Question 242 of 311

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

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

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