a
Considering symmetric elements each of length \(dl\) at \(A\) and \(B,\) we note that electric fields

perpendicular to \(PO\) are cancelled and those along \(PO\) are added. The electric field due to an

element of length \(dl\) ( \(a d \theta\) ) along \(PO\)

\(\mathrm{d} E=\frac{1}{4 \pi \mathcal{E}_{0}} \frac{\mathrm{d} q}{a^{2}} \cos \theta\)

\(=\frac{1}{4 \pi \mathcal{E}_{0}} \frac{\lambda \mathrm{d} l}{a^{2}} \cos \theta\)

\(=\frac{1}{4 \pi \mathcal{E}_{0}} \frac{\lambda(a \mathrm{d} \theta)}{a^{2}} \cos \theta\)

Net electric field at \(\mathrm{O}\)

\(E=\int_{-\pi / 2}^{\pi / 2} \mathrm{d} E=2 \int_{0}^{\pi / 2} \frac{1}{4 \pi \mathcal{E}_{0}} \frac{\lambda \cos \theta \mathrm{d} \theta}{a^{2}}\)

\(2 \cdot \frac{1}{4 \pi \mathcal{E}_{0}} \frac{\lambda}{a}[\sin \theta]_{0}^{\pi / 2}=2 \cdot \frac{1}{4 \pi \mathcal{E}_{0}} \cdot \frac{\lambda}{a} \cdot 1=\frac{\lambda}{2 \pi \mathcal{E}_{0} a}\)