physics---electricity2

Consider a radius R that is at an angle a above the horizotnal axis through P
The small arc on the semi-circle subtebded by such radius = R(da), where da is the differential angle increment at radius R.
Charge on this arc = [R(da)/pi.R].Q = Q.(da)/pi
Differential electric field at P, dE = k[Q(dq)/pi]/R^2
where k is the electrostatic constant (= 9x10^9 Nm^2/C^2)

Hence, electric field at P
E = integral { (kQ/Pi.R^2).(da)} with limits of integration from -pi/2 to pi/2
E = kQ/R^2

In figure 5.2b, the electric field E' = kQ/R^2
Hence, the two electric fields are equal.