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Charge $q$ is uniformly distributed over a thin half ring of radius $R$. The electric field at the centre of the ring is
$\frac{q}{{2{\pi ^2}{\varepsilon _0}{R^2}}}$
$\frac{q}{{4{\pi ^2}{\varepsilon _0}{R^2}}}$
$\frac{q}{{4\pi {\varepsilon _0}{R^2}}}$
$\frac{q}{{2\pi {\varepsilon _0}{R^2}}}$
Solution
(a) From figure $dl = R d\theta $;
Charge on $ dl = \lambda R d\theta $ $\left\{ {\lambda = \frac{q}{{\pi R}}} \right\}$
Electric field at centre due to $dl$ is $dE = k.\frac{{\lambda Rd\theta }}{{{R^2}}}$.
We need to consider only the component $dE\,\cos \theta ,$ as the component $dE \,sin\theta$ will cancel out because of the field at $C$ due to the symmetrical element $dl'$.
Total field at centre $ = 2\int_{\,0}^{\,\pi /2} {\,dE\,\cos \theta } $
$ = \frac{{2k\lambda }}{R}\int_{\,0}^{\,\pi /2} {\,\cos \theta \,d\theta } = \frac{{2k\lambda }}{R} = \frac{q}{{2{\pi ^2}{\varepsilon _0}{R^2}}}$
Alternate method : As we know that electric field due to a finite length charged wire on it's perpendicular bisector is given by $E = \frac{{2k\lambda }}{R}\sin \theta .$
If it is bent in the form of a semicircle then $q = 90°$
$==>$ $E = \frac{{2k\lambda }}{R}$
= $2 \times \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{q/\pi R}}{R}} \right)$
= $\frac{q}{{2{\pi ^2}{\varepsilon _0}{R^2}}}$
