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JEE Main 202529 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2025 · 29 January · Morning Shift · Physics · Electrostatics

An electric dipole of mass m, charge q, and length l is placed in a...

An electric dipole of mass m, charge q, and length ll is placed in a uniform electric field (\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \hat{i}). When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be:An electric dipole of mass m, charge q, and length l is placed in a uniform electric field ( E =E_0 i ). When the

Options

  1. A(\frac{1}{2 \pi} \sqrt{\frac{m l}{2 q E_0}})
  2. B(2 \pi \sqrt{\frac{\mathrm{~m} l}{\mathrm{qE}_0}})
  3. C(\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{~m} l}{\mathrm{qE}_0}})
  4. D(2 \pi \sqrt{\frac{\mathrm{~m} l}{2 \mathrm{qE}_0}})

Correct answer

D. (2 \pi \sqrt{\frac{\mathrm{~m} l}{2 \mathrm{qE}_0}})

Step-by-step solution

An electric dipole of mass m, charge q, and length l is placed in a uniform electric field ( E =E_0 i ). When the
τ=PE0sin⁡θ\tau=P E_0 \sin \theta If θ\theta is small τ=−(PE0)θI=m(l2)2⋅2=ml22T=2πml22⋅PE0=2πml22⋅q/E0T=2πml2qE0\begin{aligned} \tau & =-\left(P E_0\right) \theta \\ I & =m\left(\frac{l}{2}\right)^2 \cdot 2=\frac{m l^2}{2} \\ T & =2 \pi \sqrt{\frac{m l^2}{2 \cdot P E_0}}=2 \pi \sqrt{\frac{m l^2}{2 \cdot q / E_0}} \\ T & =2 \pi \sqrt{\frac{m l}{2 q E_0}} \end{aligned}
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