Motion of Electric Dipole in Uniform Electric Field Turned On and Off

To determine the correct options, let us analyze the forces, torques, and energy of the dipole system step-by-step.

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1. Center of Mass Motion (Option A)

The electric dipole consists of two equal and opposite charges, $+q$ and $-q$, each of mass $m$, placed in a uniform electric field $\vec{E} = E\hat{j}$.

The net force acting on the dipole is: $\vec{F}_{\text{net}} = q\vec{E} + (-q)\vec{E} = \vec{0}$

Since the net external force is zero, the acceleration of the center of mass is: $\vec{a}_{\text{cm}} = \frac{\vec{F}_{\text{net}}}{2m} = \vec{0}$

Given that the dipole is initially at rest, its center of mass remains at rest throughout the motion and does not deflect towards $\hat{j}$.

Thus, Option A is incorrect.

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2. Rotational Motion and Energy Conservation (Options B and C)

The center of mass of the dipole lies at the midpoint of the line joining the two charges. The distance of each mass $m$ from the center of mass is $\frac{d}{2}$.

The moment of inertia $I$ of the dipole about the axis passing through its center of mass and perpendicular to the plane of rotation is: $I = m\left(\frac{d}{2}\right)^2 + m\left(\frac{d}{2}\right)^2 = \frac{1}{2}md^2$

At any angle $\theta$ made by the dipole moment $\vec{p}$ with the $\hat{i}$-axis, the dipole moment vector is: $\vec{p} = qd (\cos\theta \hat{i} + \sin\theta \hat{j})$

The torque acting on the dipole about the center of mass due to the electric field $\vec{E} = E\hat{j}$ is: $\vec{\tau} = \vec{p} \times \vec{E} = (qd \cos\theta \hat{i} + qd \sin\theta \hat{j}) \times (E\hat{j}) = qEd\cos\theta \hat{k}$

The work done by the torque as the dipole rotates from $\theta = 0$ to $\theta = \theta_f$ is: $W = \int_{0}^{\theta_f} \tau \, d\theta = \int_{0}^{\theta_f} qEd\cos\theta \, d\theta = qEd\sin\theta_f$

By the Work-Energy Theorem, the change in kinetic energy of the dipole is: $\Delta K = K_f - K_i = \frac{1}{2}I\omega_f^2 - 0 = \frac{1}{2}\left(\frac{1}{2}md^2\right)\omega_f^2 = \frac{1}{4}md^2\omega_f^2$

Equating the work done to the kinetic energy: $\frac{1}{4}md^2\omega_f^2 = qEd\sin\theta_f \implies \omega_f^2 = \frac{4qE}{md}\sin\theta_f$

Evaluating Option B:

Given $\omega_f = \sqrt{\frac{2qE}{md}}$, we have: $\omega_f^2 = \frac{2qE}{md}$

Substituting this into the kinetic energy equation: $\frac{2qE}{md} = \frac{4qE}{md}\sin\theta_f \implies \sin\theta_f = \frac{1}{2} \implies \theta_f = \frac{\pi}{6}$

Thus, Option B is correct.

Evaluating Option C:

If $\theta_f = \frac{\pi}{3}$, the change in kinetic energy is: $\Delta K = qEd\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}qEd$

Since this does not equal $2\sqrt{3}qEd$, Option C is incorrect.

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3. Motion for $t > t_f$ (Option D)

At time $t = t_f$, the electric field is turned off ($\vec{E} = \vec{0}$).

For $t > t_f$:

• The net torque on the dipole is $\vec{\tau} = \vec{0}$.
• Consequently, the angular acceleration is $\alpha = \frac{\tau}{I} = 0$, meaning the angular velocity remains constant at $\omega = \omega_f$.
• The net force remains $\vec{F}_{\text{net}} = \vec{0}$, so the center of mass remains stationary.

Therefore, for $\theta_f = \pi/4$ (or any other angle), the dipole continues to rotate around its center of mass with a constant angular velocity after $t > t_f$.

Thus, Option D is correct.

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Conclusion

The correct options are B and D.