# 7.1 Electric potential energy  (Page 3/5)

 Page 3 / 5
$U\left(r\right)=k\frac{qQ}{r}-{U}_{\text{ref}}.$

A convenient choice of reference that relies on our common sense is that when the two charges are infinitely far apart, there is no interaction between them. (Recall the discussion of reference potential energy in Potential Energy and Conservation of Energy .) Taking the potential energy of this state to be zero removes the term ${U}_{\text{ref}}$ from the equation (just like when we say the ground is zero potential energy in a gravitational potential energy problem), and the potential energy of Q when it is separated from q by a distance r assumes the form

$U\left(r\right)=k\frac{qQ}{r}\phantom{\rule{0.2em}{0ex}}\left(z\text{ero reference at}\phantom{\rule{0.2em}{0ex}}r=\infty \right).$

This formula is symmetrical with respect to q and Q , so it is best described as the potential energy of the two-charge system.

## Potential energy of a charged particle

A $+3.0\text{-nC}$ charge Q is initially at rest a distance of 10 cm ( ${r}_{1}$ ) from a $\text{+5}\text{.0-nC}$ charge q fixed at the origin ( [link] ). Naturally, the Coulomb force accelerates Q away from q , eventually reaching 15 cm ( ${r}_{2}$ ). The charge Q is repelled by q , thus having work done on it and losing potential energy.

What is the change in the potential energy of the two-charge system from ${r}_{1}$ to ${r}_{2}?$

## Strategy

Calculate the potential energy with the definition given above: $\text{Δ}{U}_{12}=\text{−}{\int }_{{r}_{1}}^{{r}_{2}}\stackrel{\to }{\text{F}}·d\stackrel{\to }{\text{r}}.$ Since Q started from rest, this is the same as the kinetic energy.

## Solution

We have

$\begin{array}{cc}\text{Δ}{U}_{12}\hfill & =\text{−}{\int }_{{r}_{1}}^{{r}_{2}}\stackrel{\to }{\text{F}}·d\stackrel{\to }{\text{r}}=\text{−}{\int }_{{r}_{1}}^{{r}_{2}}\frac{kqQ}{{r}^{2}}dr=\text{−}{\left[-\frac{kqQ}{r}\right]}_{{r}_{1}}^{{r}_{2}}=kqQ\left[\frac{1}{{r}_{2}}-\frac{1}{{r}_{1}}\right]\hfill \\ & =\left(8.99\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}{\text{Nm}}^{2}{\text{/C}}^{2}\right)\left(5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left[\frac{1}{0.15\phantom{\rule{0.2em}{0ex}}\text{m}}-\frac{1}{0.10\phantom{\rule{0.2em}{0ex}}\text{m}}\right]\hfill \\ & =-4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}\hfill \end{array}$

## Significance

The change in the potential energy is negative, as expected, and equal in magnitude to the change in kinetic energy in this system. Recall from [link] that the change in kinetic energy was positive.

Check Your Understanding What is the potential energy of Q relative to the zero reference at infinity at ${r}_{2}$ in the above example?

It has kinetic energy of $4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}\text{J}$ at point ${r}_{2}$ and potential energy of $9.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}\text{J},$ which means that as Q approaches infinity, its kinetic energy totals three times the kinetic energy at ${r}_{2},$ since all of the potential energy gets converted to kinetic.

Due to Coulomb’s law, the forces due to multiple charges on a test charge Q superimpose; they may be calculated individually and then added. This implies that the work integrals and hence the resulting potential energies exhibit the same behavior. To demonstrate this, we consider an example of assembling a system of four charges.

## Assembling four positive charges

Find the amount of work an external agent must do in assembling four charges $+2.0\phantom{\rule{0.2em}{0ex}}\mu \text{C},$ $+3.0\phantom{\rule{0.2em}{0ex}}\mu \text{C},\phantom{\rule{0.2em}{0ex}}+4.0\phantom{\rule{0.2em}{0ex}}\mu \text{C},$ and $+5.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ at the vertices of a square of side 1.0 cm, starting each charge from infinity ( [link] ).

## Strategy

We bring in the charges one at a time, giving them starting locations at infinity and calculating the work to bring them in from infinity to their final location. We do this in order of increasing charge.

## Solution

Step 1. First bring the $+2.0\text{-}\mu \text{C}$ charge to the origin. Since there are no other charges at a finite distance from this charge yet, no work is done in bringing it from infinity,

${W}_{1}=0.$

Step 2. While keeping the $+2.0\text{-}\mu \text{C}$ charge fixed at the origin, bring the $+3.0\text{-}\mu \text{C}$ charge to $\left(x,y,z\right)=\left(1.0\phantom{\rule{0.2em}{0ex}}\text{cm},0,0\right)$ ( [link] ). Now, the applied force must do work against the force exerted by the $+2.0\text{-}\mu \text{C}$ charge fixed at the origin. The work done equals the change in the potential energy of the $+3.0\text{-}\mu \text{C}$ charge:

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