Two particles with charges -q and +q , each having mass m , are...

Let the clamped end of the rod be the origin of rotation. Since one end is clamped, the mass $m$ at this end does not contribute to the moment of inertia.

The moment of inertia of the system about the clamped end is due entirely to the mass at the free end, which is at a distance $a$:
$I = ma^2$

When the dipole is in stable equilibrium, it is aligned with the uniform electric field $E$. If the rod is given a small angular displacement $\theta$, the charge at the free end experiences an electric force $F = qE$.

The restoring torque acting on the system about the clamped end is:
$\tau = -F a \sin\theta = -qE a \sin\theta$

For small oscillations, $\sin\theta \approx \theta$. Thus, the torque becomes:
$\tau \approx -qE a \theta$

Using the relation $\tau = I\alpha$, where $\alpha$ is the angular acceleration:
$ma^2 \alpha = -qE a \theta$

$\alpha = -\dfrac{qE}{ma} \theta$

This is the equation for simple harmonic motion, $\alpha = -\omega^2 \theta$, where the angular frequency $\omega$ is:
$\omega = \sqrt{\dfrac{qE}{ma}}$

The time period of small oscillations is given by $T = \dfrac{2\pi}{\omega}$:
$T = 2\pi\sqrt{\dfrac{ma}{qE}}$