**The answer is:** **(a) -Q/4**

Charge q is in equilibrium since charges A and B exert equal and opposite forces on it.

For equilibrium of charge Q at B;

\(F_{BC}\,+\,F_{AB}=0\)

\(\implies \frac{1}{4\pi\epsilon_o}\frac{qQ}{(\frac{1}{2})^2}\) + \(\frac{1}{4\pi\epsilon_o}\frac{Q.Q}{1^2}\) = 0

\(\implies \frac{1}{4\pi\epsilon_o}\frac{Q}{(1)^2}(4q+Q)\) = 0

\(\implies\) q = \(-\frac{Q}{4}\).