11 Electric Potential Energy: Potential Difference

In the previous chapters, we learned that electric fields exert forces on charged particles. While knowing the force is important, many practical applications—from batteries and electrical circuits to nerve cells and medical devices—are better understood by considering energy rather than force alone. In this chapter, we introduce the closely related concepts of electric potential energy, electric potential, and potential difference (commonly called voltage). These quantities allow us to describe how much energy is stored in an electric field and how that energy is transferred when charges move.

From Force to Energy: The Big Picture

Previously, we described the force exerted by an electric field using

This equation tells us how strongly an electric field pushes or pulls on a charged particle. However, in many real situations we are interested in the energy transferred as charges move through an electric field. For example:

• How much energy is delivered during a defibrillator shock?
• How much energy does a sodium ion gain when it crosses a cell membrane?
• How much energy is required to move charge through an electrical circuit?

These are questions about energy, not force. When a charge moves in an electric field, the field performs work on the charge. Because the electrostatic force is a conservative force, the work done by the field is related to the change in electric potential energy by

If the electric field does positive work on a charge, the charge's electric potential energy decreases. That lost potential energy is converted into other forms of energy, such as kinetic energy.

Analogy: Gravitational Potential Energy

A useful way to understand electric potential is to compare it with gravity. Imagine a ball resting at the top of a hill. Because of its position, the ball possesses gravitational potential energy. As it rolls downhill,

• its gravitational potential energy decreases, and
• its kinetic energy increases.

The height of the ball determines how much gravitational potential energy is available per unit mass. In electricity, the analogous quantity is the electric potential, which describes how much electric potential energy is available per unit charge.

• Electric potential energy depends on the position of a charge in an electric field.
• A charge moving "downhill" in electric potential loses potential energy while gaining kinetic energy.

In this analogy, the electric field is like the slope of the hill, while the electric potential is like the height above the ground.

Electric Potential Energy (PE)

Electric potential energy (PE) is the energy a charged particle possesses because of its position in an electric field. Like all forms of energy, electric potential energy is measured in joules (J). The amount of electric potential energy a charge has depends on two things: the amount of charge and its location in the electric field. Moving the same charge to a different position changes its electric potential energy. Likewise, placing a larger charge at the same location results in a larger amount of potential energy. An electric field can therefore be viewed as a region where energy is stored. When a charged particle is released, the electric field can transfer some of this stored energy into kinetic energy by accelerating the particle. This transfer occurs because the electric field performs work on the charge. The relationship between work and electric potential energy is

where W is the work done by the electric field. The negative sign indicates that if the electric field does positive work on a charge, the charge loses electric potential energy. Conversely, if an external agent moves a charge against the electric field, work must be done on the charge and its electric potential energy increases. For example, suppose a positive charge is released in an electric field. The electric field exerts a force on the charge, causing it to accelerate. As the charge speeds up:

• The electric field does positive work.
• The charge gains kinetic energy.
• Its electric potential energy decreases by the same amount.

This is directly analogous to gravity. A ball resting at the top of a hill stores gravitational potential energy because of its position. As the ball rolls downhill, gravity does positive work, the ball speeds up, and gravitational potential energy is converted into kinetic energy. In exactly the same way, an electric field converts electric potential energy into kinetic energy as charges move through the field. This analogy is extremely useful throughout this chapter. Just as differences in height determine how much gravitational energy is available, differences in electric potential determine how much electrical energy is available to move charges. In the next section, we define the quantity that measures this energy per unit charge: electric potential, or voltage.

Electric Potential (Voltage)

Although electric potential energy is a useful quantity, it depends on the amount of charge being considered. A particle carrying twice as much charge has twice as much electric potential energy at the same location. To describe the electric field itself, independent of the particular charge placed in it, we define a new quantity called electric potential, or more commonly, voltage. Electric potential is defined as the electric potential energy per unit charge:

The SI unit of electric potential is the volt (V), where

This definition means that a voltage tells us how much electric potential energy is available for each coulomb of charge. For example, a point at a potential of 12 V provides 12 joules of electric potential energy for every coulomb of charge located there. It is important to distinguish between electric potential energy and electric potential:

• Electric potential energy depends on both the location of the charge and the amount of charge present.
• Electric potential (voltage) depends only on the location in the electric field and is independent of the test charge.

