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Step-by-Step Solution
Step 1: Understand the System
An electric dipole consists of two equal and opposite charges, each of mass $m$, separated by a distance $d$. When placed in a uniform electric field $E$, the dipole experiences a torque tending to align it with the field. We are interested in the small oscillations about this equilibrium orientation.
Step 2: Express the Torque on the Dipole
The dipole moment is $p = q \, d$. The torque on a dipole in a uniform electric field is:
$\tau = p E \sin \theta \;=\; q \, d \, E \,\sin \theta,$
where $\theta$ is the angular displacement from the equilibrium position.
Step 3: Approximate for Small Oscillations
For very small angular displacements, $\sin \theta \approx \theta$. The restoring torque can then be written as:
$\tau \approx - q \, d \, E \,\theta.$
(The negative sign indicates that the torque acts to restore the dipole to $\theta = 0$.)
Step 4: Write the Rotational Equation of Motion
The rotational equation of motion is given by:
$I \frac{d^2\theta}{dt^2} = \tau,$
where $I$ is the moment of inertia of the system about the axis through the midpoint of the dipole. Since there are two charges of mass $m$ each at a distance $\frac{d}{2}$ from the center, the total moment of inertia is:
$I = 2 \left(m \left(\frac{d}{2}\right)^2 \right) = 2 \left(m \frac{d^2}{4}\right) = \frac{m d^2}{2}.$
Step 5: Substitute into the Equation and Solve
Substituting the torque expression $\tau \approx -q \, d \, E \,\theta$ and the moment of inertia into the rotational equation of motion, we get:
$\frac{m d^2}{2} \frac{d^2 \theta}{dt^2} = - \, q \, d \, E \,\theta.$
Rearrange to obtain:
$\frac{d^2 \theta}{dt^2} = - \,\frac{2qE}{m d}\,\theta.$
This is the differential equation of simple harmonic motion of the form:
$\frac{d^2 \theta}{dt^2} + \omega^2 \theta = 0,$
where $\omega$ is the angular frequency of these small oscillations. By comparing terms, we identify:
$\omega \;=\; \sqrt{\frac{2 q E}{m d}}.$
Step 6: Final Answer
Thus, the angular frequency $\omega$ of the dipole’s small oscillations is:
$\displaystyle \omega = \sqrt{\frac{2qE}{md}}.$
