Electric field within the cavity of a hollow conductor polarized by external charges?

In summary: I think you may have too 'pictorial' view of this and there is not an equivalent to a rock pool, sitting above the level of the sea. The conductor would be continuous between your postulated hollow region where charge could be 'trapped' so there is always a path. I see, so a potential difference would always exist across the interior of the conductor, even in the absence of a field?I see, so a potential difference would always exist across the interior of the conductor, even in the absence of a field?yes, that's correct.
  • #1
tade
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Suppose we have a hollow metallic conductor, just a thin metallic shell forming a large hollow cavity.

It is then polarized by electric charges placed nearby externally.

The equilibrium electric field must be parallel to the surface normals of the shell, there must be no tangential component to the electric field.

However, is it possible for there to be a net electric field within the cavity, and if not, why not?

A second scenario is the hollow conductor not being polarized by external charges, but possessing a net amount of electric charge on itself, and i ask the same question again
 
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  • #2
Assuming that there is no charge inside the hollow of the conductor then there will be no field in the cavity, by Gauss’ law.
 
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  • #3
Dale said:
Assuming that there is no charge inside the hollow of the conductor then there will be no field in the cavity, by Gauss’ law.
what about there possibly being a net electric field with zero net enclosed electric flux? or equivalently, a field with zero divergence
 
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  • #4
tade said:
However, is it possible for there to be a net electric field within the cavity, and if not, why not?
Polarisation of the charges on the sphere will produce a minimum of potential energy (equilibrium situation) where there is no Potential Difference across the sphere (so no internal field). Any difference in potential across the inside of the sphere would result in current flow to cancel it, in the steady state.
If the sphere has finite resistivity and thickness, there can be a time lag for the charge flow to achieve the cancellation and that can detract from the 'screening properties' against RF of a metal box, made of real metal and with resistive joints and seams in it.
 
  • #5
Tade, why do you ask questions if you're just going to argue with us about the answer?

All tha charge on a conductor moves to the outside. There is no field inside the conductor, and there are no free charges. So there can't be any field in the cavity.
 
  • #6
Vanadium 50 said:
Tade, why do you ask questions if you're just going to argue with us about the answer?
I'm not sure what you mean, its not like I'm trying to force anyone to accept my opinions, I'm just trying to reason it out. I still have questions about it.
Vanadium 50 said:
All tha charge on a conductor moves to the outside. There is no field inside the conductor, and there are no free charges. So there can't be any field in the cavity.

So, Dale mentioned Gauss' Law; Gauss' Law concerns the enclosed flux or divergence of an electric field pertaining to the local charge density, it might be unable to provide a full description of an electric field.

which is what I was asking Dale about just now.

if all the charge moves to the surface (its a hollow shell anyway), how do we mathematically prove that the field inside the cavity is zero?
 
  • #7
sophiecentaur said:
Polarisation of the charges on the sphere will produce a minimum of potential energy (equilibrium situation) where there is no Potential Difference across the sphere (so no internal field). Any difference in potential across the inside of the sphere would result in current flow to cancel it, in the steady state.
If the sphere has finite resistivity and thickness, there can be a time lag for the charge flow to achieve the cancellation and that can detract from the 'screening properties' against RF of a metal box, made of real metal and with resistive joints and seams in it.
for clarity, not necessarily a sphere, but any shape

I was wondering the electric field inside a hollow shell conductor might be

Also, I understand why a conductor's surface might be an equipotential surface, with the equilibrium electric field parallel to the surface normals of the shell, with no tangential component to the electric field.

but I was thinking about whether its possible to have equilibrium solutions where there are tangential components in regions where the charge density is zero.

since the charge density is zero, there could be tangential components as they wouldn't upset the equilibrium

additionally, i was thinking about whether certain conductor shapes can make it difficult for there to be tangential electrical forces to cause charge flows, "trapping" the flow of charges and resulting in interesting and unexpected equilibrium solutions
 
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  • #8
tade said:
additionally, i was thinking about whether certain conductor shapes can make it difficult for there to be tangential electrical forces to cause charge flows, "trapping" the flow of charges and resulting in interesting and unexpected equilibrium solutions
I think you may have too 'pictorial' view of this and there is not an equivalent to a rock pool, sitting above the level of the sea. The conductor would be continuous between your postulated hollow region where charge could be 'trapped' so there is always a path. You could try to commit yourself to actually drawing the layout you have in mind. You would find that it can't be done unless you put a charged object inside your hollow shape and, in that case, the charge distribution outside the object would be the same as if it was in free space. And, in any case, it has nothing to do with the original idea.

