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(C) The dipole will oscillate about its center of mass $(CM)$.Let the mass of the positive charge be $m$ and the mass of the negative charge be $2m$.T

Vedclass · Accepted Answer

$\sqrt{\frac{3 qE }{2\,ml }}$

Vedclass · Answer

(C) The dipole will oscillate about its center of mass $(CM)$.Let the mass of the positive charge be $m$ and the mass of the negative charge be $2m$.The distance of the positive charge from the $CM$ is $r_1 = \frac{2m}{m+2m} \cdot l = \frac{2l}{3}$.The distance of the negative charge from the $CM$ is $r_2 = \frac{m}{m+2m} \cdot l = \frac{l}{3}$.The moment of inertia about the $CM$ is $I = m r_1^2 + (2m) r_2^2 = m(\frac{2l}{3})^2 + 2m(\frac{l}{3})^2 = m(\frac{4l^2}{9}) + 2m(\frac{l^2}{9}) = \frac{6ml^2}{9} = \frac{2ml^2}{3}$.For small $	heta$,the restoring torque is $	au = -pE \sin 	heta \approx -q l E 	heta$.Using $	au = I \alpha$,we have $I \alpha = -qlE 	heta$.$\frac{2ml^2}{3} \alpha = -qlE 	heta \Rightarrow \alpha = -\frac{3qE}{2ml} 	heta$.Comparing with $\alpha = -\omega^2 	heta$,we get $\omega = \sqrt{\frac{3qE}{2ml}}$.

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