$(A)\;\frac{1}{2} \bigg( \frac{qE}{Ml}\bigg)^{1/2} \\ (B)\; \bigg( \frac{2qE}{Ml}\bigg)^{1/2}\\ (C)\; \bigg( \frac{qE}{Ml}\bigg)^{1/2} \\ (D)\; \bigg( \frac{qE}{2Ml}\bigg)^{1/2} $

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The force acting at both the charged constitute a couple.

$\tau$= Force $\times$ perpendicular disatnce

=>$\tau= qE \times AB \sin \theta$

$\qquad= qEl \sin \theta$

Since , $\theta$ is small

$\therefore \sin \theta =0$

=>$ \tau =qEl \theta$

Restoring torque $\tau= -qEL \theta$

If $\alpha $ is angular acceleration

$\tau=I \alpha$

$I= \large\frac{Ml^2}{2}$

=> $\tau= \large\frac{Ml^2}{2}$$ \alpha$

=> $\alpha =-w^2 \theta$

Now, $\alpha =-w^2 \theta$

=> $w^2=\bigg(\large\frac{2qE}{Ml}\bigg)$

=> $w=\bigg(\large\frac{2qE}{Ml}\bigg)^{1/2}$

Hence B is the correct answer.

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