Gauss' Law in a dielectric material

[deep838]

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

ρ=ρ_b+ρ_f

and from there we get ∇.D=ρ_f.

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.

 

[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 $\rho_b$.

 

[deep838]

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 $\rho_b$.

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!

 

[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.

 

[deep838]

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.

. I agree to that and have understood this part.
.

Anyway, polarization comprises also surface charges which are simply a result of the medium being inhomogeneous so that div P changes at the surface..

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?

 

[DrDu]

.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?

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.

 

[Meir Achuz]

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 $\Delta{\bf D}_n=\sigma_f$.

 

[deep838]

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