# Gauss's Law and an irregular shape

1. Sep 26, 2004

### Cyrus

Gauss's Law HELP!

I need some help with Gauss's Law. In the book it states that if you have an irregular shape, then you can project each piece of that irregular shape onto a sphere. And since a sphere has field lines everywhere perpendicular to it, it makes life very simple because the flux will be the charge divided by epislon q/e. My question is this: do we draw infinitly many circles and project them for each dA, or is there one single sphere of radius r that we use when we project each little dA onto it.

If the latter is the case, then do we have an implication that the sphere we project it onto will have a radius r that is equal to the closest point on the irregualr bodies shape to the point charge within it? Because if we project onto a sphere of radius r, and there are points that lie lower than r, you could not project onto it could you. The closest point would have to be the maximum radius you could possibly use.

I attached a picture. You can see the irregular surface. Now I made a violet circle of radius r, who has a center that coincides with the point charge. Does gauss's law state that I can represent this charge having a flux through this irregular shape the same as if i calculated it through this one voilet circle?
(assuming I factor in projection scaling.) And if this is so, does that mean that the radius of my circle is restricted to the radius of the closest point from the center to the irregular body? Maybe this is a clearer explination of my question.

Last edited: Dec 26, 2005
2. Sep 26, 2004

### Cyrus

someone help please! im stuck on this and its getting late. :-(

3. Sep 26, 2004

### Tide

I'm not sure what you're asking but the flux through any closed surface exterior to the sphere will have the same total flux through it as the sphere. The flipside of that is that it doesn't matter how small you make the sphere as long as you include all of the spherically symmetric charge contained by it. It's a direct consequence of the inverse square law.

4. Sep 26, 2004

### Cyrus

Yes this is what im not seeing tide, but the flux through any closed surface exterior to the sphere will have the same total flux through it as the sphere.

Here I found the exact pictures from my text book online.

Last edited: Dec 26, 2005
5. Sep 26, 2004

### Cyrus

Now if you click on the picture you will see that they project a piece, dA, of this irregular surface onto a piece of a sphere. Two sides of the sphere will have the same length as two sides of the irregular object, but two sides on the sphere will be shortened by a factor of cos(phi). So the corresponding area of the projection on that piece of sphere will be dA cos(phi). And the book says you do this for all the piece dA and mulitply by the electric field, and you will get the flux. But this flux will be the same as the flux through a sphere. And I am assuming that the sphere they are talking about is the one we used when we did the projection? My question is, how do you then project the next piece of dA onto that same sphere? If you look at the picture, there is a common line that both the sphere and the object have, and then they make an angle with eachother. But what about the next piece. Since the top surface is not flush with the sphere of projection below it, the next piece will not have that common line. So what will be the angle between them, also, will at least two sides still be preserved, or wll all four side now be different in the spheres dA calculation?

Last edited: Sep 26, 2004
6. Sep 27, 2004

### Tide

I think it might be easier to visualize if you think about "projecting" a surface element from the sphere to the outer surface. For example, imagine the outer surface to be rigid and the sphere to be a balloon. Inflate the balloon. Eventually it will make contact with the rigid surface. When contact is made that portion of the balloon is now projected onto the surface. Continue inflating until the rest of the balloon becomes projected.

7. Sep 27, 2004

### Cyrus

So then how could I have one sphere or projection? Shouldn't I have one sphere of projection when contact is made for that one little piece, and then another sphere of projection that corresponds to the next little piece that touches, etc etc. And it would really only be that small piece that is projected onto the sphere in a sense. If you look at the first pic i attached. YOu can see a thing called dA cos (phi). Is that the projection, or is the projection the smal purple square at the bottom? To me it couldent be the small one at the bottom, because then the area calculation would not be right. When they calculated it they said that the area of the projection would be A cos phi, and only two of the sides of the area would be affected. If it were the pruple one, then all four sides would have to be smaller. So I dont think that could be the projection they are talking about could it?

