Question

An electric dipole of momentum $\overrightarrow{p}$ is placed in a uniform electric field. The dipole is rotated through a very small angle from equilibrium and is released. Prove that it executes simple harmonic motion with frequency $f=\frac{1}{2\pi }\sqrt{\frac{pE}{I}}$. Where, $I=$ moment of inertia of the dipole.

Answer 1

R.E.F image

When a dipole moment $\overrightarrow{P}$ is rotated by small angle $\theta$ from in

stable equilibrium state In an Electric field $\overrightarrow{E}$ , it experience a torque (z):-

$\overrightarrow{X}=\overrightarrow{P}\times \overrightarrow{E}$

In magnitude, $Z=PEsin\theta$

as $\theta$ is small, $sin\theta \simeq \theta \Rightarrow Z=PE\theta$

This torque this it to restore basic to equilibrium.

as restoring torque $=Z=I{\omega }^2\theta$, we compare,[I: moment of inertia of dipole]

$\Rightarrow Z=I{\omega }^2\theta =PE\theta$

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

also, $\omega =2\pi f,$ f : frequency of oscillation that gives,

$\Rightarrow f=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{PE}{I}}$ (Read)