# Confusion about electric field along a uniformly charged disk

Consider a uniformly charged non-conducting disk with surface charge density $\sigma$ and radius $R$

It is a well known result that the electric potential on the edge of the disk is given by

$V_R = \frac{\sigma \,R}{\pi \epsilon_0}$

and the potential electric potential and the center of the disk is

$V_0 = \frac{\sigma\, R}{2 \epsilon_0}$

Since $V_R - V_0 = -\int_0^RE(r)dr$ this implies that the radial component of the electric field along the surface of the disk is nonzero.

However, at the surface of the disk, doesn't it just look like an infinitely large plane of charge? This should mean that the electric field at any point on the surface is $\frac{\sigma}{2 \epsilon_0}$ perpendicular to the surface of the disk, with no radial component.

What am I missing here? Thanks!

• I have added an $R$ to each of your potential equations. – Farcher Aug 10 '17 at 15:56
• Imagine a Gaussian pillbox. By Gauss's law, it's the discontinuity in $E_\perp$ that only depends on the surface charge. This doesn't say anything about additional fields coming from outside the pillbox. – user4552 Aug 10 '17 at 17:29