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 hall as shown in figure. The electric field E at P, the center of the semicircle, is

Take PO as the x-axis and PA as the y-axis. Consider two elements EF and E'F' of width dθ at angular distance θ above and below PO, respectively. The magnitude of the field at P due to either element isdE=14πε0rdθ×Q/(πr/2)r2=Q2π2ε0r2dθResolving the fields, we find that the components along PO sum up to zero, and hence the resultant field is along PB. Therefore, field at P due to pair of elements is 2d E sinθE=∫0π/2 2dEsin⁡θ=2∫0π/2 Q2πε0r2sin⁡θdθ=Qπ2ε0r2