# Electric field from a charged disk

1. Aug 17, 2013

### Bipolarity

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

2. Aug 17, 2013

### Astrum

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.

3. Aug 17, 2013

### voko

Why do you think the integral is non-zero?

4. Aug 17, 2013

### Bipolarity

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

BiP

5. Aug 17, 2013

### WannabeNewton

Astrum answered the question, it isn't valid for $z = 0$.

6. Aug 17, 2013

### voko

When you integrate, you get $1/z - 1/\sqrt {R^2 + z^2}$. This is clearly only valid when $z > 0$. You can then talk about the limit of that at $z = 0$, but the limit of a function does not have to be equal to the value of the function.

In this case, the function is not continuous at $z = 0$. Physically this is because we are dealing with an idealized situation, where the charge is spread over a 2D domain. Obviously you can expect that the idealization works reasonably wellwhen you are far enough from the surface, so that its thickness (or non-thickness) can be neglected, but when you get very close you need a better physical model, so puzzling over why the mathematical device breaks down is rather pointless.