Electric field inside a conductor

  • #1
Count Iblis
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A cylindrical aluminium bar is placed vertically on an insulating surface. The radius of the bar is R and the length is L. The bar has a total charge of zero. Compute the electric field inside and outside the bar.

This is a Homework problem given by Count Iblis to the PF community :cool:
 

Answers and Replies

  • #2
Phlogistonian
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I can't answer that question. I could if it were an aluminum bar, but aluminium is a complete mystery to me. :confused: :cry:
 
  • #3
Domenicaccio
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Where is trick?

I thought that aluminium was a conductor, and as such at equilibrium the electric field inside is always zero. It would be so even if it had a net charge.

Maybe aluminium has some funky property I ignore? :)
 
  • #4
Redbelly98
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Count Iblis, we need more information please.

Why shouldn't the field just be zero everywhere? Is there a charge on the insulating surface, which you haven't told us about?

Is the cylindrical bar solid aluminum, or a tube? If a tube, are the ends open or capped?
 
  • #5
Count Iblis
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The bar is solid. It is resting on an insulating table.

Hint: How does gravity affect the free electron density in the metal as a function of height?
 
  • #6
ZapperZ
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Oh good grief!

Zz.
 
  • #7
Astronuc
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The bar is solid. It is resting on an insulating table.

Hint: How does gravity affect the free electron density in the metal as a function of height?
How does the force of gravity on an electron compare to the Coulomb force?
 
  • #8
Count Iblis
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You can only have equilibrium when the force on a electrons due to the electric field balances the force of gravity. If we assume that there exists a net electric charge rho(z) in the bar where z is the height, then we can write down the net electric force exerted on electrons between z and z + dz. We also know the total weight of the electrons between z and z + dz.

Now, I think Count Iblis has given so many hints that the problem has become trivial :frown:
 
  • #9
ZapperZ
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You can only have equilibrium when the force on a electrons due to the electric field balances the force of gravity. If we assume that there exists a net electric charge rho(z) in the bar where z is the height, then we can write down the net electric force exerted on electrons between z and z + dz. We also know the total weight of the electrons between z and z + dz.

Now, I think Count Iblis has given so many hints that the problem has become trivial :frown:

Except Count Iblis has ignored many important things.

When you strip down a problem to remove all the simplifying assumptions that made this a straightforward problem, you then have to explicitly outline what is included and what isn't, because if not, where do we stop?

For example, when you start considering the effect of GRAVITY on an electron gas, then other effects that are as important, if not more, should also be considered. The "electron gas" in metals are not "free", as anyone who has taken a solid state physics class can tell you. There are also many-body effects that need to be considered. These are no longer 'electrons', but rather quasiparticles in which the mass have been renormalized (i.e. they don't have the same mass as an isolated electron), etc... etc.

All those effects that I've mentioned are actually detectable and actually is a significant aspect of the physical property of a metal, which tends to imply that they are certainly more dominant than the effect of gravity. So if you are starting to include gravitational effects on a "free electron gas", you'd better include those as well.

Zz.
 
  • #10
Count Iblis
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It is not clear how these other effects are more important factors as far as the (average, coarse grained) electric field inside the metal and outside the metal is concerned. They may be important only in that they could modify the effect generated by gravity.

where do we stop?

The question is simply to compute (or estimate) the electric field inside and outside the metal bar. It is assumed that the bar is uncharged and there is no (initial) electric field in the room before the metal bar was brought in and placed in the table.

Are you saying that even in the absense of gravity you would have a pretty strong electric field in the room in due to things like "many body effects" and "renormalized mass" and that therefore we can completely ignore gravity? Surely not!
 
  • #11
ZapperZ
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It is not clear how these other effects are more important factors as far as the (average, coarse grained) electric field inside the metal and outside the metal is concerned. They may be important only in that they could modify the effect generated by gravity.

The question is simply to compute (or estimate) the electric field inside and outside the metal bar. It is assumed that the bar is uncharged and there is no (initial) electric field in the room before the metal bar was brought in and placed in the table.

Are you saying that even in the absense of gravity you would have a pretty strong electric field in the room in due to things like "many body effects" and "renormalized mass" and that therefore we can completely ignore gravity? Surely not!

But that's the "simplification" you have to make. When you first asked the question, you had none. If you had qualified the question as being simply a "statistical" electron gas and ignoring other interactions, then OK, go with that, however unrealistic that is. But if you don't, then that's why I asked where do you stop. The "mass" certainly is different, and CAN be different, in a many-body system. There are metals that exhibit charge carriers almost twice the bare mass, and thus, your computation of the gravitational effect can be quite off, and so will your E-field.

