A point particle of mass M is attached to one | vedclass

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(C) The system acts as an electric dipole in a uniform electric field. The moment of inertia $I$ of the system about the center of the rod is $I = M(L

Vedclass · Accepted Answer

$\frac{\pi }{2}\sqrt {\frac{{ML}}{{2qE}}} $

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(C) The system acts as an electric dipole in a uniform electric field. The moment of inertia $I$ of the system about the center of the rod is $I = M(L/2)^2 + M(L/2)^2 = M(L^2/4 + L^2/4) = ML^2/2$.The torque acting on the dipole is $	au = pE \sin 	heta$,where $p = qL$.For small angles,$\sin 	heta \approx 	heta$,so $	au = qLE 	heta$.The equation of motion is $I \alpha = -	au$,where $\alpha = d^2	heta/dt^2$.$ML^2/2 \cdot (d^2	heta/dt^2) = -qLE 	heta$.$d^2	heta/dt^2 = -(2qE / ML) 	heta$.This is the equation of simple harmonic motion with angular frequency $\omega = \sqrt{2qE / ML}$.The time period of oscillation is $T = 2\pi / \omega = 2\pi \sqrt{ML / 2qE}$.The time taken to move from the initial small angle to the equilibrium position (parallel to the field) is one-quarter of the time period.$t = T/4 = (1/4) \cdot 2\pi \sqrt{ML / 2qE} = \frac{\pi}{2} \sqrt{\frac{ML}{2qE}}$.

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