Question

A charge $(-q)$ and another charge $(Q)$ are kept at two points $A$ and $B$ respectively. Keeping the charge $(+Q)$ fixed at $B$, the charge $(-q)$ at $A$ is moved to another point $C$ such that $ABC$ forms an equilateral triangle of side $l$. The net work done in moving the charge $(-q)$ is

• A. $\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{l}$
• B. $\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{l^2}$
• C. $\frac{1}{4\pi {\epsilon }_0}Qql$
• D. zero

Answer 1

Answer: (D)

According to figure, potential at $A$ and $C$ are equal.
Hence work done in moving $-q$ charge from $A$ to $C$ is zero.