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

$-\frac{q}{2}$

The correct answer is option (3) : $-\frac{q}{2}$

$U=\frac{kq^2}{l}+\frac{KqQ}{l}+\frac{kqQ}{l}$

Given, $U=0$

$=\frac{kq^2}{l}+\frac{2KqQ}{l}=0$

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