Two charges \(7 \mu \mathrm{C}\) and \(-2 \mu \mathrm{C}\) are placed at \((-9,0,0) \mathrm{cm}\) and \((9,0,0) \mathrm{cm}\) respectively in an external field \(E=\frac{\mathrm{A}}{\mathrm{r}^{2}} \hat{\mathrm{r}}\), where \(A=9 \times 10^{5} \mathrm{~N} / \mathrm{C} . \mathrm{m}^{2}\). Considering the potential at infinity is 0, the electrostatic energy of the configuration is \_\_\_\_ J.

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Electrostatics

Two charges

7 μ C

and

- 2 μ C

are placed at

(- 9, 0, 0) cm

and

(9, 0, 0) cm

respectively in an external field

E = \frac{A}{r ^{2}} \overset{r}{^}

, where

A = 9 \times 1 0^{5} N / C . m^{2}

.
Considering the potential at infinity is 0, the electrostatic energy of the configuration is ____ J.

Answer: (C)

The external field

E = \frac{A}{r ^{2}}

has potential

V = \frac{A}{r}

(with

V_{\infty} = 0

). Each charge lies at

r = 9 cm = 0.09 m

from the origin, where

V = \frac{9 \times 1 0 ^{5}}{0.09} = 1 0^{7} V

. The energy of the two charges in this external field is

U_{e x t} = (q_{1} + q_{2}) V = (7 - 2) \times 1 0^{- 6} \times 1 0^{7} = 50 J

. Their mutual interaction energy, for a separation

d = 18 cm = 0.18 m

, is

U_{in t} = \frac{k q _{1} q _{2}}{d} = \frac{9 \times 1 0 ^{9} ( 7 \times 1 0 ^{- 6} ) ( - 2 \times 1 0 ^{- 6} )}{0.18} = - 0.7 J

. The total electrostatic energy is

U = 50 - 0.7 = 49.3 J

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option