A uniform rod AB of mass m and length l | vedclass

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Question

The sheet produces a uniform electric field $\mathrm{E}=\frac{\sigma}{2 \epsilon_{0}}$ towards right. The part $\mathrm{AC}$ and CB will experience el

Vedclass · Accepted Answer

$T = 2\pi \sqrt {\frac{{2m{ \in _0}}}{{3\lambda \sigma }}} $

Vedclass · Answer

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 $	au  = {m{F}}\frac{\ell }{2}\sin 	heta  \simeq \frac{{F\ell }}{2}	heta ;$

But ${m{F}} = \lambda  \cdot \frac{\ell }{2} \cdot \frac{\sigma }{{2{ \in _0}}} = \frac{{\lambda \ell \sigma }}{{4{ \in _0}}}$

$	herefore \quad 	au=\left(\frac{\lambda \ell \sigma}{8 \sigma}ight) 	heta$

Now since $	au$ is towards the mean position and $ = 	au \,\mu \,	heta $

$	herefore$ it will do $\mathrm{SHM} ightarrow$ Heme proved

and $	au  = {m{I}}\mu  = \frac{{\lambda {\ell ^2}\sigma }}{{8{ \in _0}}}	heta \quad  \Rightarrow \frac{{{m{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}}}} ight)	heta $

$	herefore {w^2} = \frac{{3n\sigma }}{{2m{ \in _0}}} \Rightarrow {\left( {\frac{{2\pi }}{T}} ight)^2}$

$ \Rightarrow {m{T}} = 2\pi \sqrt {\frac{{2{m{m}}{ \in _0}}}{{3\lambda \sigma }}} $

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