Question

Two charges of equal magnitude ‘$q$’ are placed in air at a distance ‘$2a$’ apart and third charge ‘$-2q$’ is placed at midpoint. The potential energy of the system is ($\varepsilon_0 =$ permittivity of free space)

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

Answer 1

Answer: (D)

Potential energy of the system

$U =\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q_{1} q_{2}}{r_{12}}+\frac{q_{1} q_{3}}{r_{13}}+\frac{q_{2} q_{3}}{r_{23}}\right] $
$=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q(-2 q)}{a}+\frac{q(-2 q)}{a}+\frac{q q}{2 a}\right] $
$=\frac{1}{4 \pi \varepsilon_{0}}\left[-\frac{2 q^{2}}{a}-\frac{2 q^{2}}{a}+\frac{q^{2}}{2 a}\right] $
$=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{-4 q^{2}}{a}+\frac{q^{2}}{2 a}\right]=\frac{-7 q^{2}}{8 \pi \varepsilon_{0} a} $