JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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:
- A \(\frac{1}{2 \pi} \sqrt{\frac{m l}{2 q E_0}}\)
- B \(2 \pi \sqrt{\frac{\mathrm{~m} l}{\mathrm{qE}_0}}\)
- C \(\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{~m} l}{\mathrm{qE}_0}}\)
- D \(2 \pi \sqrt{\frac{\mathrm{~m} l}{2 \mathrm{qE}_0}}\)
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…
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