Surface density charge, divergence of the electric field and gauss law

It´s known that the divergence of the electric field at a certain point is given by this formula:

$$\nabla \cdot E=\dfrac{\rho (r)}{\epsilon_{0}}$$

Being $\rho (r)$ the volume charge density at that point.

According to this, What if there is zero volume charge density but non-zero surface charge density? Would have the electric field zero divergence at that point?

In this case, How does this fact conciliate with gauss law?

The gauss law is directly related to divergence theorem:

$$∭_{V}(∇⋅E) dV=∬_{S}E\cdot dS$$

If we have a surface charge distribution inside any closed surface, the first term would be zero because the divergence of any electric field created by any surface charge distribution would be zero. Therefore, the surface integral would be zero also, and, according to gauss law:

$$∬_{S}E\cdot dS=\dfrac{Q_{int}}{\epsilon_{0}}$$

There will be no net charge inside the volume closed by the surface, which is obviously not true. I suspect that this inconsistency has to do with the fact that, for proving gauss law, when we conclude that:

$$\nabla \cdot E=\dfrac{q}{\epsilon_{0}}\delta(r)$$

Being $\delta(r)$ tri-dimensional Dirac´s delta with $r=0$ at the point where the charge $q$ is located.

We subsitute $q$ by $\rho (r)dV$, And as long as $dV$ is a third-order infinitesimal, we obtain the divergence of an electric field common expression, but if we make the same substitution in case surface or linear charge distribution, the resulting expression has a Dirac´s delta.

This suggests to me that the expression of the divergence of the electric field I learned is incomplete, and It´s only valid for only volume charge distributions.

• If you like this question you may also enjoy reading this Phys.SE post. – Qmechanic Apr 20 '16 at 11:35

"The divergence of any electric field created by any surface charge distribution would be zero."

No it's not. Consider a charged conducting sphere with uniform surface charge density and a Gaussian sphere of radius greater than the original one. The electric field is diverging through the surface of the Gaussian sphere. So the divergence cannot be zero.

Being ρ(r) the volume charge density at that point.

No. ρ(r) is the volume charge density situated some where and we are trying to find the divergence of electric field through a surface enclosing that charge, no matter what's the size of the enclosing surface, the divergence you get will be the same, provided you have at least a volume to accommodate all the charges. This is what Gauss's law tells us.

What if there is zero volume charge density but non-zero surface charge density?

If you have only a surface charge density or linear charge density, use Gauss's law in integral form. Don't stick with the differential form as the differential form of Gauss's law is applicable to volume charges only. As Gauss's divergence theorem states,

$∫E.dS$ gives you the exact measure of divergence by measuring the flux across the surface.