Explanation

(c) Net force on dipole in uniform electric field is zero but net torque is the sum of torques acting on the two charges.
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$\tau = (q)\left( \frac{d}{2} \right) \times E + (q)\left( \frac{d}{2} \right)\left( \mathbf{E} \right)$
$\ \ \ = qd \times E$
Where P=$qd$=dipole moment
$\therefore\ \ \tau = \mathbf{p} \times \mathbf{E} = pE\sin\theta\ \ \ \ldots.(i)$
If $I$ is moment of inertia and $\alpha\ $the angular acceleration, then
$\tau = I\alpha\ \ \ \ \ \ \ \ \ \ \ \ \ \ldots\ldots\ldots\ldots..(ii)$
From Eqs. (i) and (ii), we get
$I\alpha = pE\sin\theta$
Since, displacement is small $\sin{\theta \cong \theta}\ $
$\therefore\ \ \ \ \ \ I\alpha = pE\theta$
$\Longrightarrow \ \ \ \ \alpha = \frac{pE}{I}\theta$
The angular frequency of oscillation is given by
$\int_{}^{}{= \sqrt{\frac{\theta}{\alpha}} = \sqrt{\frac{\alpha I}{\alpha\ pE}} = \sqrt{\frac{I}{pE}}}$
The potential energy is lowest when the dipole moment is aligned with the field and highest when the two are antialigned.