Question

An electric dipole of mass \( m \), charge \( q \), and length \( l \) is placed in a uniform electric field \( E = E_0 \hat{i} \). When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be:

• A. \( \frac{2\pi}{\frac{q m l}{q E_0}} \)
• B. \( \frac{1}{2\pi} \frac{q^2 m l}{q E_0} \)
• C. \( \frac{1}{2\pi} \frac{q m l}{2 q E_0} \)
• D. $T = 2\pi \sqrt{\frac{ml}{2qE_0}}$

Hint:

For oscillations of a dipole in a uniform electric field, the time period depends on the moment of inertia of the dipole and the torque due to the electric field.

Answer 1

The Correct Option is D

We are tasked with analyzing the motion of an electric dipole in a uniform electric field and determining the time period $ T $ of its oscillations. The solution proceeds as follows:

1. Electric Dipole in a Uniform Field:
The dipole moment is denoted by $ \overrightarrow{P} $, and the electric field is $ \overrightarrow{E_0} $. The angle between the dipole moment and the electric field is $ \theta $.

2. Torque on the Dipole:
The torque $ \tau $ acting on the dipole is given by:

$ \tau = -(PE_0)\theta $

This equation holds when $ \theta $ is small, as the torque is proportional to the angular displacement $ \theta $.

3. Moment of Inertia:
The moment of inertia $ I $ of the dipole depends on the mass $ m $ and the length $ l $ of the dipole. For a dipole consisting of two point masses separated by a distance $ l $, the moment of inertia is:

$ I = m \left( \frac{l}{2} \right)^2 \cdot 2 = \frac{ml^2}{2} $

4. Time Period of Oscillation:
For small angular displacements, the motion of the dipole is simple harmonic. The time period $ T $ of oscillation is given by:

$ T = 2\pi \sqrt{\frac{I}{\kappa}} $

Here, $ \kappa $ is the restoring torque constant, which is equal to $ PE_0 $. Substituting $ I = \frac{ml^2}{2} $ and $ \kappa = qE_0 $, we get:

$ T = 2\pi \sqrt{\frac{\frac{ml^2}{2}}{qE_0}} $

Simplify the expression:

$ T = 2\pi \sqrt{\frac{mI^2}{2qIE_0}} $

$ T = 2\pi \sqrt{\frac{ml}{2qE_0}} $

Final Answer:
The time period of oscillation of the dipole is:

$ \boxed{T = 2\pi \sqrt{\frac{ml}{2qE_0}}} $

Answer 2

Step 1: Understand the problem setup.
An electric dipole consists of two equal and opposite charges separated by a distance \( l \). The dipole moment is given by \( p = ql \). The dipole is placed in a uniform electric field \( E = E_0 \hat{i} \). When the dipole is slightly rotated from its equilibrium position and released, it performs small angular oscillations about the equilibrium.

Step 2: Torque on the dipole.
When the dipole makes a small angle \( \theta \) with the direction of the electric field, the torque acting on it is:
\[ \tau = -pE_0 \sin\theta. \] For small angular displacements (\( \theta \) is small), we can approximate \( \sin\theta \approx \theta \). Thus,
\[ \tau = -pE_0 \theta. \] This torque provides the restoring force for angular simple harmonic motion (SHM).

Step 3: Equation of motion.
The equation of rotational motion is:
\[ I \frac{d^2\theta}{dt^2} = -pE_0 \theta, \] where \( I \) is the moment of inertia of the dipole about its center.
For a dipole consisting of two point masses of total mass \( m \) separated by a distance \( l \), the moment of inertia about its center is:
\[ I = \frac{ml^2}{2}. \] Substitute this into the equation of motion:
\[ \frac{ml^2}{2} \frac{d^2\theta}{dt^2} = -pE_0 \theta. \] Simplifying:
\[ \frac{d^2\theta}{dt^2} + \frac{2pE_0}{ml^2} \theta = 0. \] This is the standard equation for simple harmonic motion of angular frequency:
\[ \omega = \sqrt{\frac{2pE_0}{ml^2}}. \]

Step 4: Expression for time period.
The time period of small oscillations is given by:
\[ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{ml^2}{2pE_0}}. \] Substituting \( p = ql \):
\[ T = 2\pi \sqrt{\frac{ml^2}{2qEl}} = 2\pi \sqrt{\frac{ml}{2qE_0}}. \]

Final Answer:
\[ T = 2\pi \sqrt{\frac{ml}{2qE_0}}. \]