An electric dipole is formed by two equal and opposite charges q with separation d. The charges have same mass m. It is kept in a uniform electric field E. If it is slightly rotated from its equilibrium orientation, then its angular frequency $$\omega$$ is:

First of all, an electric dipole consists of two point charges, $$+q$$ and $$-q$$, separated by a distance $$d$$. The dipole moment (a vector directed from the negative to the positive charge) has the magnitude

$$p = qd.$$

When the dipole is placed in a uniform external electric field $$\vec E$$, the field exerts a torque on it. The magnitude of this torque at an angle $$\theta$$ between $$\vec p$$ and $$\vec E$$ is given by the standard relation

$$\tau = pE\sin\theta.$$

Here the equilibrium orientation is when $$\vec p$$ is parallel to $$\vec E$$, that is, $$\theta = 0$$. We now rotate the dipole slightly so that $$\theta$$ is small. For small angles we may use the approximation

$$\sin\theta \approx \theta.$$

Under this small-angle approximation, the torque becomes

$$\tau \approx pE\,\theta.$$

Because the torque always acts so as to reduce $$\theta$$, it actually points in the opposite direction to $$\theta$$, giving

$$\tau = -pE\,\theta.$$

Next we connect the torque to the rotational dynamics of the dipole. Newton’s second law for rotation states

$$\tau = I\alpha,$$

where $$I$$ is the moment of inertia of the system about the axis through the centre of the dipole (perpendicular to the dipole length) and $$\alpha = \dfrac{d^{2}\theta}{dt^{2}}$$ is the angular acceleration.

Both charges have the same mass $$m$$ and lie at distances $$\dfrac{d}{2}$$ on either side of the centre. Hence the moment of inertia is

$$I = m\left(\frac{d}{2}\right)^{2} + m\left(\frac{d}{2}\right)^{2} = 2m\left(\frac{d^{2}}{4}\right) = \frac{md^{2}}{2}.$$

Substituting $$\tau = -pE\theta$$ and $$I = \dfrac{md^{2}}{2}$$ into $$\tau = I\alpha$$, we get

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

Rearranging,

$$\frac{d^{2}\theta}{dt^{2}} + \frac{2pE}{md^{2}}\;\theta = 0.$$

The above differential equation has the standard simple-harmonic-motion form

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

where the square of the angular frequency is the coefficient of $$\theta$$. Therefore,

$$\omega^{2} = \frac{2pE}{md^{2}}.$$

But $$p = qd$$, so we substitute this value for $$p$$:

$$\omega^{2} = \frac{2(qd)E}{md^{2}} = \frac{2qE}{md}.$$

Taking the positive square root, the angular frequency is

$$\omega = \sqrt{\frac{2qE}{md}}.$$

Hence, the correct answer is Option B.