Electrostatics

Concept:The net force on an electric dipole in a uniform field is zero, but a torque acts that causes rotation; after the field is removed, the dipole rotates with constant angular velocity.Explanation:Forces on charges: $\vec{F}_{+} = qE\hat{\jmath}$, $\vec{F}_{-} = -qE\hat{\jmath}$.Net external force $\vec{F}_{\text{ext}} = 0$, so centre of mass acceleration $\vec{a}_{\text{cm}} = 0$.Initially at rest, the centre of mass remains stationary – option (A) is incorrect.Torque magnitude: $\tau = |\vec{p} \times \vec{E}| = pE\cos\theta$, where $\theta$ is the angle of dipole moment from the x–axis.Given $p = qd$, so $\tau = qdE\cos\theta$.Moment of inertia about centre of mass: $I = 2 \times m\left(\frac{d}{2}\right)^2 = \frac{1}{2}md^2$.Work done by torque from $0$ to $\theta_f$: $W = \int_{0}^{\theta_f}\tau\,d\theta = qdE\sin\theta_f$.By work–energy theorem, $\frac{1}{2}I\omega_f^2 = qdE\sin\theta_f$.Substitute $I$: $\frac{1}{4}md^2\omega_f^2 = qdE\sin\theta_f$, giving $\omega_f^2 = \frac{4qE}{md}\sin\theta_f$.For option (B): given $\omega_f = \sqrt{\frac{2qE}{md}}$. Equating: $\frac{4qE}{md}\sin\theta_f = \frac{2qE}{md}$ ⇒ $\sin\theta_f = \frac{1}{2}$ ⇒ $\theta_f = \frac{\pi}{6}$. Option (B) is correct.For option (C): $\theta_f = \frac{\pi}{3}$, then $\Delta K = qdE\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}qEd$, not $2\sqrt{3}qEd$. Option (C) is incorrect.After $t_f$, $\vec{E} = 0$ so torque $\vec{\tau} = \vec{p}\times \vec{0} = 0$.Hence angular acceleration $\vec{\alpha} = 0$, so angular velocity remains constant at $\omega_f$. Option (D) is correct.Answer:The correct options are (B) and (D).