In many textbooks, including Griffiths', they erroneously claim that a field is irrotational if and only if it is conservative (there exists a scalar potential).
This is true only if the domain of the field is simply connected, which is obviously not the case in electrostatics (we have $\mathbb{R}^3$ excluding the points where the charges are located).
Also, it seems like everyone assumes that $\vec{E}$ and $\vec{B}$ (electric and magnetic field) are continuously differentiable (which is not the case, just think about a conductor), so that they can use Stokes theorem, the divergence theorem and so on.
What am I missing?
I also have a bit of confusion about the definition of electrostatic field. By electric field I mean two things:
- The electric field generated by a finite number of charges $Q_0, \dots, Q_N$ located at $\vec{r_0}, \dots, \vec{r_N}$: $$ \vec{E}(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \sum_{i=1}^{N}\frac{Q_i}{||\vec{r}-\vec{r_i}||^3} (\vec{r}-\vec{r_i}) $$
- The electric field generated by a continuous distribution of charge $\rho$: $$ \vec{E}(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \iiint_K \frac{\rho(\vec{r'})}{||\vec{r}-\vec{r'}||^3} (\vec{r}-\vec{r'}) \ d^3\vec{r'} $$
Griffiths proves that the electrostatic field is conservative by manually calculating the line integral of a single-charge electrostatic field over an arbitrary path (the same would be for a finite number of charges); the result depends only on the end points, so $\vec{E}$ is conservative. Does this prove the fact that the electrostatic potential produced by a continuous charge distribution is conservative? I don't think so, and I cannot find a proper proof of this.