Last edited: Sep 27, 2004
8. Sep 27, 2004

### Tide

Think of the sphere/balloon as being made of a perfectly flexible and strechable type of rubber. As each pont makes contact with the rigid surface it sticks there! Essentially, you are creating a one to one correspondence between the points on the sphere with the points on the outer surface - no cuts, no tears, no holes!

9. Sep 27, 2004

### Cyrus

HMMM I THINK I GOT IT NOW! But I thought of it in different terms, maybe this is what you were saying though.

Let us say we DO use the dark purple sphere in the second picture. First of all, we have some funny irregularly shaped object. Now, it is possible for us to project this irregular patch onto some circle that is tangent at some lines, this would correspond to the first picture. Then they have a common line, and the area of the projection is indeed Acos phi. Now, we have this projection. ALSO, we can make an abitrary circle that lies entirely within the irregularly shaped body (the dark purple sphere). We can make a patch on this sphere that is in line with the patch of the projection. Im trying to describe the first picture. You can see the purple patch, in line with the larger projection patch, in line with the largest patch of the irregular shape. Now, we already know that if we have two concentric spheres, the flux through each is the same. So we can consider the irregullar shaped body as two concentric spheres, one is the purple sphere, and the other is the projection patch. So here we have two pieces of concentric spheres, so both pieces will have the same flux. And we can repeat the process with a new patch, that has a new projection patch, (but this is what was hanging me up,), yet this new projection patch does not necesarily have to be on the same radius as the previous one. It could be higher or lower. But again, all we care about is the flux through the projection patch, because it is equal to that of the corresponding purple patch. And when we do all the projection patches at their various radii, and relate thier respective projections to their corresponding little purple patches, we will map out the entire purple sphere. But the flux through a sphere is independend of the sphere, and so we have proved gauss's law for irregularly shaped bodies. The book was vague in this part of the description, but I think i got it now.

P.S. I still hate those damm funky integrals that you need a table to solve :-)

10. Sep 27, 2004

### Tide

Tables? Who uses them??

11. Sep 27, 2004

### Cyrus

I have a question concerning a problem in my text book. It says that we use gauss's law to calculate the field E at various points for an insulated sphere of uniform charge? How is this possible, i thought an insulator cannot have any charge by definition?

Oh yes by the way Tide, I have a question about this metastable buisness. I thought about it some more. If we have a conducting sphere that is already uniformly charged, and we place a charge inside the conductor, then the electric field inside the conductor due to the surface charge will be zero by definition. So it would not be metastable would it? Even if it is moved off center, then the net electric field INSIDE the conductor is STILL zero irreguardless of where we move. So even if we move off center, we should not be forced to the surface! So it should have the freedom to move any where it wants to within the sphere; however if it does so happen to reach the surface of the sphere, it will then become 'trapped' in a sense and become part of the collection as a whole. Because the moment the charge manages to find its way to the surface, it will exert a force on the charges that do happen to reside at the surface. And when they redistribute themselves, the field will now be such that there is a force to repel all of them away from eachother, and so there is this force preventing it from any migration back into the sphere. I think we were wrong in calling it metastable tide. because by DEFINITION, the net electric field inside a conductor is zero. In addition, we know that a 1/r^2 force produces no net force anywhere inside the sphere.

Last edited: Sep 27, 2004
12. Sep 27, 2004

### Tide

Regarding the insulator, it can certainly have charge. It just doesn't do a very good job of conducting that charge!

Back to metastability: Yes, we've been over the symmetry part concerning the charge on the surface and it produced zero electric field inside the sphere. However, as you displace the test charge from the (exact) center it will take some time for the neutralizing effect of the background electrons to take hold. In that time, there is a finite electric field internally and it will cause the surface charge to rearrange itself. We've already gone over the rest of the process.

This is not a problem of electrostatics because the act of perturbing the system, by definition, causes change - i.e. it introduces time dependence.

13. Sep 27, 2004

### Cyrus

Oh, I see. That is the one part I think I did not catch on to completely.
. Could you elaborate on this please. I do not see what neutralizing effecting you speak of. Even if it is displaced the smallest amount, isnt there still zero force to neutralize. I assume by neutralize you mean there will be a small force produced, but I dont see where this is comming from.

