Q 12-01-039JEE MainJEE Main 2025 (29 Jan, Shift 1)Medium
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 (treat each of the two charges as a point mass $m$)
Answer: (D) $2\pi\sqrt{\dfrac{ml}{2qE_0}}$
For a small angle $\theta$, the restoring torque is $\tau = -pE_0\sin\theta \approx -qlE_0\,\theta$.
The dipole rotates about its centre; two point masses $m$ at distance $l/2$ give
$$I = 2m\left(\frac{l}{2}\right)^2 = \frac{ml^2}{2}$$
$$T = 2\pi\sqrt{\frac{I}{pE_0}} = 2\pi\sqrt{\frac{ml^2/2}{qlE_0}} = 2\pi\sqrt{\frac{ml}{2qE_0}}$$
Solution by Sreeraj P, M.Sc Physics