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?

Just calculate the divergence of the Coulomb field you wrote and you will see that it is zero everywhere except at $r=0$. — G. Smith, Jan 23 '19 at 19:11
2 things. 1) Field and divergence of a field are not the same. 2) The electric field is not defined on the charge itself. You forget that vector = vector. Where is the vector in your RHS? You missed a unit vector $\hat{r}$. Where does it point to? It's undefined — FGSUZ, Jan 23 '19 at 19:12
@FGSUZ The divergence of a vector is a scalar. There should be a vector sign over the $\nabla$ for clarity, but not a unit vector on the right hand side. — G. Smith, Jan 23 '19 at 19:14
@G.Smith I think FGSUZ is talking about the Coulomb's law equation — BioPhysicist, Jan 23 '19 at 19:18
Related : Divergence of Electric Field Due to a Point Charge — Frobenius, Jan 23 '19 at 22:37
The answer to the question in given link may be helpful too https://physics.stackexchange.com/q/510416/227794 — Maan, Nov 17 '19 at 11:00

BioPhysicist · Accepted Answer · 2019-01-23T19:08:06.713

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).

@TaeNyFan By definition a point charge exists at a point, so the volume it occupies is $0$ — BioPhysicist, Jan 23 '19 at 19:08

score 2 · Answer 2 · answered Jan 23 '19 at 19:12

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 ($E=0$).

Isn't the OP asserting that all charge densities arise from continuum limit approximations of distributions of point charges, and therefore any physical charge distributions are sums of delta functions? — Zo the Relativist, Jan 23 '19 at 19:19
In your limit you are holding $\rho$ constant rather than $q$, so you are looking at a vanishing charge essentially. This is not the same thing as a point charge. — BioPhysicist, Jan 23 '19 at 19:20

Zecheng Gan · Answer 3 · 2019-01-23T21:20:13.520

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.

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

3 Answers3