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JEE Main 20257 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Evening Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2025 · 7 April · Evening Shift · Physics · Electrostatics

A dipole with two electric charges of 2 μ C magnitude each, with...

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 :

Options

  1. A5×10−9Nm5 \times 10^{-9} \mathrm{Nm}
  2. B5×10−3Nm5 \times 10^{-3} \mathrm{Nm}
  3. C2.5×10−12Nm2.5 \times 10^{-12} \mathrm{Nm}
  4. D2.5×10−9Nm2.5 \times 10^{-9} \mathrm{Nm}

Correct answer

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

Step-by-step solution

<br/>E=v d=55×10−4=104 V/mτ=PEsin⁡θ<br/>\begin{aligned}<br/>& \mathrm{E}=\frac{\mathrm{v}}{\mathrm{~d}}=\frac{5}{5 \times 10^{-4}}=10^4 \mathrm{~V} / \mathrm{m} \\ & \tau=\mathrm{PE} \sin \theta<br/>\end{aligned}
Where P=qa=2×10−6×5×10−7\mathrm{P}=\mathrm{qa}=2 \times 10^{-6} \times 5 \times 10^{-7}
=1×10−12C−m=1 \times 10^{-12} \mathrm{C}-\mathrm{m}
τ=1×10−12×104×12=5×10−9 N−m\tau=1 \times 10^{-12} \times 10^4 \times \frac{1}{2}=5 \times 10^{-9} \mathrm{~N}-\mathrm{m}
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