Four charges equal to $-Q$ are placed at the four corners of a square and a charge $q$ is at its centre. If the system is in equilibrium, the value of $q$ is:

For equilibrium, net force on any corner charge $(-Q)$ must be zero.Repulsive forces on $-Q$ from other three $-Q$ charges:From two adjacent corners (distance $a$): $F_1=k\frac{Q^2}{a^2}$. Their resultant $=\sqrt{2}\frac{k{Q}^2}{a^2}$ along the diagonal.From opposite corner (distance $\sqrt{2}a$): $F_2=k\frac{Q^2}{(\sqrt{2}a{)}^2}=\frac{k{Q}^2}{2a^2}$.Total repulsive force $F_R=\frac{k{Q}^2}{a^2}\left(\sqrt{2}+\frac{1}{2}\right)$.Attractive force from center charge $q$ (distance $a/\sqrt{2}$): $F_A=\frac{kQq}{(a/\sqrt{2}{)}^2}=\frac{2kQq}{a^2}$.Set $F_A=F_R$:
$$\frac{2kQq}{a^2}=\frac{k{Q}^2}{a^2}\left(\frac{2\sqrt{2}+1}{2}\right)$$
$$q=\frac{Q}{4}(1+2\sqrt{2})$$