IIT JEE , AIEEE Forum - Physics , Electricity

riku

Mast question hai!!

The symmetry of this distribution is such that half ring becomes negatively charged,, other positively charged...vector E at the centre of the ring is directed to right..

$dEcos\phi=\frac{dq cos\phi}{4\Pi\epsilon {R}^{2}}$

we will have to integrate this projection and we obtain

$E=\frac{\lambda }{4\Pi\epsilon R}\int_{0}^{2\pi} {cos}^{2}\phi d\phi= \frac{\lambda}{4\Pi\epsilon R}$

this is the electric field at center.

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