Total charge on a polarized neutral dielectric

Question

I'm going through Griffith's EM book right now and from problem 4.14 we can see that the total charge on a polarized neutral dielectric is 0 (since it is neutral) the polarization may move around some charge (bound charges) but the total is 0. This makes sense and you can prove it by simply applying the divergence theorem.

However, when introducing the auxiliary field $\vec{D}$ (in subsection 4.3.1 of Griffiths) we write the total charge density in a dielectric material as $$\rho = {\rho}_b + {\rho}_f$$ Where ${\rho}_b$ and ${\rho}_f$ are the volume bound charge density and free charge density respectively

This seems to imply that the total charge on a polarized neutral dielectric is non-zero i.e. $\rho = {\rho}_b$ (as ${\rho}_f = 0$ since it is neutral). That presents a problem when applying the regular Gauss's law for $\vec{E}$. You would get the incorrect result because the total charge in the Gaussian surface (say enclosing the entire object) is in fact 0 and not $\int {\rho}_b d \tau$

My Question: is there a more rigorous derivation of the $\vec{D}$ field that does not run into the problem I have mentioned above? Of course, this would not actually be a problem if you kept the fact that polarized neutral dielectrics have a total charge of 0 in mind.

Griffiths does mention why he left the surface bound charge density out of the derivation for Gauss's law for the $\vec{D}$ field at the end of subsection 4.3.1 I'm not sure I quite understood it but I interpreted it as "The volume bound charge blows up at the surface so we cannot apply the Gauss's law at the surface but everywhere else it works!"

Answer 1

The total flux is only associated with free charges. It’s the flux density that increases inside the dielectric because dielectric medium offers less resistance to lines of electric field and these can easily pass through dielectric medium than vacuum. This is because of the polarisation of the material. The bound volume charge density is always equal to the bound surface charge density and therefore whenever we integrate the sun becomes zero. Physically the electric flux emanating from any volume terminates on the surface because surface charge density is negative of volume charge density. These volume and surface charge densities show polarisation phenomenon and less resistance to flow of lines of electric field, therefore more lines pass through dielectric which increases D. Consider D = eps*E same as I = GV. As greater conductance increases I with same V so greater eps increases D with same E.

The volume bound charge density would always be accompanied by an equal amount of surface charge density of opposite charge, hence the flux from volume charge density would terminate at the surface charge density. I think you cannot use the bound volume charge density in isolation.

Answer 2

In your question you mention that $\rho_\text{b}=0$ doesn't make sense. It is indeed not true that $\rho_\text{b} = 0$, but the integral of it over the volume of the dielectric is: $$ \int_V\mathrm{d}V\,\rho_\text{b} = 0. $$

Sometimes this integral includes surface bound charges $\sigma_\text{b}$, which accounts for the fact that some of the bound charges actually accumulate on the boundaries of the dielectric, and it is by including these that the integral evaluates to zero as exoected. This can be taken into consideration mathematically by letting the volume and surface bound charge be given respectively by $$\rho_1=\rho'_\text{b}, \quad \rho_2=\delta(\partial V)\sigma'_\text{b},$$ where $\partial V$ is the boundary of the volume $V$, and then defining $$\rho_\text{b}\equiv\rho_1+\rho_2. $$

Then $$ \int_V\mathrm{d}V\,\rho_\text{b} = \int_V\mathrm{d}V\,\rho'_\text{b} + \oint_{\partial V}\mathrm{d}A\,\sigma'_\text{b} = 0. $$

Note that the Dirac delta function here is exactly why Griffiths says "The volume bound charge blows up at the surface so we cannot apply the Gauss's law at the surface but everywhere else it works!"

Also, your question on "deriving" the $\mathbf{D}$ field isn't clear enough, you should mention the assumptions you are starting from, since usually $\mathbf{D} = \epsilon_0\mathbf{E}+\mathbf{P}$ is taken as a definition. This definition helps some calculations, but really has not much physical meaning $^\dagger$, hence the name "auxiliary field".

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$^\dagger$ There's some discussion on whether auxiliary fields such as $\mathbf{H}$ and $\mathbf{D}$ have a consistent physical interpretation (i.e. this PSE post).