# E field using cylindrical coordinates

Can someone explain why, when I am going to calculate the $\vec{E}$ or $\vec{B}$ field of a charged ring in its axis (using cylindrical coordinates), the position of source field is $(R,0,0)$ and not $(R,\phi,0)$?

• $(R,0,0)$ is the position $R$ and $(R,\phi,0)$ the position $R^\prime$ in $\phi(R) = \frac{1}{4\pi\epsilon_0}\int \frac{\rho}{|R-R^\prime|}\,\mathrm{d}R^\prime$: this will be an integral over $\phi\in[0,\,2\,\pi)$. – WetSavannaAnimal Jun 15 '15 at 10:26

In order to 1.) simplify the calculation and 2.) make the results useful for more complicted geometries, the charged ring is uniformly charged. That is, the charge density is not a function of the angle. The density is the same at any value of $\phi$ that you choose.