Question

i had a question. A thin wire is bent into semicircle, having a radius of curvature equal to 6.40cm. It's charged uniformly with charge density 33 x 10^-9 C/m. What is the potential at the center of curvature?

Answer 1

I used the formula V=U/q

(8987742438)*(6.64E-9)/(.064)
=932.8V

Answer 2

is this the right answer

Answer 3

my version is simplified version from the Electric Potential equation