Question

An electric dipole (dipole moment = p) is placed in a uniform electric field in stable equilibrium position at rest. Now it is rotated by a small angle and released. The time after which it comes to the equilibrium position again (for first time) is t. If the moment of inertia of the dipole about the axis of rotation is $x\frac{t^{2}pE}{\pi^{2}}$, then find the value of x.

Answer 1

4

Answer 2

The torque on the dipole for a small angular displacement $\theta$ from stable equilibrium is $\tau = -pE\theta$.

This torque leads to rotational SHM with angular frequency $\omega = \sqrt{\frac{pE}{I}}$.

The time to reach equilibrium for the first time from maximum displacement is $t = \frac{\pi}{2\omega}$.

Substituting $\omega$, we get $t = \frac{\pi}{2} \sqrt{\frac{I}{pE}}$.

Rearranging for $I$, we find $I = \frac{4 pE t^2}{\pi^2}$.

Comparing this with the given expression $I = x \frac{t^{2}pE}{\pi^{2}}$, we find $x=4$.