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 $dl( = ad\theta)$ along PO.

$dE = {1 \over {4\pi {\varepsilon _0}}}{{dq} \over {{a^2}}}\cos \theta$
$\left(\because {dl = ad\theta } \right)$

$= {1 \over {4\pi {\varepsilon _0}}}{{\lambda dl} \over {{a^2}}}\cos \theta$

$= {1 \over {4\pi {\varepsilon _0}}}{{\lambda \left( {ad\theta } \right)} \over {{a^2}}}\cos \theta$ Net electric field at O

$E = \int_{ - \pi /2}^{\pi /2} {dE}$

$= 2\int_O^{\pi /2} {{1 \over {4\pi {\varepsilon _0}}}{{\lambda \left( {ad\theta } \right)} \over {{a^2}}}\cos \theta }$

$= 2.{1 \over {4\pi {\varepsilon _0}}}{\lambda \over a}\left[ {\sin \theta } \right]_o^{\pi /2}$

$= 2.{1 \over {4\pi {\varepsilon _0}}}{\lambda \over a}.1 = {\lambda \over {2\pi {\varepsilon _0}a}}$