Two charges $7 \mu \mathrm{c}$ and $-4 \mu \mathrm{c}$ are placed at $(-7 \mathrm{~cm}$, 0 , 0 ) and ( $7 \mathrm{~cm}, 0,0$ ) respectively. Given, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$, the electrostatic potential energy of the charge configuration is :

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Electrostatics

Two charges

7 μ c

and

- 4 μ c

are placed at

(- 7 cm

, 0 , 0 ) and (

7 cm, 0, 0

) respectively. Given,

ϵ_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}

, the electrostatic potential energy of the charge configuration is :

Answer: (D)

The electrostatic potential energy of a pair of point charges separated by a distance

r

U = \frac{1}{4 π ϵ _{0}} \frac{q _{1} q _{2}}{r}

The two charges lie on the x-axis at

(- 7 cm, 0, 0)

and

(7 cm, 0, 0)

, so the separation between them is

r = 7 cm + 7 cm = 14 cm = 0.14 m

Using

\frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9} N m^{2} C^{- 2}

q_{1} = 7 μ C

and

q_{2} = - 4 μ C

U = \frac{9 \times 1 0 ^{9} \times ( 7 \times 1 0 ^{- 6} ) \times ( - 4 \times 1 0 ^{- 6} )}{0.14}

U = \frac{- 252 \times 1 0 ^{- 3}}{0.14} = - 1.8 J

The electrostatic potential energy of the configuration is

- 1.8 J

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option