Question

A unirmly changes insulation de 140 cm bent into the shape of a semicircle as shown in the figure The rod has a total charge of -7.50 C. Find the magnitude and direction of the electric field at O, the center of the

Answer 1

Electric field due to the semicircle, $E=\frac{Q}{2 \epsilon_0L^2}=\frac{7.5*10^{-6}}{2*8.8510^{-12}*0.14^2}=21.61*10^6 N$

The direction of the electric field is towards the semicircle.