Question

The charges $Q$, $+q$ and $+q$ are placed at the vertices of an equilateral triangle of side $l$. If the net electrostatic energy of the system is zero, then $Q$ is equal to :

• A. $\frac{-q}{2}$
• B. $-q$
• C. $\frac{+q}{2}$
• D. Zero

Answer 1

The correct option is A $\frac{-q}{2}$

$E=2k\frac{Qq}{r}+\frac{k{q}^2}{r}=0$

$Q=\frac{-q}{2}$