Consider an electric dipole comprising two charges $+q$ and $-q$ each with mass $m$, separated by a fixed distance $d$ and initially at rest with its dipole moment pointing along $\hat{i}$. A uniform electric field $E\hat{j}$ is turned on at time $t = 0$ and it is turned off at $t = t_f$, when the dipole moment makes an angle $\theta_f$ with $\hat{i}$. Neglecting any sources of energy loss, correct option(s) is/are:

The net force on the dipole in a uniform electric field is zero, as the forces on the two charges are equal and opposite ($qE\hat{j}$ and $-qE\hat{j}$). Therefore, the center of mass remains at rest and is not deflected. Option (A) is incorrect.

The moment of inertia of the dipole about its center of mass is $I = m\left(\dfrac{d}{2}\right)^2 + m\left(\dfrac{d}{2}\right)^2 = \dfrac{md^2}{2}$.

The torque on the dipole is $\vec{\tau} = \vec{p} \times \vec{E}$. The angle between the dipole moment $\vec{p}$ and the electric field $\vec{E}$ is $90^{\circ} - \theta$. The magnitude of the torque is $\tau = pE \sin(90^{\circ} - \theta) = qdE \cos\theta$.

The work done by the electric field as the dipole rotates from $\theta = 0$ to $\theta = \theta_f$ is equal to the change in kinetic energy:

$\Delta K = \int_{0}^{\theta_f} \tau d\theta = \int_{0}^{\theta_f} qdE \cos\theta d\theta = qdE \sin\theta_f$

Equating this to the final kinetic energy $\dfrac{1}{2}I\omega_f^2$:

$\dfrac{1}{2} \left(\dfrac{md^2}{2}\right) \omega_f^2 = qdE \sin\theta_f$

$\dfrac{md^2}{4} \omega_f^2 = qdE \sin\theta_f$

For option (B), substituting $\omega_f = \sqrt{\dfrac{2qE}{md}}$:

$\dfrac{md^2}{4} \left(\dfrac{2qE}{md}\right) = qdE \sin\theta_f$

$\dfrac{qdE}{2} = qdE \sin\theta_f \Rightarrow \sin\theta_f = \dfrac{1}{2} \Rightarrow \theta_f = \dfrac{\pi}{6}$

Thus, option (B) is correct.

For option (C), if $\theta_f = \dfrac{\pi}{3}$, the change in kinetic energy is $\Delta K = qdE \sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2} qEd$. Thus, option (C) is incorrect.

For option (D), after $t > t_f$, the electric field is turned off. The net torque on the dipole becomes zero. By Newton's first law of rotational motion, the dipole will continue to rotate about its center of mass with a constant angular velocity. Thus, option (D) is correct.

Answer: If the magnitude of the final angular velocity $\omega_f = \sqrt{\dfrac{2qE}{md}}$, then $\theta_f = \dfrac{\pi}{6}$.; For $\theta_f = \pi/4$, the dipole rotates around its center of mass with a constant angular velocity after $t > t_f$.