Gauss' law on differential form.

  • #1
The differential form of Gauss' law states that
[tex]\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}[/tex].

So the divergence of the electric field is the chargedensity divided by epsilon zero.

I just wondered.. since divergence is a local or "point" property. Is the chargedensity in this law also the charge density at that point? That is to say for some x,y,z is

[tex]\nabla \cdot \vec{E}(x,y,z) = \frac{\rho (x,y,z) }{\epsilon_0}[/tex]?
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
7
I just wondered.. since divergence is a local or "point" property. Is the chargedensity in this law also the charge density at that point? That is to say for some x,y,z is

[tex]\nabla \cdot \vec{E}(x,y,z) = \frac{\rho (x,y,z) }{\epsilon_0}[/tex]?

Yes. Both divergence and charge density are local properties.
 
  • #3
Is this result then only valid within a material with a certain chargedensity?

In the space between two pointcharges, for example; we'll have divergence in the field, but the chargedensity at any point between the pointcharges is zero, isnt it?
 
  • #4
diazona
Homework Helper
2,175
7
In the space between two pointcharges, for example; we'll have divergence in the field,
No you won't. Why would you think so?
 
  • #5
You won't?

Maybe I have the wrong intuition.. My intution so far as to what divergence at a point actually mesures is that it's a mesure of "how much _field_ is going in relative to how much is going out", or alternativly, "the contraction or expansion" of the field at that point.

For the electric field i then immagine that since the strength falls off as 1/r^2, you'll have a weaker field going out than coming in in the increasing direction of r.

And is not the following calculation correct? (one dimentional case)

[tex]\nabla \cdot \vec{E} = \frac{\partial}{\partial x} \frac{1}{x^2} = -\frac{2}{x^3}[/tex]
 
  • #6
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
You won't?

Maybe I have the wrong intuition.. My intution so far as to what divergence at a point actually mesures is that it's a mesure of "how much _field_ is going in relative to how much is going out", or alternativly, "the contraction or expansion" of the field at that point.

For the electric field i then immagine that since the strength falls off as 1/r^2, you'll have a weaker field going out than coming in in the increasing direction of r.

And is not the following calculation correct? (one dimentional case)

[tex]\nabla \cdot \vec{E} = \frac{\partial}{\partial x} \frac{1}{x^2} = -\frac{2}{x^3}[/tex]

That calculation is incorrect because the electric field in 1D does not satisfy an inverse-square law. In 3D we have

[tex]\nabla \cdot \frac{\vec{r}-\vec{r}_i}{|\vec{r}-\vec{r}_i|^3} = 4\pi \delta^{(3)}(\vec{r}-\vec{r}_i)[/tex]

where [tex]\delta^{(3)}(\vec{r}-\vec{r}_i)[/tex] is the Dirac delta-function. This shows that the divergence of the electric field due to a series of point particles vanishes everywhere except the points where the charges are located.
 
  • #7
diazona
Homework Helper
2,175
7
You won't?

Maybe I have the wrong intuition.. My intution so far as to what divergence at a point actually mesures is that it's a mesure of "how much _field_ is going in relative to how much is going out", or alternativly, "the contraction or expansion" of the field at that point.
You might have the right idea, I'm not sure. It would be more accurate to say that divergence is a measure of the excess flux leaving a volume of space, or rather the limit of that quantity as the volume becomes infinitesimally small. Flux could be thought of as the number of field lines. (The field is the density of flux per unit area.)
For the electric field i then immagine that since the strength falls off as 1/r^2, you'll have a weaker field going out than coming in in the increasing direction of r.
That is true, but you also have a larger area over which the field is leaving, compared to the area over which the field is entering. Area increases in proportion to r2, which cancels out with the 1/r2 decrease in field strength to produce a constant. You can think of the divergence (at least in the radial direction) as the change in that constant, which is of course 0. (Unless there is a charge at that point)
And is not the following calculation correct? (one dimentional case)

[tex]\nabla \cdot \vec{E} = \frac{\partial}{\partial x} \frac{1}{x^2} = -\frac{2}{x^3}[/tex]
In 1 dimension, the field doesn't fall off as 1/r2; it's actually constant in space.
 
  • #8
Hmm! That is really interesting. Thanks for enlightenlig replies! :) But..

