Question

Find force on semicircular ring due to point charge $q$.

Answer 1

Consider an element at an angle $\theta$ charge $dq=\lambda Rd\theta$

$\therefore$ Force on charge $dq,df=\frac{Kqdq}{R^2}$

$=\frac{Kq\lambda Rd\theta }{R^2}$

Due to symmetry of the system we see $F_x$ gets cancelled

$\therefore \int d{f}_y=\frac{Kg\lambda }{R}{\int }_0^{\pi }\sin \theta d\theta$

$=\frac{Kq\lambda }{R}{[-\cos \theta ]}_0^{\pi }$

$F=\frac{2Kq\lambda }{R}$