Let us consider a differential element dl. charge on this element. dq=(
q
πr
)dl =
q
πr
(rdθ)( as dl=rdθ) =(
q
π
)dθ Electric field at O due to dq is dE=
1
4πE0
⋅
dq
r2
=
1
4πE0
⋅
q
πr2
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=