Four point charges Q, q, Q and q are placed at the corners of a square of side ‘a’ as shown in the figure.

Find the
(a) resultant electric force on a charge Q and
(b) potential energy of this system.

OR
(a) Three point charges q – 4q and 2q are placed at the vertices of an equilateral triangle ABC of side ‘l’ as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q.

b) Find out the amount of the work done to separate the charges at infinite distance.

(a) Force on charge Q at B due to charge q at A,
${\mathrm{F}}_1=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{n}}}\frac{\mathrm{qQ}}{{\mathrm{a}}^2} \mathrm{along} \overrightarrow{\mathrm{BY}}$

Force on charge Q at B due to charge q at c.
${\mathrm{F}}_2=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{Tr}}{{\mathrm{a}}^2}\text{ along }\mathrm{BX}$
Resultant force, F =$\sqrt{{\mathrm{F}}_1^2+{\mathrm{F}}_2^2} \mathrm{along} \overrightarrow{\mathrm{RN}}$
$\begin{array}{l}\mathit{=}\sqrt{\mathit{2}{\left(\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{a^2}\right)}^{\mathit{2}}}\\ \mathit{=}\frac{\mathit{q}\mathit{Q}\sqrt{\mathit{2}}}{\mathit{4}\mathit{\pi }{\mathit{\epsilon }}_{\mathit{0}}{\mathit{a}}^{\mathit{2}}}\end{array}$
Force due to charge Q = F₃
$=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{QQ}}{(\sqrt{2}\mathrm{a}{)}^2} \mathrm{along} \overrightarrow{\mathrm{BO}}$
Total Resultant force = $\mathrm{F}+{\mathrm{F}}_3 \mathrm{along} \overrightarrow{\mathrm{BO}}$
$\begin{array}{l}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\sqrt{2}\mathrm{Qq}}{{\mathrm{a}}^2}+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{QQ}}{2{\mathrm{a}}^2}\\ =\left[\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{Q}}{{\mathrm{a}}^2}\left(\sqrt{2}\mathrm{q}+\frac{\mathrm{Q}}{2}\right)\right]\text{ along }\overrightarrow{\mathrm{BO}}\end{array}$
(b) Total potential energy of the system,
$\begin{array}{l}\mathrm{J}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{\mathrm{Qq}}{\mathrm{a}}+\frac{\mathrm{Qq}}{\mathrm{a}}+\frac{\mathrm{Qq}}{\mathrm{a}}+\frac{\mathrm{Qq}}{\mathrm{a}}+\frac{\mathrm{QQ}}{\sqrt{2}\mathrm{a}}+\frac{\mathrm{qq}}{\sqrt{2}\mathrm{a}}\right]\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{4\mathrm{Qq}}{\mathrm{a}}+\frac{{\mathrm{Q}}^2}{\sqrt{2}\mathrm{a}}+\frac{{\mathrm{q}}^2}{\sqrt{2}\mathrm{a}}\right]\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[4\mathrm{Qq}+\frac{{\mathrm{Q}}^2+{\mathrm{q}}^2}{\sqrt{2}}\right]\end{array}$
(a) Electric force at A due to charge 2q
${\mathrm{F}}_{\mathrm{C}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}\times 2\mathrm{q}}{{\mathrm{l}}^2} \mathrm{along} \overrightarrow{\mathrm{CA}}$
Electric force at A due to charge (-4q)
${\mathrm{F}}_{\mathrm{B}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}\times (-4\mathrm{q})}{{\mathrm{l}}^2} \mathrm{along} \overrightarrow{\mathrm{AB}}$

Resultant force, 
$\begin{array}{l}\mathrm{F}=\sqrt{{\mathrm{F}}_{\mathrm{B}}^2+{\mathrm{F}}_{\mathrm{C}}^2+2{\mathrm{F}}_{\mathrm{B}}{\mathrm{F}}_{\mathrm{C}}\cos {120}^{\circ }}\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{{\mathrm{l}}^2}\sqrt{(-4{)}^2+(2{)}^2+2(-4)\left(2\right)}\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2\sqrt{7}{\mathrm{q}}^2}{{\mathrm{l}}^2}\end{array}$
(b) Work done to separate the charges to infinity:
Initial potential energy,
$\begin{array}{l}{\mathrm{U}}_{\mathrm{i}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{(-4\mathrm{q})\mathrm{q}}{\mathrm{l}}+\frac{(-4\mathrm{q})\left(2\mathrm{q}\right)}{\mathrm{l}}+\frac{\left(\mathrm{q}\right)\left(2\mathrm{q}\right)}{\mathrm{l}}\right]\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{\mathrm{l}}[-4-8+2]\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{-10{\mathrm{q}}^2}{\mathrm{l}}\right)\end{array}$
Final potential energy ${\mathbf{U}}_t=0$
 Thus, work done
$={\mathrm{U}}_{\mathrm{f}}-{\mathrm{U}}_{\mathrm{i}}=0-\left(\frac{-10{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{l}}\right)=\frac{10{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{l}}=\frac{5{\tilde{\mathrm{q}}}^2}{2{\mathrm{\pi \epsilon }}_0\mathrm{l}}$