"That is true, but you also have a larger area over which the field is leaving, compared to the area over which the field is entering. Area increases in proportion to r2, which cancels out with the 1/r2 decrease in field strength to produce a constant. You can think of the divergence (at least in the radial direction) as the change in that constant, which is of course 0. (Unless there is a charge at that point)"

I follow you here if the surface we are considering is centered at the point of origin of the charge, but what if you take this surface and put it somewhere else in the field. From the definition and what you're saying we have

[tex]\nabla \cdot \vec{E} = \lim_{V \to 0} \oint_S \frac{\vec{E}\cdot \vec{dA}}{V}[/tex]

If we now consider the halfsphere facing the charge I immagine that we'll have a stronger field, or alternativly, a denser collection of fieldlines, entering this halvsphere, then what is leaving the halfsphere facing away from the charge.

Btw: How do I quote you properly? :)
 
  • #9
Ah... never mind my previous post. Ofcourse if you have a certain numbers of fieldlines coming into a volume they must also be coming out, so the netto flux trough the surface is zero. I think I've got it now.
 
  • #10
diazona
Homework Helper
2,175
7
To quote a post you can click the "QUOTE" button at the bottom right. It comes out like this:
Code:
[quote="center o bass, post: 2901535"]Hmm! That is really interesting. Thanks for enlightenlig replies! :) But..

...

Btw: How do I quote you properly? :)[/QUOTE ]
and then you can add additional
and [/QUOTE ] tags as needed.

[tex]\nabla \cdot \vec{E} = \lim_{V \to 0} \oint_S \frac{\vec{E}\cdot \vec{dA}}{V}[/tex]
I think that's correct but it took some thinking to convince myself. It's more usually written as Stokes' theorem (or the "divergence theorem") in integral form,
[tex]\iiint_V \vec{\nabla}\cdot\vec{E}\,\mathrm{d}^3V = \iint_{\partial V}\vec{E}\cdot\mathrm{d}^2\vec{A}[/tex]
 
  • #11
I think that's correct but it took some thinking to convince myself. It's more usually written as Stokes' theorem (or the "divergence theorem") in integral form,
[tex]\iiint_V \vec{\nabla}\cdot\vec{E}\,\mathrm{d}^3V = \iint_{\partial V}\vec{E}\cdot\mathrm{d}^2\vec{A}[/tex]

Thanks diazona! :) Maybe the definition follows naturaly from the divergence theorem, but that is the definition that i learned. It is also stated at wikipedia:
http://en.wikipedia.org/wiki/Divergence

Btw it was interessting what you said about electric fields in 1D. I allways assumed that the properties would persist from 3D to 1D.
But isn't the fact that they don't a consequence of the view we have of fieldlines being the agents of force between charges?

That is [tex]F \alpha [/tex] #fieldlines acting on a body, so in 3d as the number of fieldlines falls of with r^2 because of the spherical symmetri, we will only have one simple fieldline in 1D which causes the force there to be constant. Similarly I'm then tempted to conclude that the force in 2D falls of with r because of the circular symmetry..

Have I got the idea? :)
 
  • #12
diazona
Homework Helper
2,175
7
Btw it was interessting what you said about electric fields in 1D. I allways assumed that the properties would persist from 3D to 1D.
But isn't the fact that they don't a consequence of the view we have of fieldlines being the agents of force between charges?

That is [tex]F \alpha [/tex] #fieldlines acting on a body, so in 3d as the number of fieldlines falls of with r^2 because of the spherical symmetri, we will only have one simple fieldline in 1D which causes the force there to be constant. Similarly I'm then tempted to conclude that the force in 2D falls of with r because of the circular symmetry..

Have I got the idea? :)
Yeah, that's pretty much the idea. Or instead of thinking in terms of field lines, you could also come to that conclusion from Gauss's law. If you solve
[tex]\vec{\nabla}\cdot\vec{E}(x) = 0[/tex]
in 1D, you find that [itex]E(x)[/itex] has to be a constant.
 

Related Threads on Gauss' law on differential form.

  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
8
Views
4K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
977
  • Last Post
Replies
6
Views
5K
Replies
1
Views
2K
  • Last Post
Replies
20
Views
5K
Replies
6
Views
3K
Top