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 $x\frac{t^2pE}{\pi^2}$ then find the value of x.

Answer 1

4

Answer 2

The torque on an electric dipole $\vec{p}$ in a uniform electric field $\vec{E}$ is given by $\vec{\tau} = \vec{p} \times \vec{E}$. The magnitude is $\tau = pE \sin\theta$, where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$. For stable equilibrium, $\theta=0$.

When the dipole is rotated by a small angle $\theta$ from the stable equilibrium, the restoring torque is $\tau = -pE \sin\theta$.

The equation of motion for rotation is $I\alpha = \tau$, where $I$ is the moment of inertia and $\alpha = \frac{d^2\theta}{dt^2}$ is the angular acceleration.

$I\frac{d^2\theta}{dt^2} = -pE \sin\theta$.

For small angles $\theta$, $\sin\theta \approx \theta$. The equation becomes:

$I\frac{d^2\theta}{dt^2} = -pE \theta$ $\frac{d^2\theta}{dt^2} = -\frac{pE}{I} \theta$.

This is the equation for simple harmonic motion (SHM) with angular frequency $\omega$, where $\omega^2 = \frac{pE}{I}$.

The angular frequency is $\omega = \sqrt{\frac{pE}{I}}$.

The period of oscillation is $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{I}{pE}}$.

The dipole is released from a small angular displacement. The time taken to reach the equilibrium position ($\theta=0$) for the first time is one-quarter of the period of oscillation, i.e., $t = \frac{T}{4}$.

$t = \frac{1}{4} \left(2\pi\sqrt{\frac{I}{pE}}\right) = \frac{\pi}{2}\sqrt{\frac{I}{pE}}$.

Squaring both sides:

$t^2 = \frac{\pi^2}{4} \frac{I}{pE}$.

Rearranging to solve for $I$:

$I = \frac{4t^2 pE}{\pi^2}$.

The problem states that the moment of inertia is $I = x\frac{t^2pE}{\pi^2}$.

Comparing the two expressions for $I$, we have:

$x\frac{t^2pE}{\pi^2} = \frac{4t^2pE}{\pi^2}$.

Dividing both sides by $\frac{t^2pE}{\pi^2}$ (assuming $t, p, E, \pi \neq 0$), we get:

$x = 4$.