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 [link].) 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 [link] to move a positive charge $q$ from A, the positive plate, higher potential, to B, the negative plate, lower potential, is

The potential difference between points A and B is

Entering this into the expression for work yields

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

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

where $d$ is the distance from A to B, or the distance between the plates in [link]. 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:

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

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.