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=Rdθ  Charge on dl=λRdθ  λ=qπRElectric field at centre due to di, is dE=k·λRdθR2We need to consider only the component dE cos θ as the component dE sin θ will cancel out because of the field at C due to the symmetrical element dl'. Total field at centre =2∫0π/2dEcosθ=2kλR∫0π/2cosθdθ=2kλR=q2π2ε0R2