A thin glass rod is bent into a semicircle of radius r. A charge +Q is uniformly distributed along the upper half, and a charge –Q is uniformly distributed along the lower half, as shown in the figure. The resultant electric field E at P, the center of the semicircle, is

The magnetic of the field at p due to either element isdE=14πε0rdθ×Q/πr/2r2=Q2π2ε0r2dθ where θ is the angle made by the two radius vector on both sides of  horizontal  and rdθ are the two elements on both the sides  ∴ field at P due to pair of elements is  2dEsinθi.e.,E=∫0π/22dEsinθ