Let us consider a differential element dl. charge on this element. dq=(‌
q
Ï€r
)dl =‌
q
Ï€r
(rdθ)‌‌( as dl=rdθ) =(‌
q
Ï€
)dθ Electric field at O due to dq is ‌dE=‌
1
4Ï€E0
⋅‌
dq
r2
‌=‌
1
4Ï€E0
⋅‌
q
Ï€r2
dθ The component dE‌cos‌θ will be counter balanced by another element on left portion. Hence resultant field at O is the resultant of the component dE‌sin‌θ only. ‌∴E=∫dEsin‌θ=
Ï€
∫
0
‌
q
4Ï€2r2E0
‌sin‌θdθ ‌=‌
q
4π2r2∈0
[−cos‌θ]0π ‌=‌
q
4π2r2∈0
(+1+1) ‌=‌
q
2π2r2∈0
The directions of E is towards negative y-axis. ∴