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Gauss' Law for finding E between metal plates

by collectedsoul
Tags: gauss, metal, plates
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collectedsoul
#1
Sep11-11, 12:36 PM
P: 68
I'm having some difficulty applying Gauss's Law to metal plates. From what I've studied it seems to me that Gauss's Law only really works when there is spherical symmetry, in all other cases it is an approximation. Am I correct in inferring this?

However, in a case where the distance between two charged metal plates is small compared to their length, Gauss's Law can be applied to find electric field between the plates by the equation

E = [itex]q/[/itex][itex]\epsilon[/itex]A

My question is what if the distance between the plates is now increased, according to Gauss' Law the electric field would be the same. But according to Coulomb it should go down, so what really happens in this case?
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Ken G
#2
Sep11-11, 01:51 PM
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P: 3,080
Quote Quote by collectedsoul View Post
I'm having some difficulty applying Gauss's Law to metal plates. From what I've studied it seems to me that Gauss's Law only really works when there is spherical symmetry, in all other cases it is an approximation. Am I correct in inferring this?
It depends on what you mean by "Gauss' Law"-- you seem to mean a local version of what is really a rule about a global integral. The standard approximation you are invoking to do that is the "plane parallel approximation", which treats the plates as infinite in extent. It really just applies far from the edges, for plates whose separation is much less than the width.
My question is what if the distance between the plates is now increased, according to Gauss' Law the electric field would be the same. But according to Coulomb it should go down, so what really happens in this case?
The global version of Gauss' law is just a surface integral, so it's still correct-- what breaks down is the symmetry assumption that lets you assume the surface integral has the same value over most of that surface, while the surface area doesn't change. When you don't make that assumption, you find the surface area is going up (like in the spherical symmetry), so the field strength has to go down.
collectedsoul
#3
Sep12-11, 08:22 AM
P: 68
I'm just curious about the physics of the flux law. I know Gauss' Law requires an infinitely long plane placed very close to the metal plate to work. And there are problems with fringing so it is only effective to consider the field near the centre of the plate.

My first question was a theoretical one: whether the Law works perfectly only in cases of spherical symmetry, and is only useful as an approximation in other symmetries.

As for the separation between the plates increasing slightly, my understanding is that practically if I'm using Gauss' Law the field increase is minimal since the length of the plates is much much greater than the separation. So I can claim that the field remains the same by continuing to use the formula I mentioned, if I'm not doing the surface integral. Is this correct in terms of the physics of it?

Ken G
#4
Sep12-11, 12:55 PM
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P: 3,080
Gauss' Law for finding E between metal plates

Quote Quote by collectedsoul View Post
My first question was a theoretical one: whether the Law works perfectly only in cases of spherical symmetry, and is only useful as an approximation in other symmetries.
The law itself requires no symmetry at all, it's just a law connecting the surface integral of the field to the charge inside. What makes the law useful in practice, however, is to be able to take advantage of a symmetry, and plane-parallel is just as useful as spherical, even though it does require approximation.
As for the separation between the plates increasing slightly, my understanding is that practically if I'm using Gauss' Law the field increase is minimal since the length of the plates is much much greater than the separation. So I can claim that the field remains the same by continuing to use the formula I mentioned, if I'm not doing the surface integral. Is this correct in terms of the physics of it?
Yes, and there's no contradiction with Coulomb's law there-- in effect, as you move away, each charge contributes less, but more charges contribute, because less are fighting against each other by tugging in opposite directions.
collectedsoul
#5
Sep12-11, 02:46 PM
P: 68
Okay that takes care of my doubts. Just one further thing, I was trying to find out how Gauss invented the flux law but couldn't find anything useful. Did he find it independently of Coulomb's Law? Just curious about the formulation of his law, I can't imagine what would make him relate the effect of charges and fields to the area of an enclosing surface.
BruceW
#6
Sep12-11, 03:17 PM
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Coulomb's law came waay before Gauss' law.
Gauss' law (in differential form) is one of the 4 Maxwell equations. And since Maxwell's equations are the most fundamental laws of classical electrodynamics, I would say Gauss' Law is more fundamental than Coulomb's law.

In fact, you can prove Coulomb's law If you assume Gauss' law is true.

EDIT: actually, I think you can also prove Gauss' law by assuming Coulomb's law. I think the two laws are equivalent, so I guess they are just as fundamental as each other. The reason I prefer Gauss' law (in differential form) is because it looks simpler than coulomb's law.
collectedsoul
#7
Sep12-11, 08:23 PM
P: 68
So did he find out the general divergence theorem and then apply it to electric fields? From math to physics, then?
BruceW
#8
Sep13-11, 06:54 AM
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According to wikipedia, yes he already knew about the general divergence theorem before he came up with his law for the electric field.

I also read on wikipedia - Coulomb's law is only true for stationary charges, but Gauss' law is true even for moving charges. So Gauss' law is more fundamental.


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