(a) Finding the resultant force on a charge $Q$ (b) Potential Energy of the system (a) Let us find the force on the charge $Q$ at the point C $F_1=\;\frac{1}{4 \pi \epsilon_0}\;\frac{Q^2}{(a\sqrt{2})^2}=\;\frac{1}{4 \pi \epsilon_0}\left(\;\frac{Q^2}{2 a^2}\right)$ $\;\text{ (along }\; A C \;\text{ ) }$ Force due to the charge $q$ (at $B$ ), $F_2=\;\frac{1}{4 \pi \epsilon_0}\;\frac{q Q}{a^2}\;\text{ along }\; B C$ Force due to the charge $q$ (at $D$ ), $F_3=\;\frac{1}{4 \pi \epsilon_0}\;\frac{q Q}{a^2}\;\text{ along }\; D C$ Resultant of these two equal forces Force due to the other charge $Q$ $F_{23}=\;\frac{1}{4 \pi \epsilon_0}\;\frac{q Q(\sqrt{2})}{a^2}\;\text{ (along }\; A C \;\text{ ) }\;$ $\therefore$ Net force on charge $Q$ ( at point $C$ ) $F=F_1+F_{23}=\;\frac{1}{4 \pi \epsilon_0}\;\frac{Q}{a^2}\left[\;\frac{Q}{2}+\sqrt{2} q\right]$ This force is directed along $A C$(For the charge $Q$ , at the point $A$ , the force will have the same magnitude but will be directed along $C A$ )[Note : Don't deduct marks if the student does not write the direction of the net force, $F]$(b) Potential energy of the system $\;=\;\frac{1}{4 \pi \epsilon_0}\left[4\;\frac{q Q}{a}+\;\frac{q^2}{a\sqrt{2}}+\;\frac{Q^2}{a\sqrt{2}}\right]$ $\;=\;\frac{1}{4 \pi \epsilon_0 a}\left[4 q Q+\;\frac{q^2}{\sqrt{2}}+\;\frac{Q^2}{\sqrt{2}}\right]$ OR(a) Finding the magnitude of the resultant force on charge $q$(b) Finding the work done (a) Force on charge $q$ due to the charge $-4 q$ $F_1=\;\frac{1}{4 \pi \epsilon_0}\left(\;\frac{4 q^2}{l^2}\right)\;\text{, along }\; A B$Force on the charge $q$, due to the charge $2 q$ $F_2=\;\frac{1}{4 \pi \epsilon_0}\left(\;\frac{2 q^2}{l^2}\right)\;\text{, along }\; C A$The forces $F_1$ and $F_2$ are inclined to each other at an angle of $120^{\circ}$ Hence, resultant electric force on charge $q$$\;F=\sqrt{F_1^2+F_2^2+2 F_1 F_2 \cos \theta}$$\;=\sqrt{F_1^2+F_2^2+2 F_1 F_2 \cos 120^{\circ}}$$\;=\sqrt{F_1^2+F_2^2-F_1 F_2}$$\;=\left(\;\frac{1}{4 \pi \epsilon_0}\;\frac{q^2}{l^2}\right)\sqrt{16+4-8}$$\;=\;\frac{1}{4 \pi \epsilon_0}\left(\;\frac{2\sqrt{3} q^2}{l^2}\right)$(b) Net P.E. of the system$\;=\;\frac{1}{4 \pi \epsilon_0}\cdot\;\frac{q^2}{l}[-4+2-8]$$\;=\;\frac{-10}{4 \pi \epsilon_0}\;\frac{q^2}{l}$$\;\therefore\;\text{ Work done }\;=\;\frac{10 q^2}{4 p \epsilon_0 l}=\;\frac{5 q^2}{2 \pi \epsilon_0 l}$