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(B) The torque acting on the dipole is $	au = pE \sin 	heta$. For small $	heta$,$\sin 	heta \approx 	heta$,so $	au = pE	heta$.The moment of ine

Vedclass · Accepted Answer

$t = \frac{\pi }{2}\sqrt {\frac{{mL}}{{2qE}}} $

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(B) The torque acting on the dipole is $	au = pE \sin 	heta$. For small $	heta$,$\sin 	heta \approx 	heta$,so $	au = pE	heta$.The moment of inertia of the system about the center of the rod is $I = m(\frac{L}{2})^2 + m(\frac{L}{2})^2 = \frac{mL^2}{2}$.The equation of motion is $I \alpha = -	au$,which gives $\frac{mL^2}{2} \frac{d^2	heta}{dt^2} = -qLE	heta$.This is the equation of simple harmonic motion with angular frequency $\omega = \sqrt{\frac{qLE}{I}} = \sqrt{\frac{qLE}{mL^2/2}} = \sqrt{\frac{2qE}{mL}}$.The time taken to go from angle $	heta$ to $0$ is one-quarter of the time period $T$,i.e.,$t = \frac{T}{4} = \frac{1}{4} 	imes \frac{2\pi}{\omega} = \frac{\pi}{2\omega}$.Substituting $\omega$,we get $t = \frac{\pi}{2} \sqrt{\frac{mL}{2qE}}$.

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