Surface bound charge density in dielectric displacement

Question

In Griffiths, it derived that the electric potential caused by Polarization $\bf {P}$ is written

$$V({\bf r}) = \frac{1}{4\pi \epsilon_0} \int_{\mathcal{V}} \frac{\hat{\gamma} \cdot {\bf P}{({\bf r'})}}{\gamma ^2}d\tau'$$ here $\bf{\gamma = r - r'}$, when $\bf r$ is field point and $\bf r'$ is source point.

1. Is it jus approximation? I think to get the electric potential more precisely we have to consider not only dipole but also quadrapole etc, but there is no mention about considering quadrapole.

The textbook also introduce the divergence of dielectric displacement is free charge density. However it doesn't think about surface bound charge.

$$\epsilon_0 \nabla \cdot {\bf E} = \rho = \rho_{\textrm{free}} + \rho_{\textrm{bound}}$$

I think we should add surface bound charge $\sigma_b$ in the above equation. The Griffiths wrote why he did not add it but I don't understand the logic of Griffiths.

2. Why should we omit surface bound charge $\sigma_b$?

Answer 1

The expression that you see for the potential is a dipole approximation.

As for the point about the surface charge, Griffiths only derives the expression for a gaussian surface in the meat of the dielectric volume, where $\rho = \rho_{f} +\rho_{b}$ is obviously true (the surface straddles the volume and does not lie within any gaussian surface inside). Then the steps follow logically.

The other case, where you "would have to" consider the surface charge density, is where the gaussian surface encloses it -- but this is of course outside the dielectric, and $\mathbf{P}$ drops quickly to zero, such that $- \nabla \cdot \mathbf{P}$ evaluates to zero. So we are evaluating this surface charge after all -- it is just hidden in the $\mathbf{P}$. It appears when it needs to, outside the volume, but does not "exist" otherwise (as you would expect from a heaviside function).