Question 8

Center of Mass Motion: The net force on the dipole in a uniform electric field is $\vec{F}_{net} = q\vec{E} + (-q)\vec{E} = 0$. Since the dipole starts from rest, its center of mass remains at rest and is not deflected. So, (A) is incorrect.

Angular Velocity and Energy: The torque on the dipole is $\vec{\tau} = \vec{p} \times \vec{E}$. Let $\theta$ be the angle with $\hat{i}$. $\vec{p} = qd(\cos\theta \hat{i} + \sin\theta \hat{j})$ and $\vec{E} = E\hat{j}$. $\tau = |\vec{p} \times \vec{E}| = (qd)E\cos\theta$. Work done by the field is $W = \int_0^{\theta_f} \tau d\theta = \int_0^{\theta_f} qdE \cos\theta d\theta = qdE \sin\theta_f$. By work-energy theorem, $\Delta K = W = qdE \sin\theta_f$. The moment of inertia about the center of mass is $I = m(\frac{d}{2})^2 + m(\frac{d}{2})^2 = \frac{md^2}{2}$. Kinetic energy $K = \frac{1}{2} I \omega_f^2 = \frac{1}{2} (\frac{md^2}{2}) \omega_f^2 = \frac{md^2}{4} \omega_f^2$. Setting $K = W$: $\frac{md^2}{4} \omega_f^2 = qdE \sin\theta_f \implies \omega_f = \sqrt{\frac{4qE \sin\theta_f}{md}}$. If $\omega_f = \sqrt{\frac{2qE}{md}}$, then $\frac{2qE}{md} = \frac{4qE \sin\theta_f}{md} \implies \sin\theta_f = \frac{1}{2} \implies \theta_f = \frac{\pi}{6}$. So, (B) is correct.

Kinetic Energy for $\theta_f = \pi/3$: $\Delta K = qdE \sin(\pi/3) = \frac{\sqrt{3}}{2} qdE$. Option (C) states $2\sqrt{3} qEd$, which is incorrect.

Rotation after $t_f$: For $t > t_f$, the electric field is turned off, so the torque $\vec{\tau} = 0$. In the absence of external torque and energy loss, the dipole continues to rotate with its final angular velocity $\omega_f$ around its center of mass. So, (D) is correct.