# 12.2.2 Derivation of EPE Formula

Consider two positive point charges Q1 and Q2. Let’s have Q1 bound at position $\displaystyle r=0$ and Q2 initially at rest and infinitely far away. What does it take to move Q2 from $\displaystyle r=\infty$ to some position $\displaystyle r=d$?

Obviously, Q1 exerts a rightward electrical repulsion force (Fe) on Q2. So a leftward external force (Fext) must be applied on Q2 to bring it closer to Q1. For some reason that I will explain later, let’s arrange for Fext to match Fe in magnitude at every point between $\displaystyle r=\infty$ and $\displaystyle r=d$. The total work done by Fext during the journey is thus \displaystyle \displaystyle \begin{aligned}WD&=\int_{\infty }^{d}{{{{F}_{{ext}}}dr}}\\&=\int_{\infty }^{d}{{-{{F}_{e}}dr}}\\&=\int_{\infty }^{d}{{-k\frac{{{{Q}_{1}}{{Q}_{2}}}}{{{{r}^{2}}}}dr}}\\&=\left[ {k\frac{{{{Q}_{1}}{{Q}_{2}}}}{r}} \right]_{\infty }^{d}\\&=k\frac{{{{Q}_{1}}{{Q}_{2}}}}{d}-k\frac{{{{Q}_{1}}{{Q}_{2}}}}{\infty }\\&=k\frac{{{{Q}_{1}}{{Q}_{2}}}}{d}\end{aligned}

Since Fext matched Fe in magnitude throughout the journey, Q2 did not gain or lose any KE. So $\displaystyle \displaystyle k\frac{{{{Q}_{1}}{{Q}_{2}}}}{d}$ must represent the increase in EPE between $\displaystyle r=\infty$ and $\displaystyle r=d$. But EPE is 0 at $\displaystyle r=\infty$, so the EPE at $\displaystyle r=d$ is simply $\displaystyle \displaystyle EPE=k\frac{{{{Q}_{1}}{{Q}_{2}}}}{d}$

To include the sign or not?

This formula is valid even for negative charges. So you should always include the signs of the charges when applying this formula. The formula will automatically calculate EPE to be positive when Q1 and Q2 are of the same signs, but negative when they are of opposite signs.

Example

QA, QB and QC are point charges of +1.1 pC, +2.2 pC and −3.3 pC, bound at the vertices of an equilateral triangle of side 3.0 um as shown in the diagram below. Calculate

a) total EPE of the system

b) the amount of work required to remove Qc from the system (i.e. move it to a point at infinity).

Solution

a)

EPE is stored between 3 pairs of point charges. $\displaystyle {{U}_{{AB}}}=k\frac{{{{Q}_{A}}{{Q}_{B}}}}{d}=(8.99\times {{10}^{9}})\frac{{(+1.1)(+2.2){{{({{{10}}^{{-12}}})}}^{2}}}}{{3.0\times {{{10}}^{{-6}}}}}=0.725\times {{10}^{{-8}}}\text{ J}$ $\displaystyle {{U}_{{AC}}}=k\frac{{{{Q}_{A}}{{Q}_{C}}}}{d}=(8.99\times {{10}^{9}})\frac{{(+1.1)(-3.3){{{({{{10}}^{{-12}}})}}^{2}}}}{{3.0\times {{{10}}^{{-6}}}}}=-1.088\times {{10}^{{-8}}}\text{ J}$ $\displaystyle {{U}_{{BC}}}=k\frac{{{{Q}_{B}}{{Q}_{C}}}}{d}=(8.99\times {{10}^{9}})\frac{{(+2.1)(-3.3){{{({{{10}}^{{-12}}})}}^{2}}}}{{3.0\times {{{10}}^{{-6}}}}}=-2.077\times {{10}^{{-8}}}\text{ J}$

Total EPE \displaystyle \begin{aligned}&={{U}_{{AB}}}+{{U}_{{AC}}}+{{U}_{{BC}}}\\&=[0.7252+(-1.088)+(-2.077)]\times {{10}^{{-8}}}\\&=-2.44\times {{10}^{{-8}}}\text{ J}\end{aligned}

b)

Before Qc is removed:            initial total EPE $\displaystyle ={{U}_{i}}={{U}_{{AB}}}+{{U}_{{AC}}}+{{U}_{{BC}}}=-2.44\times {{10}^{{-8}}}\text{ J}$

After Qc is removed:               final total EPE $\displaystyle ={{U}_{f}}={{U}_{{AB}}}=+0.7252\times {{10}^{{-8}}}\text{ J}$

Work required \displaystyle \begin{aligned}&={{U}_{f}}-{{U}_{i}}\\&=(0.725\times {{10}^{{-8}}})-(-2.44\times {{10}^{{-8}}})\\&=3.17\times {{10}^{{-8}}}\text{ J}\end{aligned}

Explanation Video

How to Calculate Total EPE of a System of Charges