Question

An electric dipole is situated in an electric field of uniform intensity $E$ whose dipole moment is $p$and moment of inertia is $I.$ If the dipole is displaced then the angular frequency of its oscillation is
(A) ${\left( {\dfrac{{pE}}{I}} \right)^{\dfrac{1}{2}}}$
(B) ${\left( {\dfrac{{pE}}{I}} \right)^{\dfrac{3}{2}}}$
(C) ${\left( {\dfrac{I}{{pE}}} \right)^{\dfrac{1}{2}}}$
(D) ${\left( {\dfrac{p}{{IE}}} \right)^{\dfrac{1}{2}}}$

Answer 1

When an electric dipole is placed in a uniform electric field, a torque is applied on the dipole, which aliens the dipole to the direction, parallel to the direction of the electric field. Which, in turn, gives rise to restoring couple because of which, the dipole starts to move in a simple harmonic motion. Use that equation to find angular frequency.

Complete step by step answer:
It is given in the question that,
$p$ is dipole moment of a dipole
$I$ is moment of inertia of the dipole
$E$ is the electric field intensity
When an electric dipole is placed in a uniform electric field, it generates torque. This torque tries to align the dipole in the direction parallel to the direction of the electric field with the net force zero. That torque generates restoring couple.
This restoring couple, generates simple harmonic motion of the electric dipole by a small angle $\theta$.
Now, we know that torque is equal to the dot product of electric dipole moment and the electric field in which it is placed.
$\Rightarrow \tau = \overrightarrow p .\overrightarrow E = pE\sin \theta$
And restoring couple acts equal and opposite to the torque.
$\Rightarrow C = - \tau = - pE\sin \theta$
Since, $\theta$ is very small, we can write
$\sin \theta \approx \theta$
$\Rightarrow C = - pE\theta$ . . . (1)
Also, torque is the product of inertia of a body and its angular acceleration.
i.e. $\tau = I\alpha$ . . . (2)
Therefore, from equation (1) and (2), we get
$I\alpha = - pE\theta$ . . . (3)
Now we know that angular acceleration is the rate of change of angular velocity. i.e.
$\alpha = \dfrac{{d\omega }}{{dt}}$
And angular velocity is the rate of change of angular displacement. i.e.
$\omega = \dfrac{{d\theta }}{{dt}}$
Therefore, we get
$\alpha = \dfrac{{d\omega }}{{dt}} = \dfrac{d}{{dt}}\left( {\dfrac{{d\theta }}{{dt}}} \right)$
$\Rightarrow \alpha = \dfrac{{{d^2}\theta }}{{d{t^2}}}$
Therefore, equation (3) becomes
$I\dfrac{{{d^2}\theta }}{{d{t^2}}} = - pE\theta$
$\Rightarrow \dfrac{{{d^2}\theta }}{{d{t^2}}} = - \dfrac{{pE}}{I}\theta$
By comparing it to the standard equation of simple harmonic motion. i.e.
$\dfrac{{{d^2}\theta }}{{d{t^2}}} = - {\omega ^2}\theta$
Where,
$\omega$ is angular velocity
We get
${\omega ^2} = \dfrac{{pE}}{I}$
$\Rightarrow \omega = \sqrt {\dfrac{{pE}}{I}}$
Now, we know that, angular velocity can also be written as
$\omega = 2\pi f$
Where,
$f$ is angular frequency.
$\Rightarrow f = \dfrac{\omega }{{2\pi }}$
$\Rightarrow f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{pE}}{I}}$

Therefore, from the above explanation, the correct answer is, option (B) ${\left( {\dfrac{{pE}}{I}} \right)^{\dfrac{1}{2}}}$

Note: This is a good example of how important concepts are. Nearly no information is given in this question about finding frequency. And there is no direct formula to find the angular frequency, using the information given in the question. We need to understand concepts and need to be able to use them to manipulate given information into getting what we need.