Find the electric potential at the center of curvature of a

Find the electric potential at the center of curvature of a uniformly charged plastic semicircle with a radius of ¼ m that carries a total charge of J36 nC and is embedded in a nonconductive material that has a Coulomb Constant of 20 MV·m/C.

Solution

potential at the center of curvature of arc is

V = kQ/a

V = 20*10⁶ * 36*10^-9 / (0.25 ) = 2.88 V