Question

Charges -q and +q are fixed at the ends of a light rod of length $l$. The rod is clamped at one end with axis of the dipole along the electric field. The rod is slightly displaced and then released, Neglecting gravity find the time period of small oscillations. Mass of charges is m each.

• A. $2\pi \sqrt{\frac{ml}{qE}}$
• B. $2\pi \sqrt{\frac{ml}{3qE}}$
• C. $2\pi \sqrt{\frac{ml}{2qE}}$
• D. $2\pi \sqrt{\frac{2ml}{qE}}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{ml}{qE}}$

Here the torque due to the rotational motion is equal and opposite to the torque due to the dipole. i.e,
$I\alpha =-pE\sin \theta$ where $I=m{l}^2$, moment of inertia; $\alpha$, angular acceleration; $p=ql$, dipole moment and $\theta$ , angle between electric field and dipole.
For small oscillation, $\theta$ is very small so $\sin \theta =\theta$.
now $m{l}^2\alpha =-\left(ql\right)E\theta$
$\alpha =-\frac{qE}{ml}\theta$
$\frac{d^2\theta }{d{t}^2}=-\frac{qE}{ml}\theta$
thus, ${\omega }^2=\frac{qE}{ml}$
Time period , $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{ml}{qE}}$