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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2025
  • Medium
  • MCQ

An electric dipole of mass m, charge q, and length \(l\) 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:

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Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation


\(\tau=P E_0 \sin \theta\)
If \(\theta\) is small
\(\begin{aligned}
\tau & =-\left(P E_0\right) \theta \\
I & =m\left(\frac{l}{2}\right)^2 \cdot 2=\frac{m l^2}{2} \\
T & =2 \pi \sqrt{\frac{m l^2}{2 \cdot P E_0}}=2 \pi \sqrt{\frac{m l^2}{2 \cdot q / E_0}} \\
T & =2 \pi \sqrt{\frac{m l}{2 q E_0}}
\end{aligned}\)
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