Explanation

(d) For equilibrium net electric force on any charge (say charge$\ –Q\ at\ A$) should be zero. Hence,
${\overrightarrow{F}}_{A} = {\overrightarrow{F}}_{AB} + {\overrightarrow{F}}_{AD} + {\overrightarrow{F}}_{AC} + {\overrightarrow{F}}_{AO} = \overrightarrow{O},{\overrightarrow{F}}_{AB} = \frac{1}{4\pi\varepsilon_{0}}\frac{{\ \ Q}^{2}}{a^{2}}$ along
$BA$,$\ {\overrightarrow{F}}_{AD}$=$\frac{1}{4\pi\varepsilon_{0}}.\frac{{\ Q}^{2}}{a^{2}}$ along $DA$, ${\overrightarrow{F}}_{AC}$=$\frac{1}{4\pi\varepsilon_{0}}.\frac{{\ Q}^{2}}{a^{2}}\ $along
$CA,\ $and ${\overrightarrow{F}}_{OA}$ =$- \frac{1}{4\pi\varepsilon_{0}}.\frac{{\ \ 2Qq}^{\ }}{a^{2}}$ along $AO$
Resultant of ${\overrightarrow{F}}_{AB}$ and ${\overrightarrow{F}}_{AD}$=$\frac{1}{4\pi\varepsilon_{0}}.\frac{{\ \ Q}^{2}}{\ a^{2}}\sqrt{2}$ along $COA,$
$\therefore\ {\overrightarrow{F}}_{A} = \frac{1}{4\pi\varepsilon_{0}}.\frac{{\ \ Q}^{2}}{a^{2}}\sqrt{2} + \frac{1}{4\pi\varepsilon_{0}}.\frac{{\ \ Q}^{2}}{2a^{2}} - \frac{1}{4\pi\varepsilon_{0}}.\frac{{\ 2Qq}^{\ }}{a^{2}}$= $\overrightarrow{O}$
$\Rightarrow \ \ q = \frac{Q}{4}$(1+2$\sqrt{2}$)
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