This distinction is similar to the difference between gravitational potential energy and height. A heavier object has more gravitational potential energy than a lighter one at the same height, but both objects are at the same height. Likewise, two particles with different charges placed at the same point in an electric field have different electric potential energies, but they experience the same electric potential (voltage). This analogy is one of the most useful ways to think about voltage. Just as height tells us how much gravitational energy is available per unit mass, voltage tells us how much electrical energy is available per unit charge. Throughout this chapter, we will see that voltage provides a much more convenient way to analyze electrical systems than tracking electric potential energy directly.

Potential Difference (Voltage Between Two Points)

In most practical situations, we are not interested in the electric potential at a single point. Instead, we want to know how the potential changes as a charge moves from one location to another. This change is called the potential difference, or simply the voltage between two points. If a charge moves from point A to point B, the potential difference is defined as

[latex]\Delta V=V_B-V_A[/latex]

Since electric potential is the electric potential energy per unit charge, the potential difference can also be written as

[latex]\Delta V=\frac{\Delta\text{PE}}{q}[/latex]

Rearranging this equation gives one of the most useful relationships in electricity:

[latex]\Delta\text{PE}=q\Delta V[/latex]

This equation states that the change in electric potential energy is equal to the charge multiplied by the potential difference through which it moves. A larger charge experiences a larger change in energy when moving through the same voltage. Likewise, for a given charge, a larger voltage difference produces a larger change in energy. This relationship explains why voltage is so useful. Instead of calculating electric forces at every point, we can often determine how much energy is gained or lost simply by knowing the voltage difference between two locations.

Potential Difference and Energy

The change in electric potential energy of a charge moving between two points is

[latex]\Delta\text{PE}=q\Delta V[/latex]

Remember: voltage is energy per unit charge. Multiplying the voltage difference by the charge gives the total change in electric potential energy.

Why Voltage Is Not Energy

Because voltage is measured in volts, many people casually refer to it as "electrical energy." However, voltage and energy are not the same physical quantity. Consider two familiar examples: a 12-V motorcycle battery and a 12-V car battery. Both provide approximately the same voltage, yet the car battery stores much more energy and can power a starter motor for a much longer time. The reason is that the total change in electric potential energy depends on both the voltage and the amount of charge that moves:

[latex]\Delta \text{PE}=q\Delta V[/latex]

Since both batteries have approximately the same voltage, each coulomb of charge leaving either battery gains the same amount of energy. However, the larger car battery contains much more stored charge and can move many more coulombs before becoming discharged. As a result, it can deliver much more total energy. A useful way to think about voltage is that it describes the energy available per unit charge, not the total amount of energy available. The total electrical energy depends on both the voltage and the total charge that can be transferred.

Remember

• Voltage tells you how much energy is available per coulomb of charge.
• Electrical energy depends on both the voltage and the amount of charge that moves.
• Two devices can have the same voltage but store very different amounts of energy.

Biological Example: Cell Membrane Potential

One of the most important applications of electric potential in biology is the cell membrane potential. Living cells maintain a voltage difference between the inside and outside of their membranes by actively transporting ions such as sodium (Na⁺), potassium (K⁺), and chloride (Cl⁻). In a typical resting neuron, the inside of the cell is approximately −70 mV relative to the outside. Suppose a sodium ion, which carries a single positive charge, moves across a membrane with a potential difference of 70 mV.

[latex]\Delta V=70\ \text{mV}=0.070\ \text{V}[/latex]

The charge of a sodium ion is one fundamental charge:

[latex]q=+1.60\times10^{-19}\ \text{C}[/latex]

The change in electric potential energy is therefore

[latex]\Delta\text{PE}=q\Delta V[/latex]

[latex]\Delta\text{PE}=(1.60\times10^{-19}\ \text{C})(0.070\ \text{V})[/latex]

[latex]\Delta\text{PE}=1.12\times10^{-20}\ \text{J}[/latex]

Although this amount of energy is extremely small in everyday units, it is enormous on the molecular scale. The coordinated movement of millions of ions across cell membranes produces the electrical signals responsible for nerve impulses, muscle contraction, and communication between cells.

The Electron Volt

An electron volt (eV) is a unit of energy, not voltage.

[latex]1\ \text{eV}=1.60\times10^{-19}\ \text{J}[/latex]

It is the energy gained by a particle carrying one fundamental charge when it moves through a potential difference of 1 volt.The definition leads to a simple numerical relationship. A particle carrying one elementary charge gains:

• 1 eV when accelerated through 1 V.
• 100 eV when accelerated through 100 V.
• 100 keV when accelerated through 100 kV.
• 1 MeV when accelerated through 1 million volts.

