JEE MAIN - Physics (2025 - 29th January Morning Shift - No. 9)
An electric dipole of mass $m$, charge $q$, and length $l$ is placed in a uniform electric field $\vec{E} = E_0\hat{i}$. When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be :
$ \frac{1}{2\pi} \sqrt{\frac{ml}{2qE_0}} $
$ 2\pi \sqrt{\frac{ml}{2qE_0}} $
$ 2\pi \sqrt{\frac{ml}{qE_0}} $
$\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{~m} l}{\mathrm{q} \mathrm{E}_0}}$
설명
$$\begin{aligned}
& \mathrm{I} \omega 2 \theta=\mathrm{q} \ell \mathrm{E}_0 \theta \\
& 2 \mathrm{~m}\left(\frac{\ell}{2}\right)^2 \omega^2=\mathrm{q} \ell \mathrm{E}_0 \\
& \omega^2=\frac{2 \mathrm{qE}_0}{\mathrm{~m} \ell} \\
& \mathrm{~T}=2 \pi \sqrt{\frac{\mathrm{~m} \ell}{2 \mathrm{qE}_0}}
\end{aligned}$$