Question

Find the electric field at the centre of a semicircular ring if it is uniformly positive charged.

Answer 1

W split the semicircular arc into small segments $ds$. We have point charge on each segment as $\lambda ds$. The electric field due to this small segment of charge, from Coulomb's law,
$dE=\frac{1}{4\pi {\epsilon }_0{R}^2}\frac{\lambda ds}{R^2}$
Each little portion of the arc will give $dE$, in a different direction. We must therefore take components in order to find the total field at the centre.
Here, X-component of the electric field is zero. This is the result of the fact that $d{E}_x$ shown in the figure will be cancelled by the contribution from a symmetrically placed $ds$ on the left half of the arc.
Hence we need only to compute $d{E}_y$, which is vertically downward.
$d{E}_y=\frac{\lambda ds\cos \theta }{4\pi {\epsilon }_0{R}^2}=\frac{\lambda \left(Rd\theta \right)\cos \theta }{4\pi {\epsilon }_0{R}^2}$
$E_y=\int d{E}_y=\frac{\lambda }{4\pi {\epsilon }_0{R}^2}\stackrel{\pi /2}{\underset{-\pi /2}{\int }}\cos \theta d\theta \frac{\lambda }{2\pi {\epsilon }_{0}R}$