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 (F_e) on Q₂. So a leftward external force (F_ext) must be applied on Q₂ to bring it closer to Q₁. For some reason that I will explain later, let’s arrange for F_ext to match F_e in magnitude at every point between $\displaystyle r=\infty$ and $\displaystyle r=d$ . The total work done by F_ext 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 F_ext matched F_e 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

Q_A, Q_B and Q_C 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 Q_c from the system (i.e. move it to a point at infinity).

Solution

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

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

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

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

Work required

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