An electric dipole is formed due to two particles fixed at the ends of a light rigid rod of length $l$. The mass of each particle is m and charges are –q and +q The system is suspended by a torsion less thread in an electric filed of intensity E such that the dipole axis is parallel to the filed if it is slightly displaced, the period of angular motion is

Given data: An electric dipole with charges +q and –q 

Concept used: Electric dipole

Detailed solution:

We know that the torque of a dipole is 

$$\begin{gathered} \tau =P\times E \\ \tau =-PE\sin \theta \end{gathered}$$

Also, we have  $\tau =I\alpha$

From the above equations we get 

$I\alpha =-PE\sin \theta$

For $\mathrm{\theta }$ small sin$\mathrm{\theta }$ = $\mathrm{\theta }$

$$\begin{gathered} \alpha =\frac{-P.E \theta }{I} \\ \frac{-\alpha }{\theta }=\frac{P.E}{I} \end{gathered}$$

From S.H.M we have   

$a=-{\omega }^{2}x$

$$\begin{gathered} \omega =\sqrt{\frac{-a}{x} }\left(or\right)\sqrt{\frac{-\alpha }{\theta }} \\ \Rightarrow \omega =\sqrt{\frac{P.E}{I}} \end{gathered}$$

The moment of inertia at the centre of the rod is  $I=\frac{m{l}^2}{2}$

Therefore,  $\omega =\sqrt{\frac{qlE}{\left(\frac{m{l}^2}{2}\right)}}$

$$\begin{gathered} \omega =\sqrt{\frac{2qE}{ml}} \\ \Rightarrow T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{ml}{2qE}} \end{gathered}$$