Question

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

Answer 1

Consider an element of angular width dθ, as shown in the figure.



$dq=\frac{Q}{L}Rd\mathrm{\theta }$
The net electric field has a vertical component only.
$$\begin{gathered} \mathrm{So},E_{net}=\int dE\sin \theta \\ =\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{Q}{L}\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{Rd\theta }{R^2}\sin \theta \end{gathered}$$

$$\begin{gathered} =\frac{-1}{4\mathrm{\pi }{\epsilon }_0}\frac{\mathrm{Q}}{LR}{\left[\cos \theta \right]}_0^{\mathrm{\pi }} \\ =\frac{2Q}{4\pi {\epsilon }_{0}LR} \\ =\frac{Q}{2{\epsilon }_0{L}^2}\left(\because \pi R=L\right) \end{gathered}$$