# 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?

• It's very tempting to argue by analogy from the point charge to the continuous charge distribution. But the argument can't be made. There are probably a few ways to think about it, but I think the simplest is to make the following model: The continuous is modeled by a collection of "closely spaced" point charges. The electric field is the average electric field of a region that is much much smaller than any practical region of interest, but much much larger than the spacing between the charges. There will be no singularity in that averaged field. – garyp Oct 21 '17 at 21:21

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).

• > "only holds for continuous charge distributions" I believe you intended to write "for charge distributions which have finite 3D density". Continuity of density function in 3D is not necessary. – Ján Lalinský Oct 22 '17 at 1:11
• @JánLalinský Yeah, I was just using "continuous" in the colloquial sense of "not discrete," not in the real analysis sense. – tparker Oct 22 '17 at 4:26
• Thank you for answering. Let me see whether I understood you well: if the charge density is a volume or surface density (e.g., the charged object is a ball or a hollow sphere, respectively) then the integral which gives the electric field converges and therefore it is defined everywhere. For "one dimensional" charged object, with linear charge density, Gauss' law (in both differential and integral form?) does not hold, but we can extend it if we resort to distribution and Dirac delta functions. Is that correct? – Nicol Oct 22 '17 at 11:13
• @tparker moreover, would you be so kind as to suggest some references which discuss this and similar topics with the proper mathematical formalism? Even Jackson's textbook, as far as I've read, isn't precise on these issues. – Nicol Oct 22 '17 at 11:15
• @Nicol The integral form of Gauss's law holds whenever the surface integral $\oint {\bf E} \cdot d{\bf S}$ converges. For example, it fails if the Gaussian surface passes exactly through a point charge. The differential form technically only holds for continuum volume charge densities, but in practice can often be extended using delta functions (although again, not if a point charge lies exactly on the surface). – tparker Oct 22 '17 at 15:09