Physics - Electrostatics Question with Solution | TestHub

Torque when the wire is brought in a uniform field $\mathrm{E}$. $\begin{array}{l}\tau =\mathrm{qEL}\sin \theta \\ =\mathrm{qEL}\theta \end{array}[\because \theta \text{ is very small }]$ Moment of inertia of rod AB about O $\mathrm{I}=\mathrm{m}{\left(\frac{\mathrm{L}}{2}\right)}^2+\mathrm{m}{\left(\frac{\mathrm{L}}{2}\right)}^2=\frac{{\mathrm{mL}}^2}{2}$
$\begin{array}{l}\tau =\mathrm{I}\alpha .\\ \therefore \alpha =\frac{\tau }{\mathrm{I}}=\frac{\mathrm{qEL}\theta }{\frac{{\mathrm{mL}}^2}{2}}\\ \Rightarrow {\omega }^2\theta =\frac{2\mathrm{qEL}\theta }{{\mathrm{mL}}^2}\left[\because \theta ={\omega }^2\theta \right]\\ \Rightarrow {\omega }^2=\frac{2\mathrm{qE}}{\mathrm{mL}}\end{array}$ Time period of the wire $\mathrm{T}=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{\mathrm{mL}}{2\mathrm{qE}}}$ The rod will become parallel to the field in time $\frac{T}{4}$. $\therefore \mathrm{t}=\frac{\mathrm{T}}{4}=\frac{\pi }{2}\sqrt{\frac{\mathrm{mL}}{2\mathrm{qE}}}$