Physics · Electrostatics

JEE Main 2025 — 28 January, Evening Shift — Question 65

An electric dipole of dipole moment $6 \times 10^{-6} \mathrm{Cm}$ is placed in uniform electric field of magnitude $10^{6} \mathrm{~V} / \mathrm{m}$ . Initially, the dipole moment is parallel to electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field, will be $\qquad$ J.

Answer: 12

Numerical answer — enter this value.

Step-by-step solution

$p=6 \times 10^{-6} \mathrm{Cm}$

$\mathrm{E}=10^{6} \mathrm{v} / \mathrm{m}$

$\mathrm{W}=\Delta \mathrm{U}=-\mathrm{pE}\left(\cos \theta_{\mathrm{f}}-\cos \theta_{\mathrm{i}}\right)$

$\mathrm{W}=2 \mathrm{pE}=12 \mathrm{~J}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electric Dipole