Physical Interpretation of $\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0} $

Question

The differential's form of Gauss' Law is $$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}. $$

This suggests that at every point in space, the the electric field $\vec{E}$ is determined by the charge density $\rho$ at that point.

But if charge density $\rho$ is non-zero at a point in space, it means that there is a point charge $q$ present at that point. If we now evaluate $\vec{E}$ on this point, we should get $\vec{E}=\infty $ since now we evaluating the electric field $\vec{E}$ on a point charge $q$. In other words, considering only the contribution from this point charge $q$, by Coulomb's law we get
$$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}=\frac{1}{4\pi\epsilon_0}\frac{q}{0^2}=\infty.$$

Is my interpretation of Gauss' Law wrong? How can this problem be avoided?

Answer 1

This suggests that at every point in space, the the electric field is determined by the charge density at that point.

Technically it says that the divergence of the field at a point in space is determined by the charge density at that point. We know that the actual electric field at a point in space is influenced by charges not at that point in space (for example, Coulomb's law that you give). It seems like you are thinking that the equation reads $\mathbf E=\frac{\rho}{\epsilon}$, which is not what it says.

In your point charge example, at the point charge $\rho=\infty$ (excuse my horrible math), and so it makes sense that you have an infinite divergence at the location of the point charge. (We usually get around this by using Dirac Delta functions, but this has been covered at other places on this site and in many introductory text books, so I won't go into it here).

Answer 2

Almost. The fallacy in your logic is when you say

if the charge density $\rho$ is non-zero at a point in space, it means that there is a point charge $q$ present at that point.

It is not quite true that there is a point charge $q$ present at that point, at least without $q$ being a differential quantity (a quantity so small that it is less than any real number, but still greater than zero). Thus, it would be more acurate to say that the charge is $dq$, where $dq = \rho dV$, where $dV$ is the infinitesimal volume of the point you are considering.

Let $V$ be a sphere of radius $r$. Thus, if it has a uniform charge density $\rho$, the charge contained in it is $q = (4/3)\pi r^3 \rho$. Let's compute the electric field at the surface of the sphere and then take the limit as $r \rightarrow 0$ to turn it into a "point charge" and you'll see where this all works out.

$$\lim_{r\rightarrow 0} E = \lim_{r\rightarrow 0} \frac{1}{4\pi \epsilon_0} \frac{\frac{4}{3}\pi r^3 \rho}{r^2} = \lim_{r \rightarrow 0} \frac{r\rho}{3 \epsilon_0} = 0$$

So, actually, the electric field due to an infinitesimally small charge of charge density $\rho$ is ZERO. No contradiction here. Infinitely small charges make infinitely small electric fields ($E=0$).

Answer 3

The Dirac delta function defines a point charge density, it is infinite at the source charge location and zero elsewhere.

It is indeed slightly confusing, since mathematically, the so-called Dirac delta function is not a function. A few years after Dirac proposed this, mathematicians made it more rigorous by defining it as a "generalized function".

You may want to understand the point charge density as approximately a smooth Gauss distribution around the charge location, then there is no singularity at all. But you can further take the limit for the Gauss distribution variance goes 0, in that limit, the electric field will diverge at the source charge location.