A system of two charges $7 \mu C$ and $-2 \mu C$ is placed at $(-9 ext{ cm}, 0, 0)$ and $(9 ext{ c

Question

A system of two charges $7 \mu C$ and $-2 \mu C$ is placed at $(-9	ext{ cm}, 0, 0)$ and $(9	ext{ cm}, 0, 0)$ respectively in an external electric field $E = A(1/r^2)$, where $A = 9 	imes 10^5	ext{ N C}^{-1}	ext{ m}^2$. What is the total electrostatic energy of the configuration?

Professor PC Thomas Classes · Answer

The potential due to the external field is $V(r) = \int -E dr = A/r$. The total energy $U = q_1 V(r_1) + q_2 V(r_2) + \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r_{12}}$. $U = \frac{7 	imes 10^{-6} 	imes 9 	imes 10^5}{0.09} + \frac{-2 	imes 10^{-6} 	imes 9 	imes 10^5}{0.09} - 0.7 = 70 - 20 - 0.7 = 49.3	ext{ J}$.