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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.

vector-field-with-large-positive-divergence

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?
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  • $\begingroup$ Does Griffiths claim this has to be a physical vector field resulting from a static charge density? Or is this just to give an example of a vector field with divergence? Not every vector field corresponds to an electric field that is the result of a static charge density. $\endgroup$ – Aaron Stevens Oct 28 '18 at 17:30
  • $\begingroup$ @AaronStevens He makes no claim that this is a physical vector field resulting from static charge density. But suppose we did have the setup described -- wouldn't it create this field, and what would that mean for Gauss's law? $\endgroup$ – Eli Rose Oct 28 '18 at 18:14
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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.

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  • $\begingroup$ Thanks Aaron! But I'm still confused -- 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? $\endgroup$ – Eli Rose Oct 28 '18 at 18:23
  • $\begingroup$ @EliRose see my edit. $\endgroup$ – Aaron Stevens Oct 28 '18 at 18:29
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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.

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  • $\begingroup$ But in the diagram the field becomes stronger farther away from the center. $\endgroup$ – Aaron Stevens Oct 28 '18 at 18:06
  • $\begingroup$ @AaronStevens The field strength is described by the fact the total field lines passing through an area, so if you drag this area (constant during the entire process) from very close to the charge to moving away, it's clear that field lines passing through the area are decreasing. BTW how did you correct out my LATEX? $\endgroup$ – Lakshya Sinha Oct 29 '18 at 9:36
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    $\begingroup$ I took out your indentions, which are supposed to be for when you want to display code. And I see what you are saying about field line density, but these are not field lines, these are showing the actual vectors. $\endgroup$ – Aaron Stevens Oct 29 '18 at 10:40
  • $\begingroup$ Ah! That's what it means, don't worry will soon delete my answer, if you think so. $\endgroup$ – Lakshya Sinha Oct 29 '18 at 16:41

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