A charge $Q$ is placed at each of the opposite corners of a square. A charge $q$ is placed at each of the other two corners. If the net electrical force on $Q$ is zero, then the $Q/q$ equals:

Let the side of the square be $d$. Consider charge $Q$ at corner $(d,d)$.Forces acting on it:From $q$ at $(d,0)$: $F_1=\frac{kQq}{d^2}$ (along $+x$)From $q$ at $(0,d)$: $F_2=\frac{kQq}{d^2}$ (along $+y$)From $Q$ at $(0,0)$: $F_3=\frac{k{Q}^2}{(\sqrt{2}d{)}^2}=\frac{k{Q}^2}{2d^2}$ (along $45^{\circ }$ diagonal)Resultant of $F_1$ and $F_2$ is $\sqrt{2}\frac{kQq}{d^2}$.For net force to be zero:
$$\sqrt{2}\frac{kQq}{d^2}+\frac{k{Q}^2}{2d^2}=0$$
$$\frac{kQ}{d^2}\left[\sqrt{2}q+\frac{Q}{2}\right]=0$$
$$\sqrt{2}q=-\frac{Q}{2}⟹\frac{Q}{q}=-2\sqrt{2}$$