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

From figure dl = R d θ;Charge on dl = λR dθ   Electric field at centre due to dl isWe need to consider only the component  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 Alternate method : As we know that electric field due to a finite length charged wire on it's perpendicular bisector is given by If it is bent in the form of a semicircle then θ = 90°⇒