The question asks "A thin plastic semi ring of radius R has a uniform linear positive charge density $\lambda$. a) Determine the electric potential V at point O, the center of the semi ring. b) Find the magnitude of the electric field E at point O."

I solved (a) using the equation $v=\frac{kQ}{r}$ and got $V=k\pi\lambda$

In the previous problem the way to get from V to E was to take the derivative. $ E = -\frac{dV}{dr}$ However in this problem there is no distance in the electric potential. All the "variables" are constants. So if I go to take the derivative, it equals zero and I know that is not the case.

I can go back to how we solved it in the previous chapter and use $E=-\int \frac{k \cdot dq}{R^2}$ and I can get the right answer, but I feel like I should be able to use the electric potential to calculate the electric field.

The question asks

"A thin plastic semi ring of radius $R$ has a uniform linear positive charge density $\lambda$.

a) Determine the electric potential $V$ at point O, the center of the semi ring.

b) Find the magnitude of the electric field $E$ at point O."

I solved (a) using the equation $v=\frac{kQ}{r}$ and got $V=k\pi\lambda$

In the previous problem the way to get from V to E was to take the derivative. $ E = -\frac{dV}{dr}$ However in this problem there is no distance in the electric potential. All the "variables" are constants. So if I go to take the derivative, it equals zero and I know that is not the case.

I can go back to how we solved it in the previous chapter and use $E=-\int \frac{k \cdot dq}{R^2}$ and I can get the right answer, but I feel like I should be able to use the electric potential to calculate the electric field.