Do I need to consider charges that reside at the gaussian surface?

Question

According to gauss law $$ \oint_{}^{} \vec{E} \cdot d\vec{r} = \frac{q}{\epsilon_{0}}. $$ Where we include the charges which are inside the closed surface and exclude charges which stay outside the surface. But what about the charges that reside at the surface do we include them or exclude them?

Answer 1

If a point charge $q$ happens to reside precisely on the chosen Gaussian surface $S$ with a solid angle $\Omega$ pointing to the inside (and hence a solid angle $4\pi\!-\!\Omega$ pointing to the outside), then one should include the fraction $\frac{\Omega}{4\pi}q$ on the source side of the integral form of Gauss' law. The solid angle is $\Omega=2\pi$ on a smooth surface, but not necessarily at e.g. edges or corners.

Answer 2

The OP is referring to the Gauss law, which states, according to the OP's post, that $$ \oint_{}^{} \vec{E} \cdot \hat{n}\ d\sigma = \frac{q}{\epsilon_{0}} $$ where $\vec{E}$ denotes the electric field, $\hat{n}$ denotes a unit vector perpendicular to what we call the Gaussian surface, $d\sigma$ is an infinitesimal element of that surface and $q$ is the net charge that is enclosed to that surface. The integral is a surface integral, meaning that we are integrating electric flux (perpendicular component of the electric field times a surface) along the Gaussian surface.

The Gaussian surface is not something physical, so there is no point in wandering whether or not a charge that potentially lies on the Gaussian surface (as opposed to in or out of the aforesaid surface) is inculded in the surface integral.

The reason is simple: if the OP chooses a Gaussian surface $A$ such that one or a collection of charges lies on $A$, another person can choose a Gaussian surface $B$ that surrounds the volume occupied by $A$, hence, being larger than $A$ in a way such that no charge lies on $B$, but all of the charges that used to lie on $A$, now are included in $B$.

In that way, no one has to wonder whether or not the Gauss law is sufficiently clear or not. I hope this helps.

EDIT: Due to the identity $$\oint_{S=\partial V}^{} \vec{E} \cdot \hat{n}\ d\sigma =\int_V d^3\vec{r}\ \vec{\nabla}\cdot\vec{E}$$ where $S=\partial V$ is the boundary of $V$, I would guess that the answer to your question reduces to "is the boundary of $V$ included in the $\int_Vd^3\vec{r}$ integral?" Food for thought...

Answer 3

If the surface runs through a part of space with a continous volume charge density, then the charges "on the surface" doesnt make any difference to the calculation due to continuity. Gauss law is not defined if you place the surface on discrete point charges or a surface charge density. Note that in this case, in the equivalent differential form of Gauss law ($\nabla \cdot E = \frac{\rho}{\epsilon _0}$, where $\rho$ is the volume charge density), it is clear the right side blows up.