Question

Shown as electric dipole formed by two particle fixed at the ends of a light rod of length l. The mass of each particle is m and the charges are -q and +q. The system is placed in such a way that the dipole axis is parallel to a uniform electric field E that exists in the region. The dipole is slightly rotated about its centre and released. Show that for small angular displacement, the motion is angular simple harmonic and find its time period.

Answer 1

$\begin{array}{c}Wehave\\ T=qEl\theta \\ I\alpha qEl\theta \\ \Rightarrow \alpha =\left(\frac{qEl}{I}\right)\theta \therefore 54m\\ =\left\{\frac{qEl}{\left(\frac{m{l}^2}{4}\times 2\right)}\right\}\theta \\ \alpha =\left(\frac{2qE}{ml}\right)\theta \\ =w^2\theta \\ \Rightarrow w=\sqrt{\frac{2qE}{ml}}\\ T=\frac{2\pi }{w}=2\pi \sqrt{\frac{ml}{2qE}}\end{array}$

Which is the required answer.