The given situation is shown below.
On straight wire
λ= charge per unit length
=Q∕L small charge on length element
dl is
dq=λ×dl .....(i)
Electric field at centre due to a point charge
=K×‌ When each electric line is broken into 2 components, horizontal component of electric field balance each other, only vertical components remain.
So, net electric field along vertical
dE‌vertical ‌‌=2dE‌cos‌θ=2‌‌cos‌θ‌=‌×λdl( But
dθ=‌⇒dl=rdθ)
Integrating the electric field,
E‌=‌| 2Kλ‌cos‌θ‌d‌θ |
| r |
but length of semi-circular path is
L‌=πr⇒‌‌r‌=‌∴‌‌E‌=‌=‌∴E‌=‌×‌=‌.