# 7.1 Electric potential energy  (Page 2/5)

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Check Your Understanding If Q has a mass of $4.00\phantom{\rule{0.2em}{0ex}}\mu \text{g},$ what is the speed of Q at ${r}_{2}?$

$K=\frac{1}{2}\phantom{\rule{0.2em}{0ex}}m{v}^{2},\phantom{\rule{0.2em}{0ex}}v=\sqrt{2\frac{K}{m}}=\sqrt{2\frac{4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}\text{J}}{4.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}\text{kg}}}=15\phantom{\rule{0.2em}{0ex}}\text{m/s}$

In this example, the work W done to accelerate a positive charge from rest is positive and results from a loss in U , or a negative $\text{Δ}U$ . A value for U can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.

## Electric potential energy

Work W done to accelerate a positive charge from rest is positive and results from a loss in U , or a negative $\text{Δ}U$ . Mathematically,

$W=\text{−}\text{Δ}U.$

Gravitational potential energy and electric potential energy are quite analogous. Potential energy accounts for work done by a conservative force and gives added insight regarding energy and energy transformation without the necessity of dealing with the force directly. It is much more common, for example, to use the concept of electric potential energy than to deal with the Coulomb force directly in real-world applications.

In polar coordinates with q at the origin and Q located at r , the displacement element vector is $d\stackrel{\to }{\text{l}}=\stackrel{^}{\text{r}}\phantom{\rule{0.2em}{0ex}}dr$ and thus the work becomes

${W}_{12}=\text{−}kqQ{\int }_{{r}_{1}}^{{r}_{2}}\frac{1}{{r}^{2}}\stackrel{^}{\text{r}}·\stackrel{^}{\text{r}}dr=kqQ\frac{1}{{r}_{2}}-kqQ\frac{1}{{r}_{1}}.$

Notice that this result only depends on the endpoints and is otherwise independent of the path taken. To explore this further, compare path ${P}_{1}$ to ${P}_{2}$ with path ${P}_{1}{P}_{3}{P}_{4}{P}_{2}$ in [link] .

The segments ${P}_{1}{P}_{3}$ and ${P}_{4}{P}_{2}$ are arcs of circles centered at q . Since the force on Q points either toward or away from q , no work is done by a force balancing the electric force, because it is perpendicular to the displacement along these arcs. Therefore, the only work done is along segment ${P}_{3}{P}_{4},$ which is identical to ${P}_{1}{P}_{2}.$

One implication of this work calculation is that if we were to go around the path ${P}_{1}{P}_{3}{P}_{4}{P}_{2}{P}_{1},$ the net work would be zero ( [link] ). Recall that this is how we determine whether a force is conservative or not. Hence, because the electric force is related to the electric field by $\stackrel{\to }{\text{F}}=q\stackrel{\to }{\text{E}}$ , the electric field is itself conservative. That is,

$\oint \stackrel{\to }{\text{E}}·d\stackrel{\to }{\text{l}}=0.$

Note that Q is a constant.

Another implication is that we may define an electric potential energy. Recall that the work done by a conservative force is also expressed as the difference in the potential energy corresponding to that force. Therefore, the work ${W}_{\text{ref}}$ to bring a charge from a reference point to a point of interest may be written as

${W}_{\text{ref}}={\int }_{{r}_{\text{ref}}}^{r}\stackrel{\to }{\text{F}}·d\stackrel{\to }{\text{l}}$

and, by [link] , the difference in potential energy $\left({U}_{2}-{U}_{1}\right)$ of the test charge Q between the two points is

$\text{Δ}U=\text{−}{\int }_{{r}_{\text{ref}}}^{r}\stackrel{\to }{\text{F}}·d\stackrel{\to }{\text{l}}.$

Therefore, we can write a general expression for the potential energy of two point charges (in spherical coordinates):

$\text{Δ}U=\text{−}{\int }_{{r}_{\text{ref}}}^{r}\frac{kqQ}{{r}^{2}}dr=\text{−}{\left[-\frac{kqQ}{r}\right]}_{{r}_{\text{ref}}}^{r}=kqQ\left[\frac{1}{r}-\frac{1}{{r}_{\text{ref}}}\right].$

We may take the second term to be an arbitrary constant reference level, which serves as the zero reference:

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