IIT JEE , AIEEE Forum - Physics , Electricity

ankitagg

consider a small element dq subtending angle d $\Theta$ at the center .

when u make the components dEcos $\Theta$ will be cancelled out. and all dEsin $\Theta$ components get added.

E= $\int_{0}^{180}dEsin\theta$

E= $\int_{0}^{180}kdq/r^2sin\theta$

now Q $\rightarrow$ pi.R

therefore dq $\rightarrow$ Q.Rd $\Theta$ /pi.R

dq $\rightarrow$ Q.d $\Theta$ /pi

put in the eq. and integrate it.

E= $kQ/\pi r^2\int_{0}^{180}sin\theta d\theta$

E= $2kQ/\pi r^2$

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