PS You are having a problem in reconciling the maths with the actuality. Perhaps a bit of 'faith' that it does work would help you with this - rather than being almost convinced that you have found a hole in the system. i.e. "I must be wrong" and not "I could be right". That could be a bit unsatisfactory for you but it may be a way through. :smile:
 
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  • #9
sophiecentaur said:
I think you may have too 'pictorial' view of this and there is not an equivalent to a rock pool, sitting above the level of the sea. The conductor would be continuous between your postulated hollow region where charge could be 'trapped' so there is always a path. You could try to commit yourself to actually drawing the layout you have in mind. You would find that it can't be done unless you put a charged object inside your hollow shape and, in that case, the charge distribution outside the object would be the same as if it was in free space. And, in any case, it has nothing to do with the original idea.

PS You are having a problem in reconciling the maths with the actuality. Perhaps a bit of 'faith' that it does work would help you with this - rather than being almost convinced that you have found a hole in the system. i.e. "I must be wrong" and not "I could be right". That could be a bit unsatisfactory for you but it may be a way through. :smile:
I'm not almost convinced of anything, I'm just wondering about aspects of the problem

in that case, there's a mathematical proof of this right?
 
  • #10
tade said:
I'm not almost convinced of anything, I'm just wondering about aspects of the problem

in that case, there's a mathematical proof of this right?
Bound to be. The problem will be in translating what you suggest into a valid mathematical model.
 
  • #11
sophiecentaur said:
Bound to be. The problem will be in translating what you suggest into a valid mathematical model.
i think the statement would be, for whatever shape, prove that the electric field within is always zero and that the surface is always at equipotential, for cases at equilibrium
 
  • #12
How could it not be in equilibrium? If not, then charges would flow.
 
  • #14
sophiecentaur said:
How could it not be in equilibrium? If not, then charges would flow.
yeah, I did mention "for cases at equilibrium"
 
  • #16
tade said:
right, and I asked him a question about Gauss' Law regarding the field within a conductor
which I'm still wondering about
so have you read up on Gauss's Law yet ? and searched through a few other pages yourself ?

we are here to help people, not spoonfeed them, it's nice to see people make an effort :smile:
 
  • #17
davenn said:
so have you read up on Gauss's Law yet ? and searched through a few other pages yourself ?

we are here to help people, not spoonfeed them, it's nice to see people make an effort :smile:
yes, I understand what Gauss' Law entails, this is my question to Dale about the internal electric field:
tade said:
what about there possibly being a net electric field with zero net enclosed electric flux? or equivalently, a field with zero divergence
 
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  • #18
tade said:
what about there possibly being a net electric field with zero net enclosed electric flux? or equivalently, a field with zero divergence

An electric field with zero divergence must be zero itself, by Gauss's Law.
 
  • #19
PeterDonis said:
An electric field with zero divergence must be zero itself, by Gauss's Law.
but I'm thinking, a lone point charge has an electric field some distance away from itself, and that electric field at a distance has zero divergence
 
  • #20
tade said:
i'm thinking, a lone point charge has an electric field some distance away from itself, and that electric field at a distance has zero divergence

You are thinking about it wrong. If you evaluate Gauss's Law in the vacuum region surrounding a point charge over a sphere that does not enclose the charge, then it is zero. In that sense, yes, the vacuum field has "zero divergence".

But if you evaluate Gauss's Law over a sphere that does enclose the charge, you cannot get a zero result. So if you know that Gauss's Law evaluates to zero over any sphere whatever, including a sphere just on the inner surface of the conductor in your OP, you know there cannot be any charge anywhere inside that sphere.
 
  • #21
PeterDonis said:
You are thinking about it wrong. If you evaluate Gauss's Law in the vacuum region surrounding a point charge over a sphere that does not enclose the charge, then it is zero. In that sense, yes, the vacuum field has "zero divergence".

But if you evaluate Gauss's Law over a sphere that does enclose the charge, you cannot get a zero result.

PeterDonis said:
So if you know that Gauss's Law evaluates to zero over any sphere whatever, including a sphere just on the inner surface of the conductor in your OP, you know there cannot be any charge anywhere inside that sphere.
I'm not sure if Gauss' Law does evaluate at zero over any sphere within the hollow of the conductor.

anyway, assuming that there's no charge within the hollow, i would like to know whether there could be an electric field within the hollow
 
  • #22
tade said:
I'm not sure if Gauss' Law does evaluate at zero over any sphere within the hollow of the conductor.

