Question

An electric dipole of dipole moment p is placed in a uniform electric field E in stable equilibrium position. Its moment of inertia about the centroidal axis is I. If it is displaced slightly from its mean position find the period of small oscillations.
\eqalign{ & A.\;{\text{ 2}}\pi \sqrt {\dfrac{I}{{2pE}}} \cr & B.\;{\text{ 2}}\pi \sqrt {\dfrac{{2I}}{{pE}}} \cr & C.\;{\text{2}}\pi \sqrt {\dfrac{I}{{pE}}} \cr & D.\;{\text{ }}\pi \sqrt {\dfrac{{2I}}{{2pE}}} \cr}

Answer 1

Torque is the rotational equivalent of linear force which causes an angular acceleration and angular acceleration is the rate of change of the angular velocity with respect to the time. Here first find the equation for the torque and the angular acceleration. Find the correlation between the two, compare and find the required period of small oscillations.

Complete step by step answer:
Let, an electric dipole have –
The dipole moment = p
The electric field = E
The centroid axis = I

The above figure shows that when an electric dipole at mean position is slightly displaced, initially it was at the stable position we have displaced it with the angle $\theta$
The restoring torque is the property which tends to come at its initial position.
Let it be $= \tau$
$\therefore \tau = - pE\sin \theta$
Given that there is slight displacement, therefore in case of the angular displacement –
$\sin \theta \approx \theta$ (By rule)
$\therefore \tau = - pE\theta \,{\text{ }}......{\text{(1)}}$
Also, the angular acceleration is
$\alpha = \dfrac{\tau }{I}$
Place value from equation $(1)$
$\alpha = \dfrac{{ - pE\theta }}{I}$
Re-writing the above equation –
$\alpha = \left( {\dfrac{{ - pE}}{I}} \right)\theta {\text{ }}.......{\text{(2)}}$
We know that, $\alpha = - {\omega ^2}\theta \,{\text{ }}......{\text{(3)}}$
Compare equations $(2){\text{ and (3)}}$
$- {\omega ^2}\theta \, = \left( {\dfrac{{ - pE}}{I}} \right)\theta {\text{ }}$
Taking negative $\theta$ from both the sides and therefore they are removed.
${\omega ^2}\, = \left( {\dfrac{{pE}}{I}} \right){\text{ }}$
Take square-roots on both the sides of the equations
$\sqrt {{\omega ^2}} \, = \sqrt {\left( {\dfrac{{pE}}{I}} \right)} {\text{ }}$
Square and square-roots cancels each other on left hand side of the equation-
${\omega ^{}}\, = \sqrt {\left( {\dfrac{{pE}}{I}} \right)} {\text{ }}......{\text{(4)}}$
Now, time period is $T = \dfrac{{2\pi }}{\omega }$
Place value from the equation $(4)$
$T = \dfrac{{2\pi }}{{\sqrt {\left( {\dfrac{{pE}}{I}} \right)} {\text{ }}}}$
Rearrange the above equation –
$T = 2\pi \sqrt {\left( {\dfrac{I}{{PE}}} \right)}$
Therefore, the required solution is the period of the small oscillations is $T = 2\pi \sqrt {\left( {\dfrac{I}{{PE}}} \right)}$
Hence, from the given multiple choices, option C is the correct answer.

Note: Torque is required to create an angular acceleration of any object. Torque is also referred to as the moment, moment of force, the rotational force or the turning effect. The angular acceleration is directly proportional to the torque in case of the constant rotational inertia of the object.