Question

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

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

Answer 1

(a) Force on charge Q at B due to charge q

$F_2=\frac{1}{4\pi {ϵ}_0}\frac{qQ}{a^2}\text{along}\underset{BY}{\to }$


$F_2=\frac{1}{4\pi {ϵ}_0}\frac{qQ}{a^2}\text{along}\overrightarrow{BX}$

Resultant force, $F=\sqrt{F_1^2+F_2^2}\text{along}\overrightarrow{BO}$

$=\sqrt{2{\left(\frac{1}{4\pi {ϵ}_0}\frac{qQ}{a^2}\right)}^2}=\frac{qQ\sqrt{2}}{4\pi {ϵ}_0{a}^2}$

Force due to charge $Q=F_3=\frac{1}{4\pi {ϵ}_0}\frac{QQ}{\left(\sqrt{2}{a}^2\right)}\text{along}\overrightarrow{BO}$

Total resultant force = $F+F_3\text{along}\overrightarrow{BO}$

$=\frac{1}{4\pi {ϵ}_0}\frac{\sqrt{2}Qq}{a^2}+\frac{1}{4\pi {ϵ}_0}\frac{QQ}{2a^2}$

$=\left[\frac{1}{4\pi {ϵ}_0}\frac{Q}{a^2}(\sqrt{2}q+\frac{Q}{2})\right]\text{along}\overrightarrow{BO}$

(b) Total potential energy of the system,

$U=\frac{1}{4\pi {ϵ}_0}[\frac{Qq}{a}+\frac{Qq}{a}+\frac{Qq}{a}+\frac{Qq}{a}+\frac{Qq}{a}+\frac{QQ}{\sqrt{2}a}+\frac{qq}{\sqrt{2}a}]$

$=\frac{1}{4\pi {ϵ}_0}[\frac{4Qq}{a}+\frac{Q^2}{\sqrt{2}a}+\frac{q^2}{\sqrt{2}a}]$

$=\frac{1}{4\pi {ϵ}_0}[4Qq+\frac{Q^2+q^2}{\sqrt{2}}]$