Figure a shows a nonconducting rod with a uniformly distribu

Figure (a) shows a nonconducting rod with a uniformly distributed charge +Q. The rod forms a 10/23 of circle with radius P and produces an electric field of magnitude E_arc at its center of curvature P. If the arc is collapsed to a point at distance R from P (see Figure (b)), by what factor is the magnitude of the electric field at P multiplied?

Solution

electric field due to arc at centre = 2k (lambda) / R = 2k(23Q/10piR) / R

Earc = 46kQ / 10piR^2

for figure b) E = kQ / R^2

now ration = Earc / E = 46/10pi = 1.46