Question

A rod of length L has a total charge Q distributed uniformly along its length. It is bent in th shape of a semicircle. Find the magnitude of the electric field at the centre of curvatere of he semicircle.

Answer 1

$dq=\frac{Q}{L}R.d\theta$
Net electric field have only vertical component
$=\int dE\cos \theta =\frac{1}{4\pi {ϵ}_0}\frac{Q}{L}{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{Rd\theta }{R^2}\cos \theta$

$=\frac{1}{4\pi {ϵ}_0}\frac{Q}{LR}[\sin \theta {]}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}=\frac{1}{4\pi {ϵ}_0}\frac{Q}{LR}[1-0(-1)]=\frac{2Q}{4\pi ϵLR}=\frac{Q}{2{ϵ}_0{L}^2}(\because \pi R=L)$