Question

An electric dipole is formed by 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 torsionless thread in an electric field of intensity $E$ such that the dipole axis is parallel to the field. If it is slightly displaced, the period of angular motion is :

• A. $\frac{1}{2\pi }\sqrt{\frac{2qE}{ml}}$
• B. $2\pi \sqrt{\frac{ml}{qE}}$
• C. $2\pi \sqrt{\frac{ml}{2qE}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{ml}{4qE}}$

Answer 1

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

We know torque on a dipole is $\tau =P\times E$
When the dipole is displaced by $\theta ,\tau =PEsin\theta$
The electric field tries to align the dipole in the same direction as that of the field.

$\therefore \theta$ and $\alpha$ (angular acceleration) are always in opposite directions.

we know $\tau =I\alpha$

$I\alpha =-PEsin\theta$

$sin\theta =\theta$ for small $\theta$

$\Rightarrow \alpha =\frac{-PE\theta }{I}$

we know in S. H. M ,

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

$\therefore \omega =\sqrt{\frac{PE}{I}}$

I at center of rod $=\frac{m{l}^2}{2}$

$\Rightarrow \omega =\sqrt{\frac{2\cdot q\cdot l\cdot E}{m{l}^2}}$

$\therefore T=\frac{2\pi }{\omega }$

and frequency $f=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{2qE}{ml}}$

$T=2\pi \sqrt{\frac{ml}{2qE}}$