Charges Q, q, Q, q are placed at the comers A, B, C, D of a square, respectively. If the resultant force on the charge Q is zero due to other charges, what is the relation between Q and q?

Different forces on Q are shown in figure. a is the side of square.

${\mathrm{F}}_{\mathrm{CA}}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}.\frac{{\mathrm{Q}}^2}{2{\mathrm{a}}^2}\mathrm{along}\mathrm{AC}$

If the resultant force on C is to be zero, the forces on C due to D and B must be along CD and CB, respectively. Thus, q must be negative. Resultant force, in the direction of AC is given by

$\begin{array}{l}\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}.\frac{\mathrm{Qq}}{{\mathrm{a}}^2}.\frac{1}{\sqrt{2}}+\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}.\frac{\mathrm{Qq}}{{\mathrm{a}}^2}.\frac{1}{\sqrt{2}}=\sqrt{2}.\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}.\frac{\mathrm{Qq}}{{\mathrm{a}}^2}\\ \Rightarrow \sqrt{2}.\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}.\frac{\mathrm{Qq}}{{\mathrm{a}}^2}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}.\frac{\mathrm{Qq}}{2{\mathrm{a}}^2}\\ \left|\mathrm{Q}\right|=2\sqrt{2}\left|\mathrm{q}\right|\\ \mathrm{As}\mathrm{q}\mathrm{i}\mathrm{negative},\mathrm{Q}=-2\sqrt{2}\mathrm{q}\end{array}$