The electron volt is especially useful in biology and medicine because many important microscopic processes occur within this energy range.

• Chemical bond energies are typically a few eV.
• Visible and ultraviolet photons have energies of a few electron volts.
• X-rays typically have energies of thousands of electron volts (keV).
• Gamma rays and particles emitted during nuclear decay often have energies of millions of electron volts (MeV).

For example, if breaking a particular chemical bond requires about 5 eV, then a particle carrying 30 keV of energy has enough energy, in principle, to break thousands of such bonds:

[latex]\frac{30\,000\ \text{eV}}{5\ \text{eV}}=6000[/latex]

In reality, not all of the particle's energy is deposited into breaking chemical bonds. Nevertheless, this comparison illustrates why ionizing radiation can damage biological molecules such as DNA and why the electron volt is such a convenient unit for describing interactions at the atomic and molecular scales.

Example 11.1: Calculating Electrical Energy from a Battery

A 12.0 V motorcycle battery can deliver a total charge of 5000 C, while a 12.0 V car battery can deliver 60,000 C. Calculate the total electrical energy supplied by each battery.

Strategy

The energy transferred by a battery depends on both its voltage and the total amount of charge it moves. The relationship between electrical energy, charge, and potential difference is

[latex]\Delta\text{PE}=q\Delta V[/latex]

Since both batteries have the same voltage, the battery capable of moving the greater amount of charge will supply more total energy.

Solution

Motorcycle battery

[latex]\Delta\text{PE}=q\Delta V[/latex]

[latex]\Delta\text{PE}=(5000\ \text{C})(12.0\ \text{V})[/latex]

[latex]\Delta\text{PE}=(5000\ \text{C})(12.0\ \text{J/C})[/latex]

[latex]\Delta\text{PE}=6.00\times10^{4}\ \text{J}[/latex]

Car battery

[latex]\Delta\text{PE}=q\Delta V[/latex]

[latex]\Delta\text{PE}=(60\,000\ \text{C})(12.0\ \text{V})[/latex]

[latex]\Delta\text{PE}=7.20\times10^{5}\ \text{J}[/latex]

Discussion

This example illustrates an important concept: voltage alone does not determine how much energy a battery stores. Although both batteries have the same voltage (12.0 V), the car battery stores twelve times more energy because it can transfer twelve times more charge. This distinction explains why batteries with the same voltage can have very different capacities. A small motorcycle battery and a much larger car battery can both provide 12 V, yet the larger battery can power electrical devices for a much longer time because it contains a much greater amount of stored charge.

In the previous example, we focused on the amount of electrical energy transferred. However, the sign of the change in electric potential energy is just as important because it tells us whether a charge gains or loses energy as it moves. Recall the relationship between electric potential energy and potential difference:

[latex]\Delta \text{PE}=q\Delta V[/latex]

The sign of q and the sign of ΔV together determine whether the potential energy increases or decreases. Consider the 12-V battery connected to a headlight shown in Figure 11.2. The battery maintains a potential difference of 12 V between its terminals, with the positive terminal at the higher electric potential. By definition,

[latex]\Delta V=V_{\text{positive}}-V_{\text{negative}}=+12.0\ \text{V}[/latex]

In the external circuit, the moving charges are electrons, each with charge

[latex]q=-1.60\times10^{-19}\ \text{C}[/latex]

Electrons move from the negative terminal toward the positive terminal. Although they move toward a region of higher electric potential, they carry a negative charge. Substituting these values into the equation for electric potential energy gives

[latex]\Delta \text{PE}=(-1.60\times10^{-19}\ \text{C})(+12.0\ \text{V})[/latex]

[latex]\Delta \text{PE}=-1.92\times10^{-18}\ \text{J}[/latex]

The negative sign indicates that the electrons lose electric potential energy as they move through the external circuit. That lost electrical energy is converted into other forms of energy—in this case, primarily light and thermal energy produced by the headlight. Figure 11.2 shows a 12-V battery connected to a headlight, with electrons moving through the external circuit from the negative terminal to the positive terminal.

Example 11.2: How Many Electrons Pass through a Headlight Each Second?

A 12.0 V car battery powers a single 30.0 W headlight. How many electrons pass through the headlight each second?

Strategy

Power is the rate at which energy is transferred. Since the headlight consumes 30.0 W, it converts 30.0 J of electrical energy into light and heat every second.

