Question

A thin glass rod is bent into a semi circle of radius R a charge Q is uniformly distributed along the upper-half and a charge −Q is uniformly distributed along the lower-half. Find the electric field E~ at the center of the semicircle. 

Answer 1

The semi circle should look like the one shown below

considering the center P of the semicircle, and taking two elements of width $d\theta$ the magnitude of the field at P is;

$dE=\frac{1}{4\pi \epsilon_{o}} \times \frac{rd\theta \times Q(\pi r/2)}{r^2}=\frac{Q}{2\pi ^2\epsilon_{o}r^2} d\theta$

field at x due to pair of electrons$=2dEsin\theta$

$E=_{0}\int^{\pi /2} 2dEsin\theta=2\times _{0}\int^{\pi /2}\frac{Q}{2\pi \epsilon_{o} r^2}sin\theta d\theta=\frac{Q}{\pi ^2 \epsilon_{o}r^2}$