$(a)$ Determine the electrostatic potential energy of a system consisting of two charges $7 \; \mu C$ and $-2 \; \mu C$ (with no external field) placed at $(-9 \; cm, 0, 0)$ and $(9 \; cm, 0, 0)$ respectively.
$(b)$ How much work is required to separate the two charges infinitely away from each other?
$(c)$ Suppose that the same system of charges is now placed in an external electric field $E = A(1/r^2)$; $A = 9 \times 10^5 \; C \cdot m^{-2}$. What would the electrostatic energy of the configuration be?

(N/A) The electrostatic potential energy $U$ is given by $U = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}$.
Here,$q_1 = 7 \times 10^{-6} \; C$,$q_2 = -2 \times 10^{-6} \; C$,and $r = 18 \; cm = 0.18 \; m$.
$U = 9 \times 10^9 \times \frac{(7 \times 10^{-6})(-2 \times 10^{-6})}{0.18} = -0.7 \; J$.
$(b)$ Work required to separate the charges to infinity is $W = U_{\infty} - U = 0 - (-0.7) = 0.7 \; J$.
$(c)$ The potential $V(r)$ due to the field $E = A/r^2$ is $V(r) = \int E \cdot dr = A/r$.
The total energy is $U_{total} = q_1 V(r_1) + q_2 V(r_2) + \frac{q_1 q_2}{4 \pi \varepsilon_0 r_{12}}$.
$U_{total} = (9 \times 10^5) \left( \frac{7 \times 10^{-6}}{0.09} \right) + (9 \times 10^5) \left( \frac{-2 \times 10^{-6}}{0.09} \right) - 0.7$.
$U_{total} = 70 - 20 - 0.7 = 49.3 \; J$.