[latex]P=\frac{\Delta E}{t}[/latex]

We first determine the amount of charge that must flow each second using the relationship between electrical energy and potential difference,

[latex]\Delta\text{PE}=q\Delta V[/latex]

Finally, we divide the total charge by the charge carried by a single electron to determine how many electrons move through the headlight each second.

Solution

Since the headlight has a power of 30.0 W, it uses

[latex]\Delta\text{PE}=30.0\ \text{J}[/latex]

of electrical energy every second. Using

[latex]\Delta\text{PE}=q\Delta V[/latex]

the amount of charge transferred each second is

[latex]q=\frac{\Delta\text{PE}}{\Delta V}=\frac{30.0\ \text{J}}{12.0\ \text{V}}=2.50\ \text{C}[/latex]

Each electron carries a charge with magnitude

[latex]e=1.60\times10^{-19}\ \text{C}[/latex]

Therefore, the number of electrons passing through the headlight each second is

[latex]n=\frac{2.50\ \text{C}}{1.60\times10^{-19}\ \text{C/electron}}[/latex]

[latex]n=1.56\times10^{19}\ \text{electrons/s}[/latex]

Discussion

Although each electron carries an extremely small amount of charge, an enormous number of electrons move through an electrical circuit every second. This example also highlights the relationship between power, voltage, and charge transfer: a device operating at a fixed voltage requires a greater flow of charge each second as its power consumption increases. The smooth operation of everyday electrical devices is the result of the coordinated motion of vast numbers of electrons rather than the motion of individual particles.

Conservation of Energy in Electric Fields

Electric forces, like gravitational forces, are conservative forces. This means that the work done by the electric field depends only on the initial and final positions of a charge, not on the path taken between them. As a result, energy is conserved as a charged particle moves through an electric field. If no energy is lost to friction, collisions, or other nonconservative forces, the total mechanical energy remains constant:

[latex]\text{KE}+\text{PE}=\text{constant}[/latex]

or, equivalently,

[latex]\text{KE}_i+\text{PE}_i=\text{KE}_f+\text{PE}_f[/latex]

As a charged particle moves, its electric potential energy and kinetic energy are continually converted into one another. When the electric potential energy decreases, the kinetic energy increases by the same amount. Conversely, if the particle moves against the electric field, its kinetic energy decreases while its electric potential energy increases. This conservation of energy approach is often much simpler than calculating the electric force at every point along the particle's path. Once the potential difference between two points is known, the particle's speed can often be determined directly.

Example 11.3: Electrical Potential Energy Converted to Kinetic Energy

An electron starts from rest and is accelerated through a potential difference of 100 V. Determine its final speed.

Strategy

Because the electric force is conservative, the decrease in electric potential energy equals the increase in kinetic energy. Since the electron starts from rest, its initial kinetic energy is zero, so we apply conservation of energy:

[latex]q\Delta V=\frac{1}{2}mv^2[/latex]

Solving this equation for the final speed gives the desired result.

Solution

Rearranging the energy equation gives

[latex]v=\sqrt{\frac{2q\Delta V}{m}}[/latex]

Using the magnitude of the electron charge, [latex]|q|=1.60\times10^{-19}\ \text{C}[/latex], the electron mass, [latex]m=9.11\times10^{-31}\ \text{kg}[/latex], and a potential difference of [latex]\Delta V=100\ \text{V}[/latex],

[latex]v=\sqrt{\frac{2(1.60\times10^{-19}\ \text{C})(100\ \text{V})}{9.11\times10^{-31}\ \text{kg}}}[/latex]

[latex]v=5.93\times10^{6}\ \text{m/s}[/latex]

Discussion

Even a modest potential difference of only 100 V accelerates an electron to nearly 6 million meters per second because electrons have such a small mass. This principle is used in many scientific and medical instruments, including electron microscopes, X-ray tubes, and particle accelerators. At much higher voltages, the calculated speed approaches a significant fraction of the speed of light, and relativistic effects must be included.

Section Summary

• Electric potential energy is the energy a charge has because of its position in an electric field.
• Electric potential, or voltage, is the electric potential energy per unit charge:
  [latex]V=\frac{\text{PE}}{q}[/latex]
• The potential difference (voltage) between two points is related to the change in electric potential energy by
  [latex]\Delta V=\frac{\Delta\text{PE}}{q}[/latex]

  or equivalently,

  [latex]\Delta\text{PE}=q\Delta V[/latex]
• The electron volt (eV) is a convenient unit of energy for atomic and molecular processes:
  [latex]1\ \text{eV}=1.60\times10^{-19}\ \text{J}[/latex]
• Electric forces are conservative, so the total mechanical energy of an isolated system remains constant:
  [latex]\text{KE}+\text{PE}=\text{constant}[/latex]

  A decrease in electric potential energy corresponds to an equal increase in kinetic energy, and vice versa.

