Is the static electric field $\vec{E}$ defined and finite where charges are located? Suppose you have a continuous charge distribution, e.g. a wire, a disk, or a plane, represented by a subset $D\subset \mathbb{R^3}$. The charge density is $\rho: \mathbb{R^3}\rightarrow \mathbb{R}$. Call $E$ the electric field it generates.
Now, I thought that the the domain of the field $E$ was $\mathbb{R^3}-D$, because by Coulomb law you'd get a division by zero if you tried to calculate the force exerted by the charge...on the charge. And if you calculate the electric field for some charge distributions like an infinite wire you get exactly that: the $\vec{E}$ field is undefined on the wire. But this openly contradicts Gauss' law in differential form: $$\nabla\cdot \vec{E}=4\pi \rho,$$ which tells us that the divergence of the electric field (which has the same domain as the electric field itself) is null in $\mathbb{R^3}-D$ and equal to the charge density in $D$. So the electric field has to be defined on the charged wire or disk etc. But how can this be so? Am I missing something?
 A: Strictly speaking, Gauss's law in differential form only holds for continuous charge distributions - i.e. the charge's support must be full-dimensional, codimension-0 sets. In this case, the electric field is perfectly finite and well-defined everywhere, because (heuristically) the contribution to the nearby field from a tiny patch of volume dV is
$$dE \sim \frac{\rho(x)\, dV}{r^2} \sim \rho(x) \frac{dx^3}{dx^2} \sim \rho(x)\, dx$$
remains finite (indeed, goes to zero) as $dx \to 0$.
For charge distributions confined to lower-dimensional manifolds, the volume charge density is not well-defined at the charges and Gauss's law technically does not apply. In practice, we replace the charge density function with a charge density distribution containing Dirac delta functions, which allows us to extend Gauss's law to apply to such distributions, but the math is a bit subtle. To figure out where the field remains finite and well-defined, you need to carefully think about when the integral ${\bf E}({\bf x}) = \int d^3x' \frac{\rho(x')}{|{\bf x - x'}|}\hat{\bf r}$ converges.
The field remains finite (though discontinuous) near a codimension-1 charge distribution like a 2D sheet, because we have
$$dE \sim \frac{\sigma(x)\, dA}{r^2} \sim \sigma(x) \frac{dx^2}{dx^2} \sim \sigma(x)$$
is finite. For codimension-2 and higher distributions (lines and point particles), the electric field diverges near the charges. So the answer is that field is defined everywhere except at the location of charge distributions that are codimension-2 or higher (i.e. except where the charges lie along lines or points).
