Charge on \(d \ell=\lambda \mathrm{Rd} \theta\)

where \(\lambda=\) linear charge density.

Electric field at centre due to \(d \ell\)

\(\mathrm{dE}=\mathrm{k} \cdot \frac{\lambda \mathrm{Rd} \theta}{\mathrm{R}^{2}}\)

We need to consider only the component \(dE\) \(\cos \theta,\) as the component \(dE\) sin \(\theta\) will cancel out.

\(\therefore  \) Total field at centre \(=2 \int_{0}^{\pi / 2} \mathrm{dE}\, \cos \theta\)

\(=2 \int_{0}^{\pi / 2} \frac{\mathrm{k} \lambda \mathrm{R}\, \cos \theta}{\mathrm{R}^{2}} \mathrm{d} \theta=\frac{2 \mathrm{k} \lambda}{\mathrm{R}} \int_{0}^{\pi / 2} \cos\, \theta \mathrm{d} \theta\)

\(=\frac{9}{2 \pi^{2} \epsilon_{0} R^{2}} \quad\left(\text { since } \lambda=\frac{q}{\pi R}\right)\)