Question

Find the electric field at the centre of an arc of linear charge density $\lambda$, radius $R$ subtending angle $\varphi$ at the centre.

Answer 1

Consider two small elements subtending angle $d\theta$ at centre symmetrically at angle $\theta$ with symmetry line. Electric field due to each of them
$d{E}^{\prime }=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda Rd\theta }{R^2}=\frac{1}{4\pi {\epsilon }_0}\frac{\lambda }{R}d\theta$
Net field due to both the symmetrical elements $dE=2d{E}^{\prime }\cos \theta$
$E=\int dE=\stackrel{+/2}{\underset{0}{\int }}2\frac{1}{4\pi {\epsilon }_0}\frac{\lambda }{R}\cos \theta d\theta =\frac{1}{4\pi {\epsilon }_0}\frac{2\lambda }{R}\sin \left(\frac{\varphi }{2}\right)$