An electric dipole of mass m, charge q, and length l is placed in a uniform electric field $\overrightarrow{\mathrm{E}}={\mathrm{E}}_0\hat{\mathrm{i}}.$ When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be :

The torque acting on a dipole in a uniform electric field is:

• $$\tau = \mathbf{p} \times \mathbf{E}$$ $$\tau = pE_0 \sin\theta$$

Since $p = ql$,

• $$\tau = ql E_0 \sin\theta$$

For small angles, $\sin\theta \approx \theta$, so:

• $$\tau \approx ql E_0 \theta$$

Using Newton’s second law for rotational motion:

$$I \alpha = \tau$$

• $\alpha = \frac{d^2\theta}{dt^2}$ is the angular acceleration.

For a dipole (two point masses $q$ separated by $l$), the moment of inertia about the center is:

$$I = m \left(\frac{l}{2}\right)^2 + m \left(\frac{l}{2}\right)^2 = 2m \frac{l^2}{4} = \frac{ml^2}{2}$$

So the equation of motion becomes:

$$\frac{ml^2}{2} \frac{d^2\theta}{dt^2} = - ql E_0 \theta$$

 $$\frac{d^2\theta}{dt^2} + \frac{2 q E_0}{m l} \theta = 0$$

This is the standard form of simple harmonic motion:

$$\frac{d^2\theta}{dt^2} + \omega^2 \theta = 0$$

where:

$$\omega^2 = \frac{2 q E_0}{m l}$$

Thus, the angular frequency is:

$$\omega = \sqrt{\frac{2 q E_0}{m l}}$$

The time period of oscillation is given by:

$$T = \frac{2\pi}{\omega}$$

 $$T = 2\pi \sqrt{\frac{m l}{2 q E_0}}$$