Question

Two charges of equal magnitude '$q$' are placed in air at a distance ${}^{\prime }2a^{\prime }$ apart and third charge $-2q^{\prime }$ is placed at midpoint. The potential energy of the system is $({ϵ}_0=$permittivity of free space) :

• A. $-\frac{q^2}{8\pi {ϵ}_{0}a}$
• B. $-\frac{3q^2}{8\pi {ϵ}_{0}a}$
• C. $-\frac{5q^2}{8\pi {ϵ}_{0}a}$
• D. $-\frac{7q^2}{8\pi {ϵ}_{0}a}$

Answer 1

The correct option is D $-\frac{7q^2}{8\pi {ϵ}_{0}a}$

Potential energy of the system $U=\frac{-2q^2}{4\pi {ϵ}_{o}a}+\left(\frac{-2q^2}{4\pi {ϵ}_{o}a}\right)+\frac{q^2}{4\pi {ϵ}_o\left(2a\right)}$

$\therefore$ $U=\frac{-2q^2}{4\pi {ϵ}_{o}a}+\left(\frac{-2q^2}{4\pi {ϵ}_{o}a}\right)+\frac{q^2}{4\pi {ϵ}_{o}2a}$

Or $U=\frac{q^2}{4\pi {ϵ}_{o}a}[-2-2+\frac{1}{2}]$

$⟹$ $U=\frac{-7q^2}{8\pi {ϵ}_{o}a}$