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A uniform rod $AB$ of mass $m$ and length $l$ is hinged at its mid point $C$ . The left half $(AC)$ of the rod has linear charge density $-\lambda $ and the right half $(CB)$ has $+\lambda $ where $\lambda $ is constant . A large non conducting sheet of unirorm surface charge density $\sigma $ is also .present near the rod. Initially the rod is kept perpendicular to the sheet. The end $A$ of the rod is initially at a distance $d$ . Now the rod is rotated by a small angle in the plane of the paper and released. The time period of small angular oscillations is
$T = 2\pi \sqrt {\frac{{m{ \in _0}}}{{3\lambda \sigma }}} $
$T = 2\pi \sqrt {\frac{{2m{ \in _0}}}{{\lambda \sigma }}} $
$T = 2\pi \sqrt {\frac{{4m{ \in _0}}}{{3\lambda \sigma }}} $
$T = 2\pi \sqrt {\frac{{2m{ \in _0}}}{{3\lambda \sigma }}} $
Solution
The sheet produces a uniform electric field $\mathrm{E}=\frac{\sigma}{2 \epsilon_{0}}$ towards right. The part $\mathrm{AC}$ and CB will experience electric force $\mathrm{F}$ as shown. They can be considered to be acting at the mid points of those parts respectively. The. rod. will experience torque about the point $\mathrm{'e'}$ in the anticlockwise direction
whose magnitude is $\tau = {\rm{F}}\frac{\ell }{2}\sin \theta \simeq \frac{{F\ell }}{2}\theta ;$
But ${\rm{F}} = \lambda \cdot \frac{\ell }{2} \cdot \frac{\sigma }{{2{ \in _0}}} = \frac{{\lambda \ell \sigma }}{{4{ \in _0}}}$
$\therefore \quad \tau=\left(\frac{\lambda \ell \sigma}{8 \sigma}\right) \theta$
Now since $\tau$ is towards the mean position and $ = \tau \,\mu \,\theta $
$\therefore$ it will do $\mathrm{SHM} \rightarrow$ Heme proved
and $\tau = {\rm{I}}\mu = \frac{{\lambda {\ell ^2}\sigma }}{{8{ \in _0}}}\theta \quad \Rightarrow \frac{{{\rm{m}}{\ell ^2}\sigma }}{{12}}\alpha = \frac{{\lambda {\ell ^2}\sigma }}{{8{ \in _0}}}\alpha $
$ \Rightarrow a = \left( {\frac{{3\lambda \sigma }}{{2m{ \in _0}}}} \right)\theta $
$\therefore {w^2} = \frac{{3n\sigma }}{{2m{ \in _0}}} \Rightarrow {\left( {\frac{{2\pi }}{T}} \right)^2}$
$ \Rightarrow {\rm{T}} = 2\pi \sqrt {\frac{{2{\rm{m}}{ \in _0}}}{{3\lambda \sigma }}} $
