1. Restoring torque on a dipole in a uniform electric field

For a dipole of moment $p = ql,$ placed at a small angle $\theta$ with a uniform electric field $E$, the torque is $\tau = -pE\sin\theta.$ For very small oscillations, $\sin\theta \approx \theta$, so $\tau \approx -pE\theta = -qlE\theta.$

This is the restoring torque for angular SHM.

2. Angular frequency formula

For rotational SHM, $I\,\ddot{\theta} = -qlE\theta,$ so $\ddot{\theta} + \frac{qlE}{I}\theta = 0.$ Hence, $\omega = \sqrt{\frac{qlE}{I}}.$

So we need the moment of inertia $I$ of the dipole about its center of mass.

3. Masses of the two particles

Given:

• mass of positive charge $= m$
• mass of negative charge $= 2m$

because the positive charged particle has half the mass of the negative charged particle.

The separation between them is $l$.

4. Position of center of mass

Let the positive charge $+q$ of mass $m$ be at one end and the negative charge $-q$ of mass $2m$ at the other end.

If distances of the two masses from the center of mass are $r_1$ and $r_2$, then $r_1 + r_2 = l$ and $m r_1 = 2m r_2.$ Thus, $r_1 = 2r_2.$ Using $r_1+r_2=l$, $2r_2+r_2=l \Rightarrow 3r_2=l \Rightarrow r_2=\frac{l}{3},$ $r_1=\frac{2l}{3}.$

So:

• mass $m$ is at distance $\frac{2l}{3}$ from CM,
• mass $2m$ is at distance $\frac{l}{3}$ from CM.

5. Moment of inertia about the center of mass

$I = m\left(\frac{2l}{3}\right)^2 + 2m\left(\frac{l}{3}\right)^2.$

Compute: $I = m\cdot \frac{4l^2}{9} + 2m\cdot \frac{l^2}{9}$ $I = \frac{4ml^2}{9} + \frac{2ml^2}{9} = \frac{6ml^2}{9} = \frac{2ml^2}{3}.$

6. Substitute into angular frequency expression

$\omega = \sqrt{\frac{qlE}{I}} = \sqrt{\frac{qlE}{\frac{2ml^2}{3}}}.$

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

7. Match with the options

This corresponds to: $\boxed{\sqrt{\frac{3 q E}{2 m l}}}$ which is Option A.