Gauss' Law and gaussian surface

In summary, the equation for Gauss' Law is that the net flux through a gaussian surface multiplied by the Electric Field strength at the gaussian surface is equal to the total enclosed charge by the surface all over epsilon_o. However, I'm having trouble understanding how the law works intuitively and I'm wondering if anyone can help me out.
  • #1
EEWannabe
32
0
Hey there, just had a question about Gauss' law, should be relativity simple however the explanation we were given was quite poor and only seems to apply well to the examples we were given. (This isn't homework).

I (think I ) know the equation for Gauss' Law and what it means, that basically, the net flux through a gaussian surface multiplied by the Electric Field strength at the gaussian surface is equal to the total enclosed charge by the surface all over epsilon_o.

However I'm having problems intuitively with that;

If you had a shape like this;

http://i27.photobucket.com/albums/c171/Chewbacc0r/gauss.jpg

Surely at that point where the Gaussian surface, there's obviously so charge enclosed but is there really no electric field? The only explanation I can think of is by thinking that charge wouldn't be uniformly distributed locally at that point, as it's much closer together, so the charge around the surface would be weaker?

That's possible, but I've got a feeling I've just misinterpreted the law somehow, if anyone could help that'd be great.

Edit; imagine the little structure at the bottom is at the middle of the box.
 
Last edited:
Physics news on Phys.org
  • #2
EEWannabe said:
Hey there, just had a question about Gauss' law, should be relativity simple however the explanation we were given was quite poor and only seems to apply well to the examples we were given. (This isn't homework).

I (think I ) know the equation for Gauss' Law and what it means, that basically, the net flux through a gaussian surface multiplied by the Electric Field strength at the gaussian surface is equal to the total enclosed charge by the surface all over epsilon_o.

However I'm having problems intuitively with that;

If you had a shape like this;

http://i27.photobucket.com/albums/c171/Chewbacc0r/gauss.jpg

Surely at that point where the Gaussian surface, there's obviously so charge enclosed but is there really no electric field? The only explanation I can think of is by thinking that charge wouldn't be uniformly distributed locally at that point, as it's much closer together, so the charge around the surface would be weaker?

That's possible, but I've got a feeling I've just misinterpreted the law somehow, if anyone could help that'd be great.

I'm not sure I understand your sketch. The key point about Gauss' law is that there will be no NET flux out of a closed surface unless there is some net charge enclosed by the surface. You can have lots of flux and an E-field, but the flux in one side of the surface equals the flux going out of the other side of the enclosed surface, unless net charge is enclosed.
 
  • #3
berkeman said:
I'm not sure I understand your sketch. The key point about Gauss' law is that there will be no NET flux out of a closed surface unless there is some net charge enclosed by the surface. You can have lots of flux and an E-field, but the flux in one side of the surface equals the flux going out of the other side of the enclosed surface, unless net charge is enclosed.

Yeah I understand that, i'll re-draw the sketch as best I can, basically it's a 3d box where in the middle there's some sort of structure protruding into the middle of the box, at the red point, sureley if the charge is evenly distributed around the whole shape the E field at the red point would be non-zero, come to think of it, in almost any non-spherical shape this would be the case.

gaus1.jpg
My question is that is Gauss' law true that there is infact no E-field inside? And would this follow for the inside of any hollow 3d shape? ( providing the charge is distributed on the outside and non inside) Or am I interpreting it wrongly? - Maybe is it something to do with how the charge is re-distributed to make the E field inside 0?

I'm really struggling to picture this and any tips on picturing it would be great.
 
  • #4
EEWannabe said:
My question is that is Gauss' law true that there is infact no E-field inside?
Gauss' law doesn't say that the field inside the volume will be zero. It says that the flux through the closed surface will be zero since there's no charge enclosed.
 
  • #5
Doc Al said:
Gauss' law doesn't say that the field inside the volume will be zero. It says that the flux through the closed surface will be zero since there's no charge enclosed.

Hmmm, I think I see what you're getting at,

So all gauss' law tells us is the relationship between the next flux and the enclosed charge? Does that then mean that really unless some clever symmetry is used Gauss' law can't be used to find the electric field stregnth inside any hollow 3d shape?

thanks for the replies
 
  • #6
EEWannabe said:
So all gauss' law tells us is the relationship between the next flux and the enclosed charge?
Right.
Does that then mean that really unless some clever symmetry is used Gauss' law can't be used to find the electric field stregnth inside any hollow 3d shape?
Exactly! While Gauss' law always applies, it's only useful for finding the electric field in special situations that have high symmetry. For an arbitrary shape, it's useless.
 
  • #7
Doc Al said:
Right.
While Gauss' law always applies, it's only useful for finding the electric field in special situations that have high symmetry. For an arbitrary shape, it's useless.

I think I understand now;

My teacher whenever they showed an example he jumped straight from the integral to the EA = Qenclosed / E_o, taking it for granted every time, since he said he "didn't want to get into the integration."

