Two point charges -4 μ c and 4 μ c , constituting an electric dipole, are placed at (-9,0,0) cm and (9,0,0) cm in a uniform electric field of strength 10 4 NC -1 . The work done on the dipole in rotating it from the equilibrium through 180 ° is :

Physics · Electrostatics

JEE Main 2025 — 23 January, Evening Shift — Question 58

Two point charges $-4 \mu \mathrm{c}$ and $4 \mu \mathrm{c}$ , constituting an electric dipole, are placed at $(-9,0,0) \mathrm{cm}$ and $(9,0,0) \mathrm{cm}$ in a uniform electric field of strength $10^{4} \mathrm{NC}^{-1}$ . The work done on the dipole in rotating it from the equilibrium through $180^{\circ}$ is :

Option A:
14.4mJ
Correct
Option B:
18.4mJ
Option C:
12.4mJ
Option D:
16.4mJ

Answer: A

Step-by-step solution

$\mathrm{U}=-\mathrm{PE} \cos \theta$

$\mathrm{w}_{\text {ext }}=\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=-\mathrm{PE} \cos 180^{\circ}+\mathrm{PE} \cos 0^{\circ}$

$\mathrm{w}_{\text {ext }}=2 \mathrm{PE}$

$=2 \times\left(4 \times 10^{-6}\right)(18) \times 10^{4}$

$=144 \times 10^{-2}$

$=14.4 \mathrm{~mJ}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electric Dipole

Two charges 7 mu c and -4 mu c are placed at ( -7 cm , 0,0) and ( 7 cm, 0,0 ) respectively. Given, epsilon 0=8.85 × 10^-12 C^2 N^-1 m^-2 …

Answer: A

Step-by-step solution

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