19.2 Electric Potential in a Uniform Electric Field

In the previous section, we explored the relationship between voltage and energy. In this section, we will explore the relationship between voltage and electric field. For example, a uniform electric field $\mathbf{\text{E}}$ is produced by placing a potential difference (or voltage) $\mathrm{\Delta }V$ across two parallel metal plates, labeled A and B. (See Figure 19.5.) Examining this will tell us what voltage is needed to produce a certain electric field strength; it will also reveal a more fundamental relationship between electric potential and electric field. From a physicist’s point of view, either $\mathrm{\Delta }V$ or $\mathbf{\text{E}}$ can be used to describe any charge distribution. $\mathrm{\Delta }V$ is most closely tied to energy, whereas $\mathbf{\text{E}}$ is most closely related to force. $\mathrm{\Delta }V$ is a scalar quantity and has no direction, while $\mathbf{\text{E}}$ is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field strength, a scalar quantity, is represented by $E$ below.) The relationship between $\mathrm{\Delta }V$ and $\mathbf{\text{E}}$ is revealed by calculating the work done by the force in moving a charge from point A to point B. But, as noted in Electric Potential Energy: Potential Difference, this is complex for arbitrary charge distributions, requiring calculus. We therefore look at a uniform electric field as an interesting special case.

The work done by the electric field in Figure 19.5 to move a positive charge $q$ from A, the positive plate, higher potential, to B, the negative plate, lower potential, is

$$W=\mathrm{–\Delta }\text{PE}=-q\mathrm{\Delta }V.$$

The opposite of the potential difference between points A and B, which we denote as $V_{\text{AB'}}$, is

$$-\Delta V=-(V_{\text{B}}-V_{\text{A}})=V_{\text{A}}-V_{\text{B}}=V_{\text{AB}}.$$

Entering this into the expression for work yields

$$W={\text{qV}}_{\text{AB}}.$$

Work is $W=\text{Fd}\text{cos}\theta$; here $\text{cos}\theta =1$, since the path is parallel to the field, and so $W=\text{Fd}$. Since $F=\text{qE}$, we see that $W=\text{qEd}$. Substituting this expression for work into the previous equation gives

$$\mathrm{qEd}={\text{qV}}_{\text{AB}}.$$

The charge cancels, and so the voltage between points A and B is seen to be

$$\begin{array}{c}\left.\begin{array}{c}{V}_{\text{AB}}=\mathrm{Ed}\\ E=\frac{V_{\text{AB}}}{d}\end{array}\right\}\text{(uniform}E\text{- field only),}\end{array}$$

where $d$ is the distance from A to B, or the distance between the plates in Figure 19.5. Note that the above equation implies the units for electric field are volts per meter. We already know the units for electric field are newtons per coulomb; thus the following relation among units is valid:

$$\text{1 N}/\mathrm{C}=\text{1 V}/\mathrm{m}.$$

In more general situations, regardless of whether the electric field is uniform, it points in the direction of decreasing potential, because the force on a positive charge is in the direction of $\mathbf{\text{E}}$ and also in the direction of lower potential $V$. Furthermore, the magnitude of $\mathbf{\text{E}}$ equals the rate of decrease of $V$ with distance. The faster $V$ decreases over distance, the greater the electric field. In equation form, the general relationship between voltage and electric field is

$$E=-\frac{\mathrm{\Delta }V}{\mathrm{\Delta }s},$$

where $\mathrm{\Delta }s$ is the distance over which the change in potential, $\mathrm{\Delta }V$, takes place. The minus sign tells us that $\mathbf{\text{E}}$ points in the direction of decreasing potential. The electric field is said to be the gradient (as in grade or slope) of the electric potential.

For continually changing potentials, $\mathrm{\Delta }V$ and $\mathrm{\Delta }s$ become infinitesimals and differential calculus must be employed to determine the electric field.

Adapted from College Physics 2e by OpenStax (openstax.org), licensed under CC BY-NC-SA 4.0. Changes were made. License: CC-BY-NC-SA-4.0.