At each  of the four corners of a square of side a , a charge + q is placed freely . What charge should be placed at the centre of the square  so that the whole system is in equilibrium ..?

Dear Student,

The given system consist 4 point charges +q placed at 4 corners A, B, C and D of  a square of side a and a charge Q at the centre of the square. Any system of charges is in equilibrium if its total electric potential energy is equal to zero.

$$\begin{gathered} \mathrm{Hence}, {\mathrm{U}}_{\mathrm{system}}=\mathrm{PE} \mathrm{due} \mathrm{to} \mathrm{all} \mathrm{those} \mathrm{pairs} \mathrm{in} \mathrm{which} \mathrm{charges} \mathrm{placed} \mathrm{at} \mathrm{the} \mathrm{corners} \mathrm{of} \mathrm{the} \mathrm{sides} \\ \mathrm{of} \mathrm{square}+\mathrm{PE} \mathrm{due} \mathrm{to} \mathrm{all} \mathrm{those} \mathrm{pairs} \mathrm{in} \mathrm{which} \mathrm{charges} \mathrm{are} \mathrm{placed} \mathrm{at} \mathrm{the} \mathrm{ends} \mathrm{of} \mathrm{diagonal} \mathrm{of} \\ \mathrm{square}+\mathrm{PE} \mathrm{due} \mathrm{to} \mathrm{pairs} \mathrm{formed} \mathrm{by} \mathrm{any} \mathrm{charge} \mathrm{at} \mathrm{the} \mathrm{corner} \mathrm{and} \mathrm{charge} \mathrm{at} \mathrm{the} \mathrm{centre}. \\ {\mathrm{U}}_{\mathrm{system}}=4\times \frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{{\mathrm{q}}^2}{\mathrm{a}}+2\times \frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{{\mathrm{q}}^2}{\mathrm{a}\sqrt{2}}+4\times \times \frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{qQ}}{{\displaystyle \raisebox{1ex}{$\mathrm{a}$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}} \\ {\mathrm{U}}_{\mathrm{system}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{4\mathrm{q}}{\mathrm{a}}\left[\mathrm{q}+\frac{\mathrm{q}}{2\sqrt{2}}+\mathrm{Q}\sqrt{2}\right] \\ \mathrm{The} \mathrm{system} \mathrm{will} \mathrm{be} \mathrm{in} \mathrm{equilibrium} \mathrm{if} \\ {\mathrm{U}}_{\mathrm{system}}=0 \\ \frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{4\mathrm{q}}{\mathrm{a}}\left[\mathrm{q}+\frac{\mathrm{q}}{2\sqrt{2}}+\mathrm{Q}\sqrt{2}\right]=0 \\ \mathrm{q}+\frac{\mathrm{q}}{2\sqrt{2}}+\mathrm{Q}\sqrt{2}=0 \\ \mathrm{Q}\sqrt{2}=-\left(\mathrm{q}+\frac{\mathrm{q}}{2\sqrt{2}}\right) \\ \mathrm{Q}=-\left(\frac{\mathrm{q}}{\sqrt{2}}+\frac{\mathrm{q}}{4}\right) \\ \mathrm{Q}=-\mathrm{q}\left(\frac{2\sqrt{2}+1}{4}\right) \\ \mathrm{Q}=-\frac{\mathrm{q}}{4}\left(2\sqrt{2}+1\right) \end{gathered}$$