# A large sheet carries uniform surface charge density sigma.A rod of length 2l has a linear charge density lamda on one half and -lama on the other.The rod is hinged at mid point O and makes theta with normal to the sheet.The torque experienced by the rod is

Hi,

Elementary length. = Eλdx

Now, torque = Eλdxxsinθ

$\tau ={\int }_{0}^{l}E\lambda xdx\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}⇒\tau =\frac{\lambda E\mathrm{sin}\theta {l}^{2}}{2}$

Thus, we get λEsinθl2/2

similarly for lower part torque will be λEsinθl2/2

so net torque = λEsinθl2
E=σ/2ε0

so τ =  λEsinθl2/2ε0

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