A dipole comprises of two charged particles of identical magnitude $$q$$ and opposite in nature. The mass $$m$$ of the positive charged particle is half of the mass of the negative charged particle. The two charges are separated by a distance $$l$$. If the dipole is placed in a uniform electric field $$\vec{E}$$, in such a way that dipole axis makes a very small angle with the electric field, $$\vec{E}$$. The angular frequency of the oscillations of the dipole when released is given by:

We need to find the angular frequency of small oscillations of a dipole in a uniform electric field, where the positive charge has mass $$m$$ and the negative charge has mass $$2m$$.

Taking the positive charge at the origin and the negative charge at a distance $$l$$, the center of mass coordinate is given by

$$x_{CM} = \frac{m \cdot 0 + 2m \cdot l}{m + 2m} = \frac{2l}{3}$$

Hence the positive charge lies at a distance $$\frac{2l}{3}$$ from the center of mass and the negative charge at a distance $$\frac{l}{3}$$.

Using these distances, the moment of inertia about the center of mass becomes

$$I = m\left(\frac{2l}{3}\right)^2 + 2m\left(\frac{l}{3}\right)^2 = \frac{4ml^2}{9} + \frac{2ml^2}{9} = \frac{6ml^2}{9} = \frac{2ml^2}{3}$$

When the dipole is displaced by a small angle $$\theta$$ in a uniform electric field $$E$$, the restoring torque is

$$\tau = -pE\sin\theta \approx -pE\theta$$

where the dipole moment is $$p = ql$$.

Applying Newton’s second law for rotation gives

$$I\alpha = -pE\theta$$

Substituting $$I = \frac{2ml^2}{3}$$ and $$p = ql$$ yields

$$\frac{2ml^2}{3}\,\alpha = -qlE\,\theta$$

from which

$$\alpha = -\frac{3qE}{2ml}\,\theta$$

This equation describes simple harmonic motion with angular frequency

$$\omega = \sqrt{\frac{3qE}{2ml}}$$

Therefore, the angular frequency is $$\sqrt{\dfrac{3qE}{2ml}}$$.