Electric potential (energy)

Electric potential energy is similar to gravitational potential energy, except it can be positive or negative since the electric force can repel as well as attract. Electric potential energy is given by the formula

$E_{\text{el}}=k\frac{q_1{q}_2}{r}\text{.}$

From this we can derive a formula for a change in electric potential energy:

$\mathrm{\Delta }{E}_{\text{el}}=k{q}_1{q}_2(\frac{1}{r_2}-\frac{1}{r_1})\text{.}$

Electric pontential is, confusingly, not the same as electric potential energy. While the latter is measured in joules, electric potential is measured in joules per coulomb. It represents the $E_{\text{el}}$ that a unitary point charge would have at a particular point. It is denoted with $V$ and calculated with

$V=\frac{E_{\text{el}}}{q_2}=\frac{k{q}_1}{r}\text{.}$

$V$ is to $E_{\text{el}}$ as $\overrightarrow{ϵ}$ is to $F_{\text{el}}\text{.}$ They differ only by an $r$ in the denominator:

$\left|\overrightarrow{ϵ}\right|=\frac{V}{r}\text{.}$

Most of the time, we talk not about electric potential but about electric potential difference, also known as voltage:

$\mathrm{\Delta }V=kq(\frac{1}{r_2}-\frac{1}{r_1})\text{.}$

In general, we can relate work to voltage with

$W=\mathrm{\Delta }{E}_{\text{el}}=q_2\mathrm{\Delta }V\text{.}$

For parallel plates, we have a more specific equation. The work done by the electric force to move a charge $q_2$ from one plate to the other is given by

$W=\mathrm{\Delta }{E}_{\text{el}}=q_2\left|\overrightarrow{ϵ}\right|d\text{,}$

where $d$ is the distance between the plates in metres (m). Why were we able to replace $\mathrm{\Delta }V$ by $\left|\overrightarrow{ϵ}\right|d$? Because, for parallel plates,

$\left|\overrightarrow{ϵ}\right|=\frac{\mathrm{\Delta }V}{d}\text{.}$

This is not true for point charges. It looks similar to the general equation $\left|\overrightarrow{ϵ}\right|=V/r$ mentioned above, but that $\mathrm{\Delta }$ makes a big difference. With point charges, the amount of work done is strictly related to the values of $r_1$ and $r_2\text{.}$ Not so with parallel plates—since the electric field is uniform, it doesn’t matter where the particle is between the plates.