14. Sep 27, 2004

### Tide

In the other discussion someone (I forget who it was) mentioned shielding. Basically, if you place a test charge into a conductor it WILL produce an electric field. However, being a conductor there is an ocean of mobile charges surrounding it and they will respond to that electric field. Specifically, they will rearrange themselves in such a way that they neutralize the electric field produced by the test charge.

For example, if the test charge is positive the background electrons will be attracted toward it leaving behind them a region of locally positive charge. If the conductor is large then electrons further out will tend to fill in. The effect is that very far from the test charge the electric field has been nullified and the only net electric field is very close to the test charge. This is called shielding.

In electrostatics problems you usually just say the electric field is zero everywhere in a conductor. However, it takes a finite time for the shielding to be established and you cannot ignore that when you create a disturbance by placing a test charge in the middle of an otherwise field free conductor.

You must also realize that the problem you're posing is somewhat contrived since charges don't just simply appear. We've skipped over the problem of just how you get the test charge into the center of the conductor in the first place. It has to be moved from point A to point B and all the while the sea of electrons and any excess charge on the surface are responding to the moving intruder.

15. Sep 27, 2004

### Cyrus

First, you say the background electrons? I assume you mean by this ALL the outer valance electrons of the sphere. Maybe we can step through a situation in which we know i metastable, the center charge case, and then go to an off center case.

In the metastable case, I place a test charge in the center. Now according to what you are saying, all the valence electrons will move in towards this center electron. So doesnt that mean we know have a small region near the surface that has gained some partial amount of + charge in addition to the + charge it had to begin with ( since we stated that the sphere already had a positive charge to it).

Now you say " If the conductor is large then electrons further out will tend to fill in." I am under the impression that this is the partial + charge at the surface that I just described, that results from the electrons moving into the center from all directions.

"The effect is that very far from the test charge the electric field has been nullified and the only net electric field is very close to the test charge."

I dont know what you mean by this? Can you look at what I described and see if there are any mistakes. It would be helpful if you described the center chrage case please.

16. Sep 27, 2004

### Tide

When I say "background" electrons I'm talking about the "free" electrons. Remember it's a conductor. As to the rest - will take a look later - gotta run.

17. Sep 27, 2004

### Cyrus

ok thanks tide.

18. Sep 28, 2004

### ehild

It was in the thread "charge in a conductor" and you still haven't answer my question..

You said that it is the extra charge itself that migrates out to the surface, as it goes off center first by thermal motion and then it redistributes the surface charges, and this produces an attracting field for the extra charge in the conductor so as it is driven out to surface. You also said to Cyrus that one extra electron does not go out to the surface, which means that the charge inside a metal in equilibrium can be not only zero, but also 1 elementary charge.

That was my question: Consider a metal sphere with a hole in the center. There is vacuum in the hole. We place a little uniformly charged isolating sphere at the middle of this hole, its radius being considerably smaller than that of the hole and it is fixed to the center so it can not move, and it is isolated from the metal bulk, so no charge transfer is possible between this charged body and the metal. The charge of the metal sphere was previously Q, all this charge at the surface already (or you can say inside the surface layer of Debye length which is a negligible distance with respect to the radius of the sphere). The little sphere in the center carries charge q. How much charge will be at the surface of the big sphere when equilibrium is established? According to your arguments, it should stay Q as the charge of the little sphere can not move outward. Am I right?

ehild

19. Sep 28, 2004

### Tide

ehild,

Thanks for reminding me who it was and sorry for not getting back to you.

The charge on the outer sphere will not change. It will redistribute itself until the electric fields at the surface (inner and outer) are completely normal. And there will be electric fields in the void.

20. Sep 29, 2004

### ehild

I think it must be clear for everybody that the charge of the metal sphere can not change if the additional charge in the center is isolated from it.

But I have asked about the surface charge, not about the charge of the metal sphere itself.

You certainly remember the relation of D and the surface charge density? How much would the magnitude of the electric displacement change just outside the outer surface of a metallic sphere (of radius R) if an additional immobile charge q is placed in a hole in its the center?

ehild