# 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 $\overrightarrow{\u03f5}$ is to ${F}_{\text{el}}\text{.}$ They differ only by an $r$ in the denominator:

$\left|\overrightarrow{\u03f5}\right|=\frac{V}{r}\text{.}$

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

$W=\mathrm{\Delta}{E}_{\text{el}}={q}_{2}\left|\overrightarrow{\u03f5}\right|d\text{,}$

where $d$ is the distance between the plates in metres (m). Why were we able to replace $\mathrm{\Delta}V$ by $\left|\overrightarrow{\u03f5}\right|d$? Because, for parallel plates,

$\left|\overrightarrow{\u03f5}\right|=\frac{\mathrm{\Delta}V}{d}\text{.}$

This is *not* true for point charges. It looks similar to the general equation $\left|\overrightarrow{\u03f5}\right|=V/r$ 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.