Physics · Electrostatics

JEE Main 2025 — 4 April, Morning Shift — Question 55

Two small spherical balls of mass 10 g each with charges $-2 \mu \mathrm{C}$ and $2 \mu \mathrm{C}$ ,are attached to two ends of very light rigid rod of length 20 cm . The arrangement is now placed near an infinite nonconducting charge sheet with uniform charge density of $100 \mu \mathrm{C} / \mathrm{m}^{2}$ such that length of rod makes an angle of $30^{\circ}$ with electric field generated by charge sheet. Net torque acting on the rod is:

(Take $\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}$ )

Option A:
1.12 Nm
Correct
Option B:
2.24 Nm
Option C:
112 Nm
Option D:
11.2 Nm

Answer: A

Step-by-step solution

$\tau=P E \sin \theta$

$=q d E \sin \theta$

$=2 \times 10^{-6} \times 0.2 \times \frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \times \frac{1}{2}$

$\tau=1.12 \mathrm{Nm}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 4 April, Morning Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electric Dipole