Electric field from a charged disk

This is rather strange and has been bugging me, from chapter 22 (Electric fields) of Resnick & Halliday:

So suppose you have a charged disk of radius ##R## (the disk has no thickness and charge is uniformly spread on the surface of the disk). If you are at a distance ##z## away from the center of the disk (and orthogonal to the disk), the net electric field you experience is:

## E = \frac{\displaystyle zσ}{\displaystyle 4ε_{0}}\int^{R}_{0}(z^{2}+r^{2})^{-3/2}(2r)dr ##

Now if ## z = 0 ##, then it is clear that the field will be 0. This makes sense, because if you are at the center of the disk, then symmetry will ensure that the field induced by all charge is cancelled out by the charge diametrically opposite to it on the disk.

But if you actually evaluate the integral, you get
## E = \frac{\displaystyle σ}{\displaystyle 2ε_{0}}(1- \frac{\displaystyle z}{\displaystyle \sqrt{z^{2}+R^{2}}} )##

Now the problem with this is that if you set z=0 here, you end up something nonzero! What is the issue here? Thanks!

BiP

The equation is only valid for ##z > 0##

The electric field (and magnetic field of a surface current) due to a surface charge is discontinuous when you cross the surface.

Why do you think the integral is non-zero?

Why do you think the integral is non-zero?

Plugging in z=0 for the second expression yields a nonzero value.

BiP

WannabeNewton