# 4.2: Electric Potential Energy - Potential Difference (2023)

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##### Learning objectives

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

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

When a free positive charge $$q$$ is accelerated by an electric field, as shown in Figure $$\PageIndex{1}$$, it is given kinetic energy. The process is analogous to an object accelerated by a gravitational field. It is as if the charge descends an electric hill where its electrical potential energy is converted to kinetic energy. Let's explore the work done on a charge $$q$$ by the electric field in this process, so that we can 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. This is exactly analogous to the gravitational force in the absence of dissipative forces like 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 electrical potential energy, which has units of joules (J). The change in potential energy, $$\Delta \mathrm{PE}$$, is crucial, since the work done by a conservative force is the negative of the change in potential energy; that is, $$W=-\Delta \mathrm{PE}$$. For example, the work $$W$$ done to accelerate a positive charge from rest is positive and results from a loss of PE, or a negative $$\Delta \mathrm{PE}$$. There must be a minus sign in front of $$\Delta \mathrm{PE}$$ for $$W$$ to be positive. PE can be found at any point by taking one point as a reference and calculating the work required to move a charge to the other point.

##### POTENTIAL ENERGY

$$W=-\Delta \mathrm{PE}$$. For example, the work $$W$$ done in accelerating a positive charge from rest is positive and results from a loss of PE, or negative $$\Delta \mathrm{PE}$$. There must be a minus sign in front of $$\Delta \mathrm{PE}$$ to make $$W$$ positive. PE can be found at any point by taking one point as a reference and calculating the work required to move a charge to the other point.

Gravitational potential energy and electrical potential energy are quite analogous. Potential energy represents the work done by a conservative force and provides additional information about energy and energy transformation without the need to deal directly with the force. It is much more common, for example, to use the concept of voltage (related to electrical potential energy) than to directly address the Coulomb force.

Calculating the work directly is generally difficult, since $$W=Fd\cos \theta$$ and the direction and magnitude of $$F$$ can be complex for multiple charges, for irregularly shaped objects, and along arbitrary paths. . But we do know that, since $$F=qE$$, the work, and therefore $$\Delta \mathrm{PE}$$, is proportional to the proof load $$q$$ To have a physical quantity that is independent of the test load, we defineelectric potential$$V$$ (or simply potential, since electrical is understood) is the potential energy per unit charge:

$V=\dfrac{\mathrm{PE}}{q}.$

##### ELECTRIC POTENTIAL

This is the electrical potential energy per unit charge.

$V=\dfrac{\mathrm{PE}}{q}$

Since PE is proportional to $$q$$, the dependency on $$q$$ cancels. Therefore $$V$$ does not depend on $$q$$. The change in potential energy $$\Delta \mathrm{PE}$$ is crucial, so we are concerned with the potential difference or potential difference $$\Delta V$$ between two points, where

$\Delta V =V_{B}-V_{A}=\dfrac{\Delta \mathrm{PE}}{q}.$

Hepotential differencebetween points A and B, $$V_{B}-V_{A}$$, is defined as the change in potential energy of a charge $$q$$ moved from A to B, divided by the charge. The units of potential difference are joules per coulomb, named volt (V) after Alessandro Volta.

$1\mathrm{V}=1\mathrm{\dfrac{J}{C}}$

##### POTENTIAL DIFFERENCE

The potential difference between points A and B, $$V_{B}-V_{A}$$, is defined as the change in potential energy of a charge $$q$$ moved from A to B, divided by load . The units of potential difference are joules per coulomb, named volt (V) after Alessandro Volta.

$1\mathrm{V}=1\mathrm{\dfrac{J}{C}}$

the familiar termVoltageis the common name for the potential difference. It must be taken into account that whenever a voltage is quoted, it is understood that it is the potential difference between two points. For example, each battery has two terminals, and their voltage is the potential difference between them. More fundamentally, the point you choose as zero volts is arbitrary. This is analogous to the fact that gravitational potential energy has an arbitrary zero, like sea level or perhaps the floor of a conference room.

In short, the relationship between potential difference (or voltage) and electric potential energy is given by

$\Delta V=\dfrac{\Delta \mathrm{PE}}{q}\: \mathrm{and}\: \Delta \mathrm{PE}=q\Delta V.$

##### POTENTIAL DIFFERENCE AND ELECTRICAL POTENTIAL ENERGY

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

$\Delta =\dfrac{\Delta \mathrm{PE}}{q}\: \mathrm{y}\: \Delta \mathrm{PE}=q\Delta V.$

The second equation is equivalent to the first.

Voltage is not the same as energy. Voltage is energy per unit charge. Therefore, a motorcycle battery and a car battery can have the same voltage (more precisely, the same potential difference between the battery terminals), but one stores much more energy than the other since $$\Delta PE=q\Delta V$$. The car battery can move more charge than the motorcycle battery, although both are 12 V batteries.

