Charge q is uniformly distributed over a thin half ring of radius R. The electric field at the centre of the ring is 

From figure  $dl=Rd\text{θ}$
Charge on  $\text{F = }\frac{\text{1}}{{\text{4πε}}_{\text{0}}}\text{.}\frac{\text{Qq}}{\left({\text{x}}^{\text{2}}\text{+}\frac{{\text{d}}^{\text{2}}}{\text{4}}\right)}$
Electric field at center due to dl is $\text{dE = k.}\frac{\text{λRdθ}}{{\text{R}}^{\text{2}}}$.
We need to consider only the component  $\text{dE cos\hspace{0.05em}θ}$, as the component $\text{dE cos\hspace{0.05em}θ}$ will cancel out because of the field at C  

due to the symmetrical element dl’ Total field at centre                              
$\begin{array}{l}\text{= 2}{\int }_{\text{0}}^{\text{π/2}}\text{dE\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}cos\hspace{0.17em}\hspace{0.17em}θ}\\ \text{= }\frac{\text{2kλ}}{\text{R}}{\int }_{\text{0}}^{\text{π/2}}\text{dE\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}cos\hspace{0.17em}\hspace{0.17em}θ}\text{=}\frac{\text{2kλ}}{\text{R}}\text{=}\frac{\text{q}}{{\text{2π}}^{\text{2}}{\text{ε}}_{\text{0}}{\text{R}}^{\text{2}}}\end{array}$