Why not?

tade said:
assuming that there's no charge within the hollow, i would like to know whether there could be an electric field within the hollow

No. Do you understand that Gauss's Law evaluates the electric field over a sphere?
 
  • #23
PeterDonis said:
Why not?
cos that might need to be mathematically proven

PeterDonis said:
No. Do you understand that Gauss's Law evaluates the electric field over a sphere?
I understand that Gauss' Law evaluates the net closed-surface electric flux, so I think it might not be able to provide a full description of an electric field
 
  • #24
tade said:
cos that might need to be mathematically proven

It already is. See below.

tade said:
I understand that Gauss' Law evaluates the net closed-surface electric flux

Yes. And we have already seen that, if there is no charge inside a surface, the net flux through that surface must be zero. So, since we have already assumed that there is no charge inside the hollow of the conductor, Gauss's Law evaluated over any surface inside the hollow must be zero.

tade said:
I think it might not be able to provide a full description of an electric field

The flux through a surface only tells you about the field normal to the surface; it tells you nothing about the field tangent to the surface. Yes, that is true.

But now consider: suppose we pick a surface just at the inner surface of the conductor. We know the flux is zero through that surface, therefore the field normal to the surface is zero. But we also know that the field tangent to that surface is zero, since that must be true at any surface of a conductor. So we know the entire field is zero on that surface, and therefore must be zero inside it as well.
 
  • #25
PeterDonis said:
An electric field with zero divergence must be zero itself, by Gauss's Law.
The Coulomb field
$$\vec{E}=\frac{q}{4 \pi r^3} \vec{r}$$
has zero divergence (and also zero curl btw) for all ##\vec{r} \neq 0## but is nowhere 0 ;-)).
 
  • #26
vanhees71 said:
The Coulomb field

Yes, I covered this case in post #20.
 
  • #27
PeterDonis said:
But now consider: suppose we pick a surface just at the inner surface of the conductor. We know the flux is zero through that surface, therefore the field normal to the surface is zero. But we also know that the field tangent to that surface is zero, since that must be true at any surface of a conductor. So we know the entire field is zero on that surface, and therefore must be zero inside it as well.

from my post #7, "but I was thinking about whether its possible to have equilibrium solutions where there are tangential components in regions where the charge density is zero.

since the charge density is zero, there could be tangential components as they wouldn't upset the equilibrium"

also, for the field normals, is it possible that they are non-zero? but some stick in and others stick out such that the net amount is zero
 
  • #28
If you have a conductor in an electrostatic field, there must not be tangential components of the electric field on its surface (because otherwise there'd be a current along the surface), which implies that the electrostatic potential is constant along this surface. If there's no charge distribution inside the conducting surface a solution of the electrostatic equations inside the conductor is ##\vec{E}=0##, i.e., ##\Phi=\text{const}##, which also fulfills the boundary condition that the surface must be an equipotential surface. Then theorems about potential theory (i.e., the Laplace equation, ##\Delta \Phi=0##) tells you that this is the unique solution of the boundary-value problem of the Laplace equation.

Charges outside the conductor of course shift the electrons along the surface such that it becomes an equipotential surface in the static equilibrium state and the electric-field components normal to the surface are discontinuous there with the jump equal (or in SI units proportional) to the induced surface-charge density.
 
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  • #29
vanhees71 said:
If there's no charge distribution inside the conducting surface a solution of the electrostatic equations inside the conductor is ##\vec{E}=0##, i.e., ##\Phi=\text{const}##

what's this theorem called?
 
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  • #30
I'm not sure, whether it has a specific name.
 
  • #31
vanhees71 said:
I'm not sure, whether it has a specific name.
what terms should i search for to find more info about it?
 
  • #32
See, e.g., A. Sommerfeld, Lectures on theoretical physics, vol. 3, Sect. 5.8.
 
  • #33
vanhees71 said:
See, e.g., A. Sommerfeld, Lectures on theoretical physics, vol. 3, Sect. 5.8.
thanks, I'm reading it, and however, it seems to be about the conservation of energy, which is not involved in my issue
 
  • #34
As you'll see, it is involved! It's a very elegant proof, and it's important physics too!
 
  • #35
vanhees71 said:
As you'll see, it is involved! It's a very elegant proof, and it's important physics too!
damn, I'm too stupid to understand it

could you help adapt it to my problem? :oldbiggrin:
 

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