# 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

$\displaystyle E_{\text{el}} = k \frac{q_{1} q_{2}}{r}$.

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

$\displaystyle \Delta{} E_{\text{el}} = k q_{1} q_{2} \left ( \frac{1}{r_{2}} - \frac{1}{r_{1}} \right )$.

*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
$\displaystyle E_{\text{el}}$
that a unitary point charge would have at a particular point. It is denoted with $\displaystyle V$ and calculated with

$\displaystyle V = \frac{E_{\text{el}}}{q_{2}} = \frac{k q_{1}}{r}$.

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

$\displaystyle \left \lvert \vec{\epsilon} \right \rvert = \frac{V}{r}$.

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

$\displaystyle \Delta{} V = k q \left ( \frac{1}{r_{2}} - \frac{1}{r_{1}} \right )$.

In general, we can relate work to voltage with

$\displaystyle W = \Delta{} E_{\text{el}} = q_{2} \Delta{} V$.

For parallel plates, we have a more specific equation. The work done by the electric force to move a charge $\displaystyle q_{2}$ from one plate to the other is given by

$\displaystyle W = \Delta{} E_{\text{el}} = q_{2} \left \lvert \vec{\epsilon} \right \rvert d$,

where $\displaystyle d$ is the distance between the plates in metres (m). Why were we able to replace $\displaystyle \Delta{} V$ by $\displaystyle \left \lvert \vec{\epsilon} \right \rvert d$? Because, for parallel plates,

$\displaystyle \left \lvert \vec{\epsilon} \right \rvert = \frac{\Delta{} V}{d}$.

This is *not* true for point charges. It looks similar to the general equation
$\displaystyle \left \lvert \vec{\epsilon} \right \rvert = V / r$
mentioned above, but that $\displaystyle \Delta{}$ makes a big difference. With point charges, the amount of work done is strictly related to the values of
$\displaystyle r_{1}$
and
$\displaystyle r_{2}$.
Not so with parallel plates—since the electric field is uniform, it doesn’t matter where the particle is between the plates.