##### Example $$\PageIndex{1}$$: Calculate 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

Saying that we have a 12.0 V battery means that its terminals have a potential difference of 12.0 V. When a battery of this type moves charge, it subjects it to a potential difference of 12.0 V, and the charge it is given a change in potential energy equal to $$\Delta PE=q\Delta V$$.

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

Solution

For the motorcycle battery, $$q=5000 \mathrm{C}$$ and $$\Delta =12.0\mathrm{V}$$. The total energy delivered by the motorcycle battery is

$\Delta \mathrm{PE}_{ciclo}=(5000\mathrm{C})(12.0\mathrm{V})$

$=(5000\mathrm{C})(12.0\mathrm{J/C})$

$=6.00\times 10^{4}\mathrm{J}.$

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

$\Delta \mathrm{PE}_{car}=(60.000\mathrm{C})(12,0\mathrm{V})$

$=7.20\times 10^{5}\mathrm{J}.$

Discussion

Although voltage and power are related, they are not the same. The battery voltages are identical, but the power supplied by each is quite different. Also note that when a battery discharges, some of its power is used internally and the voltage across its terminals drops, such as when headlights dim due to a low car battery. The power supplied by the battery is still calculated as in this example, but not all power is available for external use.

Note that the energies calculated in the example above are absolute values. The potential energy change of the battery is negative, since it loses energy. These batteries, like many electrical systems, actually move negative charge, particularly electrons. Batteries repel electrons from their negative (A) terminals through whatever circuitry is involved and attract them to their positive (B) terminals, as shown in Figure $$\PageIndex{2}$$. The change in potential is $$\Delta V =V_{B}-V_{A}=+12\mathrm{V}$$ and the charge $$q$$ is negative, so $$\Delta \mathrm { PE}=q\Delta V$$ is negative, which means that the potential energy of the battery has decreased as $$q$$ has moved from A to B.

##### Example $$\PageIndex{2}$$: How many electrons move through a lighthouse every second?

When a 12.0 V car battery powers 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 by the equation $$\Delta \mathrm{PE}=q\Delta V$$. A 30.0 W lamp uses 30.0 joules per second. Since the battery loses energy, we have $$\Delta \mathrm{PE}=-30.0J$$ and, since the electrons go from the negative to the positive terminal, we see that $$\Delta V=+12.0V$$.

Solution

To find the charge $$q$$ moved, we solve the equation $$\Delta \mathrm{PE}=q\Delta V$$:

$q=\dfrac{\Delta \mathrm{PE}}{\Delta V}.$

Inputting the values ​​for $$\Delta PE$$ and $$\Delta V$$, we get

$q=\dfrac{-30.0\mathrm{J}}{+12.0\mathrm{V}}=\dfrac{-30.0\mathrm{J}}{+12.0\mathrm{J/C}}=-2.50 \mathrm{C}.$

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

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

Discussion

This is a very large number. It's no wonder we don't normally observe individual electrons, since there are so many present in ordinary systems. In fact, electricity had been in use for many decades before moving charges in many circumstances were determined to be negative. Positive charge moving in the opposite direction of negative charge often produces identical effects; this makes it difficult to determine which one is moving or if both are moving.

## The electron volt

The energy per electron is very small in macroscopic situations like the one in the previous example: a small fraction of a joule. But on a submicroscopic scale, that energy per particle (electron, proton, or ion) can be of great importance. For example, even a small fraction of a joule can be enough for these particles to destroy organic molecules and damage living tissue. The particle can cause direct collision damage, or it can create damaging X-rays, which can also cause damage. It is useful to have a unit of energy related to submicroscopic effects. The figure $$\PageIndex{3}$$ shows a situation related to the definition of said unit of energy. An electron is accelerated between two charged metal plates as it would in a television tube or an old model oscilloscope. The electron is given kinetic energy which is then converted into 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 $$\Delta PE=q\Delta V$$, we can think of the joule as a coulomb-volt.

On the submicroscopic scale, it is more convenient to define a unit of energy calledelectronvoltio(eV), which is the energy given to a fundamental charge accelerated through a potential difference of 1 V. In equation form,

$1\mathrm{ev}=(1.60\times 10^{-19}\mathrm{C})(1\mathrm{V})=(1.60\times 10^{-19}\mathrm{C}) (1\mathrm{J/C})$

$=1.60\times 10^{-19}J.$

##### ELECTRON VOLT

On the submicroscopic scale, it is more convenient to define a unit of energy called an electron volt (eV), which is the energy given to a fundamental charge accelerated through a potential difference of 1 V..In equation form,

$1 \mathrm{eV}=(1,60\times 10^{-19} \mathrm{C})(1 \mathrm{V})=(1,60\times 10^{-19} \mathrm{C}) (1\mathrm{J/C})$

$=1.60\times 10^{-19} \mathrm{C}$

An electron accelerated through a potential difference of 1 V receives 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 receive 200 eV of energy. These simple relationships between the accelerating voltage and the charges of the particles make the electron volt a simple and convenient unit of energy in such circumstances.

