# Question Concerning Gauss' Law

1. Aug 23, 2009

### jslam

I'm reading the book "Electricity and Magnetism: Vol. II" by Edward M. Purcell, in which he describes Gauss' law as stating that the flux through a closed surface is 4π times the enclosed charge (he uses the Centimeter-Gram-Second system).
Later, he refers to Gauss' law as stating that the change in field from one side of a layer to the other must be 4πσ, where σ is the charge density in the layer.
I don't see the relation. Maybe it's an obvious corollary of Gauss' law, but in any case I don't understand how.

The offending text can be found here:
http://rapidshare.com/files/270469221/offending_passage.pdf
The bad news actually only starts in section 1.14 (on the second page), but the first page is included because it get reffered to in the one of the two offending passage.

Last edited by a moderator: Apr 24, 2017
2. Aug 23, 2009

### Born2bwire

What is this "layer" that you are talking about?

3. Aug 23, 2009

### jslam

Sorry. It's a layer of charge. The cases referred to here are an infinite flat sheet of charge with uniform charge density, a spherical shell of charge with uniform charge density and then finally any old layer of charge, saying in each case that from one side of the layer of charge to the other, there has to be a difference in field of 4πσ. I don't know if you read the attached PDF (or if you weren't able to for some reason, in which case let me know), but it explains it all.

4. Aug 23, 2009

### Born2bwire

Oh well that's a correct description then, except it only applies to the component of the fields normal to the surface. The net flux over any closed surface is proportional to the amount of charge in the enclosed volume, via Gauss' law. So if I enclose a sheet of charge of charge density \sigma, then the net flux will be proportional to the product of the enclosed area of the charge sheet times \sigma. So, just take the surface integral to the limit case where the thickness of the volume (thickness is normal to the charge sheet) goes to zero. Now, the enclosed "volume" is the charge sheet, the flux is now the area times the difference between the normal electric fields on the opposite sides of the sheets, this is still proportional to the area times \sigma. So the areas cancel out and you are left with the relationship that the difference between components of an electric field normal to a surface on either side of the surface are proportional to the charge density of any bound charges on that surface.

Another note though is that the general case deals with the electric flux density, sometimes called the displacement field, and not the electric field. This case deals with bound charges that collect on the surface between different materials. The case where we are talking strictly about the electric field only applies to a homogeneous medium.

This is one of what are known as the boundary conditions and are discussed in most electromagnetic textbooks.

http://www.amanogawa.com/archive/docs/EM5.pdf