Question

Two equal charges $q$ are placed at a distance $2a$ and a third charge $-2q$ is placed at the midpoint. The potential energy of the system is

• A. $\frac{9q^2}{8\pi {ϵ}_{0}a}$
• B. $\frac{q^2}{8\pi {ϵ}_{0}a}$
• C. $\frac{-7q^2}{8\pi {ϵ}_{0}a}$
• D. $\frac{6q^2}{8\pi {ϵ}_{0}a}$

Answer 1

The correct option is C $\frac{-7q^2}{8\pi {ϵ}_{0}a}$

Given two equal charges ${}^{\prime }{q}^{\prime }$ are at a distance $2a$.

A third charge $-2q$ is placed at the mid-point. We have to find the potential energy of the system.

Let $O$ be the midpoint of the charges.

So, potential energy of the system is

$=\frac{1}{4\pi {\epsilon }_0}[\frac{(-2q)\left(q\right)}{a}-\frac{\left(2q\right)\left(q\right)}{a}+\frac{\left(q\right)\left(q\right)}{2a}]$

$=\frac{1}{4\pi {\epsilon }_0}[\frac{-2q^2}{a}-\frac{2q^2}{a}+\frac{q^2}{2a}]$

$=\frac{1}{4\pi {\epsilon }_0}[\frac{-4q^2}{a}+\frac{q^2}{2a}]$

$=\frac{1}{4\pi {\epsilon }_0}\left[\frac{-8q^{2}a+q^{2}a}{2a^2}\right]$

$=\frac{1}{4\pi {\epsilon }_0}\left[\frac{-7q^{2}a}{2a^2}\right]$

$=\frac{1}{4\pi {\epsilon }_0}\left[\frac{-7q^2}{2a}\right]$

$=\frac{-7q^2}{8\pi {\epsilon }_{0}a}$