But I think I understand now that it the reason is related to the integral;

[tex]\oint E.dA[/tex]

For the gaussian surface the Electric field is constant, for example in the sphere because the electric field is only dependant on distance from the centre, which is the same for the whole surface, and therefore can be taken out of the integral. To give EA = Qenclosed/Eo.

Whereas for a non-symmetrical example, such as the one above, by no means would it be constant, and you'd need the electrical field strength as a function of area or something else in order to evaluate it!

So is that right? Gauss' law can only be used where the electrical field strength is constant for the entire surface? - Or can at least be split up into numerous examples where that's the case?

Thanks again!
 
  • #8
EEWannabe said:
So is that right? Gauss' law can only be used where the electrical field strength is constant for the entire surface? - Or can at least be split up into numerous examples where that's the case?
You have it exactly right.

Here are the usual examples where Gauss' law can be used to find the field, since over the Gaussian surfaces used the electric field is constant (or zero): http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html#c4"
 
Last edited by a moderator:
  • #9
Thanks so much! You should be our teacher = D
 
  • #10
Doc Al said:
You have it exactly right.

Here are the usual examples where Gauss' law can be used to find the field, since over the Gaussian surfaces used the electric field is constant (or zero): http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html#c4"

With all due respect I disagree. The main point on Gauss Law, related to this possible application, is that, in a certain number of cases, you are able to infer the field in one point from the knowledge of the flux of the electric field through some cleverly choosed surface. This does not imply that the field must be constant on the surface, although these situations are the most frequently found in exercises and the easiest to solve. The use of Gauss Law to provide the field value in one point may come, for example, from a situation where this geometrical point is a kind of point of mean electric field. So having the flux and the area, we may also deduce the value of E at the desired point.

Hoping to have reached some nice degree of clarity,

Best wishes

DaTario
 
Last edited by a moderator:
  • #11
DaTario said:
With all due respect I disagree. The main point on Gauss Law, related to this possible application, is that, in a certain number of cases, you are able to infer the field in one point from the knowledge of the flux of the electric field through some cleverly choosed surface. This does not imply that the field must be constant on the surface, although this situations are the most frequently found in exercises and the easiest to solve.
Sure. As long as you can solve the integral to get the field, you're good. As you say, in most simple situations where Gauss' law is applied, the field is constant over the surface. Of course it doesn't have to be.
But the use of Gauss Law to provide the field value in one point may come, for example, from a situation where this geometrical point is a kind of point of mean electric field. So having the flux and the area, we may also deduce the value of E at the desired point.
Why don't you give an example?
 
  • #12
I will try, but it may take a week to prepare...

Best wishes

DaTario
 
  • #13
we may go first on some artificial setup, like this:

a point like charge Q is positioned in the origin of a catesian system of coordinates. Containing this charge there is a closed surface which passes through the point (1,2,5). We want to determine the field E0 at this point. We also know that the scalar product between the field and the normal unitary vector may be written as follows:

gif.latex?\vec{E}\cdot%20\hat{n}%20=%20\frac{E0\,%20cos(\phi)\,%20\sin(\theta)}{\rho^2}.gif


where E0, is the field value in (1,2,5).

Perhaps this example is good enough.

Best wishes

DaTario
 

Attachments

  • gif.latex?\vec{E}\cdot%20\hat{n}%20=%20\frac{E0\,%20cos(\phi)\,%20\sin(\theta)}{\rho^2}.gif
    gif.latex?\vec{E}\cdot%20\hat{n}%20=%20\frac{E0\,%20cos(\phi)\,%20\sin(\theta)}{\rho^2}.gif
    946 bytes · Views: 489
Last edited:

1. What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the enclosed electric charge.

2. What is a gaussian surface?

A gaussian surface is an imaginary closed surface that is used to apply Gauss' Law. It is chosen to simplify calculations and is often a symmetrical shape such as a sphere or cylinder.

3. How is Gauss' Law useful in electrostatics?

Gauss' Law allows us to calculate the electric field at a point due to a known charge distribution. This is especially useful for symmetrical charge distributions, as the electric field can be found using only the total charge enclosed by the gaussian surface.

4. Can Gauss' Law be applied to non-uniform electric fields?

Yes, as long as the electric field is known at each point on the gaussian surface, Gauss' Law can be used to calculate the total enclosed charge. However, this can be more complex and may require integration.

5. What are some real-world applications of Gauss' Law?

Gauss' Law is used in various fields, including electrical engineering, physics, and astronomy. It is applied in the design of electronic circuits, in the study of planetary magnetic fields, and in the calculation of electric fields in charged particle accelerators.

Similar threads

Replies
8
Views
742
  • Classical Physics
Replies
5
Views
2K
  • Classical Physics
Replies
5
Views
1K
  • Classical Physics
Replies
5
Views
1K
Replies
1
Views
651
  • Introductory Physics Homework Help
Replies
10
Views
597
  • Introductory Physics Homework Help
Replies
11
Views
275
  • Classical Physics
Replies
5
Views
979
  • Classical Physics
Replies
2
Views
4K
  • Electromagnetism
3
Replies
83
Views
3K
Back
Top