How does Gauss's Law work with this charge density setup?

Question

My friend and I are self-studying electrodynamics. In Griffiths, Introduction to Electrodynamics (1999), the concept of divergence is introduced mathematically and the following vector field is drawn.

Griffiths states that this vector field has "a large positive divergence".

When trying to imagine what physical scenarios could give rise to this vector field, we imagined a thin shell of very strong negative charge surrounding the center point. We thought this would create an attractive force which gets stronger as you move towards it, just as in the diagram.

But we are troubled, because later Gauss's law is introduced:

$$\mathbf{\nabla} \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$$

Our questions:

• (Less importantly) What does Griffiths mean by "the vector field has a large positive divergence"? Surely the divergence only has a value at a certain point? I'm assuming he means that it is positive at all points, but wanted to call out that assumption specifically.
• (More importantly) How is our "thin shell" scenario compatible with Gauss's law? Gauss's law seems to state that if there is no charge density at a point, the divergence at that point must be zero. But in our case, there is no charge density anywhere but at the edges. How, then, does this field have a large positive divergence?

Answer 1

Since we aren't given what the field actually is, let's make one up. Let's say the field is radially symmetric (only depends on $r$, the distance from the center), and let's say it grows linearly with $r$ (It looks like this is what is happening here. The outer vectors are about three times as far from the center as the inner vectors, and they appear to be about three times as long).

Then $$\vec E=E_0r\hat r$$

And so $$\nabla\cdot\vec E=\frac {1}{r^2}\frac{\partial(r^2E_r)}{\partial r}=3E_0=\frac{\rho}{\epsilon_0}$$

Therefore $$\rho=3\epsilon_0E_0$$

So a constant charge density in space could yield this electric field, and the divergence of the field is constant in all of space.

Notice how we didn't have to imagine up a charge density once we had the field. Gauss's law tells us what the charge density is.

From a comment

would our setup (no charge anywhere except around the edges) yield this field as well, and if so how does that square with Gauss's law?

The field inside a shell of charge is $0$. This is easily seen using the integral form of Gauss's law. The process can be found in many introductory physics text books.

Answer 2

I wasn't able to understand much about the divergence topic but I was able to understand something about Gauss Law.

Here's what I learnt, Gauss Law in electrostatics tell us the the number of field line (electric, of course) passing through a given surface, qualitatively.

The mathematical equation tells us actually the charge responsible for such effects

$$\oint {\overrightarrow{E}}\cdot d{\overrightarrow{S}}$$

Where dS is the small area vector from which the Electric field is passing.

Now remember the Coulomb's Law of Electrical force, from which we derive the Electric field at a distance r due to a point charge as $$E = \frac{q}{4\pi \epsilon_{o}r^{2} }$$

Now in order to imagine the field lines due to positive charge, assuming E and r are in same direction and assume dS to be independent of r (vague??)

$$\int E\cdot dS = \frac{q}{ \epsilon_{o}} \\ \int \frac{q}{4\pi \varepsilon_{o}r^2 }.dS = \frac{q}{ \epsilon_{o}} \\ \int dS = 4 \pi r^{2}$$

Remember this?? Meaning you get a same electric field pattern has a spherical field pattern.

So all the field lines are moving radially outward for a positive point charge.