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Physics · Electrostatics

JEE Main 2025 — 7 April, Evening Shift — Question 48

A dipole with two electric charges of 2μC2 \mu \mathrm{C} magnitude each, with separation distance 0.5μ m0.5 \mu \mathrm{~m}, is placed between the plates of a capacitor such that its axis is parallel to an electric field established between the plates when a potential difference of 5 V is applied. Separation between the plates is 0.5 mm . If the dipole is rotated by 30∘30^{\circ} from the axis, it tends to realign in the direction due to a torque. The value of torque is :

  1. Option A:

    5×10−3Nm5 \times 10^{-3} \mathrm{Nm}

  2. Option B:

    2.5×10−12Nm2.5 \times 10^{-12} \mathrm{Nm}

  3. Option C:

    5×10−9Nm5 \times 10^{-9} \mathrm{Nm}

    Correct
  4. Option D:

    22.5×10−9Nm2.5 \times 10^{-9} \mathrm{Nm}

Answer: C

Step-by-step solution

E=50.5×10−3=104 N/CE=\frac{5}{0.5 \times 10^{-3}}=10^{4} \mathrm{~N} / \mathrm{C}

P=10−12C⋅ mP=10^{-12} \mathrm{C} \cdot \mathrm{~m}

U=PEsin⁡θU=P E \sin \theta =5×10−9 N.m=5 \times 10^{-9} \mathrm{~N} . \mathrm{m}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
7 April, Evening Shift
Subject
Physics
Chapter
Electrostatics
Topic
Electric Dipole

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