# Gauss' Law in a dielectric material

1. Sep 16, 2014

### deep838

This is what we have in text-books and in Wikipedia:

ρ=ρbf

and from there we get ∇.D=ρf.

But I am unable to understand why we are not considering the bound surface charge in deriving this equation.

Can anyone explain this to me.

2. Sep 17, 2014

### Meir Achuz

It is usually clear from the steps of the derivation. At one point,
$\nabla\cdot{\bf E}=4\pi\rho_f-4\pi\nabla\cdot{\bf P}$
(in Gaussian units). Then D is defined as as ${\bf E}+4\pi{\bf P}$,
and $-\nabla\cdot{\bf P}$as $\rho_b$.

Last edited: Sep 17, 2014
3. Sep 18, 2014

### deep838

This part is alright, what's bothering me is that we are nowhere bringing the surface charge density in this derivation. Why is that? Or is it hiding somewhere!

4. Sep 18, 2014

### DrDu

You shouldn't distinguish between bound and free charges, rather between charges from the medium and external charges (controlled by the observer) although this is kind of a convention and is treated differently from field to field. In quantum mechanics, you can't distinguish between bound and free charges. Anyway, polarization comprises also surface charges which are simply a result of the medium being inhomogeneous so that div P changes at the surface.

5. Sep 18, 2014

### deep838

. I agree to that and have understood this part.
.
This is what I'm talking about. Of course we have polarization charges on the surface and its the normal component of P... So why do we not bring it in the divergence equations?

6. Sep 18, 2014

### DrDu

Of course it is in the divergence equations. That is the whole trick behind introducing P or D: Replace the surface charges by some equivalent polarization. Instead of surface charges which form at the surface of the material you consider a polarization (a dipole density in the simplest cases) which stands in a more or less local relationship with the inducing fields.

7. Sep 18, 2014

### clem

There is bound surface charge, given by $\sigma_b={\bf{\hat n}\cdot\bf P}$, but this affects only E, not D.
Applying Gauss's law across a surface gives the discontinuity in E as $\Delta{\bf E}_n=\sigma_f+\sigma_b$, and the discontinuity in D as $\Delta{\bf D}_n=\sigma_f$.

Last edited: Sep 18, 2014
8. Sep 19, 2014

### deep838

Okay. That was helpful. Thank you everyone for helping me with this. I have a better understanding now.