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. 

\[dq = \frac{Q}{L}Rd\theta\] 

The net electric field has a vertical component only.

\[\text{So},    E_{net}  = \int dE  \sin\theta\] 

\[                         = \frac{1}{4\pi \epsilon_0}\frac{Q}{L} \int\limits_0^\pi \frac{Rd\theta}{R^2}\sin\theta\]

\[= \frac{- 1}{4\pi \epsilon_0}\frac{Q}{LR} \left[ \cos\theta \right]_0^\pi \] 

\[ = \frac{2Q}{4\pi \epsilon_0 LR}\] 

\[ = \frac{Q}{2 \epsilon_0 L^2}    \left( \because \pi R = L \right)\]