Conceptual Questions

1. Voltage is the common word for potential difference. Which term is more descriptive, voltage or potential difference?
2. 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.
3. What is the relationship between voltage and energy? More precisely, what is the relationship between potential difference and electric potential energy?
4. Voltages are always measured between two points. Why?
5. How are units of volts and electron volts related? How do they differ?

Problems & Exercises

1. 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 [latex]1.67\times10^{-27}\,\text{kg}[/latex].
2. An evacuated tube uses an accelerating voltage of 40 kV to accelerate electrons to hit a copper plate and produce x rays. Non-relativistically, what would be the maximum speed of these electrons?
3. A bare helium nucleus has two positive charges and a mass of [latex]6.64\times10^{-27}\,\text{kg}[/latex]. (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?
4. Integrated Concepts Singly charged gas ions are accelerated from rest through a voltage of 13.0 V. At what temperature will the average kinetic energy of gas molecules be the same as that given these ions?
5. Integrated Concepts The temperature near the center of the Sun is thought to be 15 million degrees Celsius [latex](1.5\times10^{7}\,^{\circ}\text{C})[/latex]. 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?
6. Integrated Concepts (a) What is the average power output of a heart defibrillator that dissipates 400 J of energy in 10.0 ms? (b) Considering the high-power output, why doesn’t the defibrillator produce serious burns?
7. Integrated Concepts A lightning bolt strikes a tree, moving 20.0 C of charge through a potential difference of [latex]1.00\times10^{2}\,\text{MV}[/latex]. (a) What energy was dissipated? (b) What mass of water could be raised from [latex]15^{\circ}\text{C}[/latex] 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.
8. Integrated Concepts A 12.0 V battery-operated bottle warmer heats 50.0 g of glass, [latex]2.50\times10^{2}\,\text{g}[/latex] of baby formula, and [latex]2.00\times10^{2}\,\text{g}[/latex] of aluminum from [latex]20.0^{\circ}\text{C}[/latex] to [latex]90.0^{\circ}\text{C}[/latex]. (a) How much charge is moved by the battery? (b) How many electrons per second flow if it takes 5.00 min to warm the formula? (Hint: Assume that the specific heat of baby formula is about the same as the specific heat of water.)
9. Integrated Concepts A battery-operated car utilizes a 12.0 V system. Find the charge the batteries must be able to move in order to accelerate the 750 kg car from rest to 25.0 m/s, make it climb a [latex]2.00\times10^{2}\,\text{m}[/latex] high hill, and then cause it to travel at a constant 25.0 m/s by exerting a [latex]5.00\times10^{2}\,\text{N}[/latex] force for an hour.
10. Integrated Concepts Fusion probability 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 [latex]1.00\times10^{-12}\,\text{m}[/latex] by finding the voltage of one at that distance and multiplying by the charge of the other. (b) At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?
11. Unreasonable Results (a) Find the voltage near a 10.0 cm diameter metal sphere that has 8.00 C of excess positive charge on it. (b) What is unreasonable about this result? (c) Which assumptions are responsible?
12. Construct Your Own Problem Consider a battery used to supply energy to a cellular phone. Construct a problem in which you determine the energy that must be supplied by the battery, and then calculate the amount of charge it must be able to move in order to supply this energy. Among the things to be considered are the energy needs and battery voltage. You may need to look ahead to interpret manufacturer’s battery ratings in ampere-hours as energy in joules.

Glossary

electric potential

The electric potential energy per unit charge at a given point in an electric field. It is commonly called voltage and is measured in volts (V).

potential difference (voltage)

The difference in electric potential between two points. It is equal to the change in electric potential energy per unit charge:

[latex]\Delta V=\frac{\Delta\text{PE}}{q}[/latex]

electron volt (eV)

A unit of energy equal to the energy gained by a particle carrying one elementary charge when it moves through a potential difference of 1 V. One electron volt equals [latex]1.60\times10^{-19}\ \text{J}[/latex].

mechanical energy

The total energy of a system due to its motion and position. It is the sum of the kinetic energy and potential energy:

[latex]\text{KE}+\text{PE}[/latex]

For a system acted on only by conservative forces, the total mechanical energy remains constant.