(a) Finding the resultant force on a charge $\mathit{Q}$
(b) Potential Energy of the system
(a) Let us find the force on the charge $\mathit{Q}$ at the point C
${\mathit{F}}_1=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\frac{{\mathit{Q}}^2}{{\left(\mathit{a}\sqrt{2}\right)}^2}=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\left(\frac{{\mathit{Q}}^2}{2{\mathit{a}}^2}\right)$
Force due to the charge $\mathit{q}$ (at $\mathit{B}$ ),

Force due to the charge $\mathit{q}$ (at $\mathit{D}$ ),

Resultant of these two equal forces

Force due to the other charge $\mathit{Q}$

$\therefore$ Net force on charge $\mathit{Q}$ ( at point $\mathit{C}$ )
$\mathrm{F}={\mathit{F}}_1+{\mathit{F}}_{23}=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\frac{\mathrm{Q}}{{\mathit{a}}^2}\left[\frac{\mathit{Q}}{2}+\sqrt{2}\mathit{q}\right]$
This force is directed along $\mathit{A}\mathit{C}$
(For the charge $\mathrm{Q}$ , at the point $\mathrm{A}$ , the force will have the same magnitude but will be directed along $\mathit{C}\mathit{A}$ )
[Note : Don't deduct marks if the student does not write the direction of the net force, $\mathit{F}]$
(b) Potential energy of the system
$=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\left[4\frac{\mathit{q}\mathit{Q}}{\mathit{a}}+\frac{{\mathit{q}}^2}{\mathit{a}\sqrt{2}}+\frac{{\mathit{Q}}^2}{\mathit{a}\sqrt{2}}\right]$
$=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0\mathit{a}}\left[4\mathit{q}\mathit{Q}+\frac{{\mathit{q}}^2}{\sqrt{2}}+\frac{{\mathit{Q}}^2}{\sqrt{2}}\right]$
OR
(a) Finding the magnitude of the resultant force on charge $\mathit{q}$
(b) Finding the work done
(a) Force on charge $\mathit{q}$ due to the charge $-4\mathit{q}$

Force on the charge $\mathit{q}$, due to the charge $2\mathit{q}$

The forces ${\mathit{F}}_1$ and ${\mathit{F}}_2$ are inclined to each other at an angle of ${120}^{\circ }$
Hence, resultant electric force on charge $\mathit{q}$
$\mathit{F}=\sqrt{{\mathit{F}}_1^2+{\mathit{F}}_2^2+2{\mathit{F}}_1{\mathit{F}}_2\cos \mathit{\theta }}$
$=\sqrt{{\mathit{F}}_1^2+{\mathit{F}}_2^2+2{\mathit{F}}_1{\mathit{F}}_2\cos {120}^{\circ }}$
$=\sqrt{{\mathit{F}}_1^2+{\mathit{F}}_2^2-{\mathit{F}}_1{\mathit{F}}_2}$
$=\left(\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\frac{{\mathit{q}}^2}{{\mathit{l}}^2}\right)\sqrt{16+4-8}$
$=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\left(\frac{2\sqrt{3}{\mathit{q}}^2}{{\mathit{l}}^2}\right)$
(b) Net P.E. of the system
$=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\cdot \frac{{\mathit{q}}^2}{\mathit{l}}\left[-4+2-8\right]$
$=\frac{\left(-10\right)}{4\mathit{\pi }{\mathit{\epsilon }}_0}\frac{{\mathit{q}}^2}{\mathit{l}}$