objects to each other?

An object has a mass of 1000 kg and a charge of 50.0 µC, and is initially stationary. Another object has a mass of 1.00 kg, a charge of 10.0 µC, and is initially traveling directly at the first point charge at 10.0 m/s from very far away.

Figure 2.6 A system consisting of two point charges initially has the smaller charge moving toward the larger charge

Note that the internal energy of this two-charge system will not change, due to an absence of external forces acting on the system. Initially, the internal energy is equal to the kinetic energy of the smaller charge, and the potential energy is effectively zero due to the enormous distance between the two objects. Conservation of energy tells us that the internal energy of this system will not change. Hence the distance of closest approach will be when the internal energy is equal to the potential energy between the two charges, and there is no kinetic energy in this system.

The initial kinetic energy may be calculated as 50.0 J. Applying Equations (19.38) and (19.2), we find a distance of 9.00 cm. After this, the mutual repulsion will send the lighter object off to infinity again. Note that we did not include potential energy due to gravity, as the masses concerned are so small compared to the charges that the result will never come close to showing up in significant digits. Furthermore, the first object is much more massive than the second. As a result, any motion induced in it will also be too small to show up in the significant digits.