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 $f=\frac{1}{2\pi }\sqrt{\frac{pE}{1}}$. Where, $I=$ moment of inertia of the dipole.

Answer 1

When a dipole of dipole moment $vecp$ is rotated by small sngle $\theta$ from is stable equilibrium state in an Electric field $\overrightarrow{E}$, it experiences a torque (z) :-

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

In magnitude, $Z=PE\sin \theta$

as $\theta$ is small, $\sin \theta \simeq \theta \Rightarrow z=pE\theta$.

This torque tries it to restore back to equilibrium.

as Restoring torque = $z=I{w}^2\theta$, we compare, [I : moment of inertia of dipole]

$\Rightarrow z=I{w}^2\theta =pE\theta$

$\Rightarrow w=\sqrt{\frac{pE}{I}}$

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

$\Rightarrow f=\frac{w}{2\pi }=\frac{1}{2\pi }\sqrt{\frac{pE}{I}}$.