Physics - Electrostatics Question with Solution | TestHub

When the wire is brought in a uniform field E, then the torque is given by

$\tau =qEL\sin \theta$

$=qEL\theta$$\left[\because \theta is very small\right]$

The moment of inertia of rod AB about O is

$I=m{\left(\frac{L}{2}\right)}^2+m{\left(\frac{L}{2}\right)}^2=\frac{m{L}^2}{2}$

As$\tau =I\alpha$



So,$\alpha =\frac{\tau }{I}=\frac{\mathrm{2q}EL\theta }{m{L}^2}$

$\Rightarrow {\omega }^2\theta =\frac{2qEL\theta }{m{L}^2}$$\left[\because \alpha ={\omega }^2\theta \right]$

$\Rightarrow {\omega }^2=\frac{2qE}{mL}$

The time period of the wire is

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m{L}^2}{2qE}}$

The rod will become parallel to the field in time$\frac{T}{4}$

So,$t=\frac{T}{4}=\frac{\pi }{2}\sqrt{\frac{mL}{2qE}}$