Given that, linear charge density of non-conducting ring λ=λ0‌cos‌φ, and Azimuthal angle =φ Consider two elements of length (dl=Rdφ) are taken symmetrically at angle φ on both sides as shown in figure. Here, dE1=dE2 and sin‌φ components cancelled out each other. Charge on the elements, dq1=λdl=(λ0‌cos‌φ)rdφ . . . (i)
Now, net field at centre of ring due to both elements, dE=2dE1‌cos‌φ⇒
E
∫
0
dE=
Ï€
∫
0
2(‌
dq
4πε0r2
)‌cos‌φ =
Ï€
∫
0
‌
2
4πε0r2
(λ0‌cos‌φ)(rdφ)‌cos‌φ‌‌[∵ from Eq. (i) ] =‌