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!

  • 1
    $\begingroup$ I have added an $R$ to each of your potential equations. $\endgroup$ – Farcher Aug 10 '17 at 15:56
  • $\begingroup$ 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. $\endgroup$ – user4552 Aug 10 '17 at 17:29

I believe that what you say holds when you're near the center of the disk. But as you move radially outwards, you start to lose that "infinite plane" approximation, and no longer have rotational symmetry. So the field will also acquire a component parallel to the disk, pointing in the direction that you're moving in, since you've got more disk (and therefore charge) behind you than in front of you.

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Consider what would happen if the non-conducting disc suddenly became conducting.
The charges on the disc would migrate towards the edge of the disc because each of the charges would feel a repulsion due to the other charges on the disc as for each of the charges there would be more charges on one "side" than on the other.
Thus the charge distribution would no longer be uniform.
There is a "hand-waving" explanation of what happens in this answer.

The electric field would only be perpendicular to the surface at the centre of the disc and so with an electric field not orthogonal to the disc there must be a radial component of the electric field as shown in this answer.

This question has a graph of electric potential against position and an answer to explains the divergence of the electric field at the edge.

So although you may think that being on the surface near the middle of the disc is equivalent to being on an infinite plane that is not correct as shown by the fact that there is a radial electric field acting.

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