It is required to hold four equal point charges each having a charge $\text{Q=}\frac{8}{7}(1-2\sqrt{2})\mathrm{C}$ in equilibrium at the corners of a square. Find the point charge, in coulomb, that will do this if placed at the center of the square.

Let us consider the equilibrium of charges Q placed at 2. Then for equilibrium .

${\overrightarrow{F}}_2=\overrightarrow{0}$
$$\begin{gathered} \begin{array}{l}\Rightarrow F_{2x}\widehat{i}+F_{2y}\widehat{j}=\overrightarrow{0}\\ \Rightarrow F_{2x}=0\\ \Rightarrow F_{21}+F_{24}\cos 45+F_{25}\cos 45+F_{23}\cos 90=0 \\ \begin{array}{l}\Rightarrow \frac{Q^2}{4\pi {\epsilon }_0{a}^2}+\frac{Q^2}{4\pi {\epsilon }_0(\sqrt{2}a{)}^2}\frac{1}{\sqrt{2}}+\frac{Qq}{4\pi {\epsilon }_0{\left(\frac{a}{\sqrt{2}}\right)}^2}\frac{1}{\sqrt{2}}+0=0\\ \Rightarrow Q+\frac{Q}{2\sqrt{2}}+\sqrt{2}q=0\\ \Rightarrow Q(2\sqrt{2}+1)+4q=0 \\ \Rightarrow q=-\frac{Q}{4}(2\sqrt{2}+1) \\ \begin{array}{l}\text{ Since }Q=\frac{8}{7}(1-2\sqrt{2})\\ \Rightarrow q=-\frac{1}{4}\left[\frac{8}{7}(1-2\sqrt{2})(1+2\sqrt{2})\right]\\ \Rightarrow q=-\frac{1}{4}\left(\frac{8}{7}\right)(-7)=2C\end{array}\end{array}\end{array} \end{gathered}$$