##### CONNECTIONS: POWER UNITS

The electron volt (eV) is the most common unit of energy for submicroscopic processes. This will be particularly noticeable in the chapters on modern physics. Energy is so important to so many issues that there is a tendency to define a special unit of energy for each major issue. 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 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 can break up to 6,000 of these molecules ( $$30,000 \mathrm{eV} \div 5\mathrm{eV}$$ per molecule $$=6000$$ molecules). Nuclear decay energies are on the order of 1 MeV (1,000,000 eV) per event and therefore can cause significant biological damage.

## energy conservation

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 constant.

Mechanical energyis the sum of the kinetic energy and the potential energy of a system; that is, $$KE + PE=\: \mathrm{constant}$$. A loss of PE from a charged particle is converted into an increase to its KE. Here PE is the electric potential energy. The conservation of energy is expressed in equation form as

$\mathrm{KE}+\mathrm{PE}=\mathrm{constante}$

o

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

where i and f represent the initial and final conditions. As we have discovered many times before, considering energy can give us ideas and make problem solving easier.

##### Example $$\PageIndex{3}$$: Electrical potential energy converted into kinetic energy

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

Strategy

We have a system with only conservative forces. Assuming that the electron is accelerated in a vacuum and neglecting the gravitational force (we will verify this assumption later), all electrical potential energy is converted to kinetic energy. We can identify the initial and final forms of energy as $$\mathrm{KE}_{i}=0,\mathrm{KE}_{f}=\dfrac{1}{2}mv^{2}, \mathrm {PE}_{i}=qV,\: \mathrm{y}\: \mathrm{PE}_{f}=0$$.

Solution

The conservation of energy indicates that

$\mathrm{KE}_{i}+\mathrm{PE}_{i}=\mathrm{KE}_{f}+\mathrm{PE}_{f}$

Entering the forms identified above, we obtain

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

We solve this for $$v$$:

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

Inputting values ​​for $$q,\: V,\: \mathrm{y}\: m$$ yields

$v=\sqrt{\dfrac{2(-1.60\times 10^{-19}\mathrm{C})(-100 \mathrm{J/C})}{9.11\times 10^{-31} \mathrm{kg}}}$

$=5.93\times 10^{6} \mathrm{m/s}.$

Discussion

Note that both the charge and the initial voltage are negative, as inCipher. From the discussions inElectric charge and electric field., we know that the electrostatic forces on small particles are generally very large compared to the gravitational force. The high final speed confirms that the gravitational force here is negligible. The high velocity also indicates how easy it is to accelerate electrons with small voltages due to their very small mass. In electron guns voltages much higher than 100 V are often used in this problem. Those higher voltages produce such great electron velocities that relativistic effects must be taken into account. That is why in this example it is considered (accurately) a low voltage.

## Summary

• Electric potential is potential energy per unit charge.
• The potential difference between points A and B, $$V_{\mathrm{B}}-V_{\mathrm{A}}$$, defined as the change in potential energy of a moved charge $$q$$ from A to B, is equal to the change in potential energy divided by the charge. The potential difference is commonly called voltage, represented by the symbol $$\Delta V$$.

$$\Delta V= \dfrac{\Delta \mathrm{PE}}{q}\: \mathrm{y}\: \Delta \mathrm{PE}=q\Delta V.$$

• One electron volt is the energy given to a fundamental charge accelerated through a potential difference of 1 V. In equation form,

$$1\mathrm{eV}=(1,60\times 10^{-19}\mathrm{C})(1 \mathrm{V})=(1,60\times 10^{-19}\mathrm{C}) (1 \mathrm{J/C})$$

$$=1.60\times 10^{-19}\mathrm{J}.$$

• Mechanical energy is the sum of the kinetic energy and the potential energy of a system, that is, $$\mathrm{KE}+\mathrm{PE}$$ This sum is a constant.

## Glossary

electric potential
potential energy per unit charge
potential difference (or voltage)
change in potential energy of a charge moved from one point to another, divided by the charge; The units of potential difference are joules per coulomb, known as a volt.
electronvoltio
the energy given to a fundamental charge accelerated through a potential difference of one volt
mechanical energy
sum of the kinetic energy and the potential energy of a system; this sum is a constant
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