Electric Potential Energy: Potential Difference

Electric Potential Energy: Potential Difference

When a free positive charge \(q\) is accelerated by an electric field, such as shown in this figure, 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.

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, \(\text{Δ}\text{PE}\), is crucial, since the work done by a conservative force is the negative of the change in potential energy; that is, \(W=\text{–Δ}\text{PE}\). For example, work \(W\) done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative \(\text{Δ}\text{PE}\). There must be a minus sign in front of \(\text{Δ}\text{PE}\) 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.

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 \(W=\text{Fd}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \) 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 \(F=\text{qE}\), the work, and hence \(\text{ΔPE}\), is proportional to the test charge \(\mathrm{q.}\) 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:

\(V=\cfrac{\text{PE}}{q}\text{.}\)

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

\(\Delta V={V}_{\text{B}}-{V}_{\text{A}}=\cfrac{\Delta \text{PE}}{q}\text{.}\)

The potential difference between points A and B, \({V}_{\text{B}}\phantom{\rule{0.25em}{0ex}}–\phantom{\rule{0.25em}{0ex}}{V}_{\text{A}}\), 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.

\(\text{1 V = 1}\phantom{\rule{0.25em}{0ex}}\cfrac{J}{C}\)

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

\(\Delta V=\cfrac{\text{ΔPE}}{q}\phantom{\rule{0.25em}{0ex}}\text{and ΔPE =}\phantom{\rule{0.25em}{0ex}}q\Delta V\text{.}\)

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 \(\text{ΔPE =}\phantom{\rule{0.25em}{0ex}}q\Delta V\). The car battery can move more charge than the motorcycle battery, although both are 12 V batteries.

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 this figure. The change in potential is \(\Delta V={V}_{\text{B}}{\mathrm{–V}}_{\text{A}}=\text{+12 V}\) and the charge \(q\) is negative, so that \(\text{ΔPE}=q\Delta V\) is negative, meaning the potential energy of the battery has decreased when \(q\) has moved from A to B.