There are other complications. Since this free electron gas is inside a metal, to what extent is the E-field produced by such charge distribution would escape the metal's surface, considering that under electrostatic condition (and the skin depth of a metal), the very same free charges will try to shield themselves. We already know that shielding effect is quite pronounced in many systems, including large atoms. This definitely can affect the resulting E-field!

Zz.
 
  • #12
DavidWhitbeck
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Iblis, in physics you choose your model as the simplest one that incorporates only the necessary features. Gravitational force is so much weaker than electrical force (by many orders of magnitude) that you do not include it in your model. You may be technically correct to say that gravity is present, but you are still foolish to include it. If you include really small effects like that, then you will quickly become overwhelmed. Not only would you have to include everything that Zapper said, but you would also have to question every assumption and assertion in the theory that you studied before then.
 
  • #13
ws0619
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And what is the actual answer please?
 
  • #14
Domenicaccio
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The bar is solid. It is resting on an insulating table.

Hint: How does gravity affect the free electron density in the metal as a function of height?

Show us your results about how much effect really has gravity in polarizing the bar... :uhh:

If this exercise was included in a school test, I would call it a very bad teacher.
 
  • #15
ws0619
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Wow!this is the most difficult physics question that I have seen before.I am so sorry.I don't know how to answer it.
 
  • #16
Count Iblis
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Iblis, in physics you choose your model as the simplest one that incorporates only the necessary features. Gravitational force is so much weaker than electrical force (by many orders of magnitude) that you do not include it in your model. You may be technically correct to say that gravity is present, but you are still foolish to include it. If you include really small effects like that, then you will quickly become overwhelmed. Not only would you have to include everything that Zapper said, but you would also have to question every assumption and assertion in the theory that you studied before then.


David, that's simply not true. If you leave gravity out then the (coarse grained) electric field is exactly zero. The other effects will only modify the results somewhat. Free electron gas + gravity is the simplest model, what Zapper says is important if you want to compute the effect very precisely.
 
  • #17
ZapperZ
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David, that's simply not true. If you leave gravity out then the (coarse grained) electric field is exactly zero. The other effects will only modify the results somewhat. Free electron gas + gravity is the simplest model, what Zapper says is important if you want to compute the effect very precisely.

Actually, it isn't, because if you are constrained by the fact that the whole tube is neutral, then the lack of electrons at the "top" would expose the metal's ions, creating a net positive charge "distribution". Yet, in your "hints", the only thing you care about are the "electron gas" in metal.

Zz.
 
  • #18
Count Iblis
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Actually, it isn't, because if you are constrained by the fact that the whole tube is neutral, then the lack of electrons at the "top" would expose the metal's ions, creating a net positive charge "distribution". Yet, in your "hints", the only thing you care about are the "electron gas" in metal.

Zz.

That should be obvious. You focus on the electron distribution, but the overall charge is zero, of course, due to the positive ions. Anyway, the original problem had no hints, so everyone is free to ignore it. I actually rarely give hints to my students for their practice problems. If they misinterpret things or go in a direction that I didn't intent, then that's no problem. In the real world you have some problem, or you want to compute some effect and then there is no one to ask for hints.

In this case, the problem is simply to compute (or just estimate) the the electric field in and around the metal bar. The fact that you need to consider gravity is not part of the original question, that's something you should figure out while solving the problem. But if people think that I'm wrong and gravity is not important, then that's fine with me. Just compute/estimate the electric field and show your work. :smile:
 
  • #19
Domenicaccio
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I want to know if the resulting electric field is more relevant to a probe charge in the vicinity of the tube than the effect of Van-der-Waals forces.
 
  • #20
Ulysees
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Compute the electric field inside and outside the bar.

So how much does the electric field work out to according to your initial model in the centre? Maybe your integral cannot be done analytically.

Once we have your value for the E-field, maybe then we can calculate additional effects and see how they compare with this.
 
  • #21
kanato
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Are you looking for an answer like F_g = -F_e => mg = eE, so E = -mg/e everywhere inside the bar? But that's wrong. Any internal electric field caused by a displacement of electrons would affect the ionic cores. Gravity would also affect the ionic cores. Both these effects would attract the ionic cores back to the electrons. When the dust settles (equilibrium conditions), on a coarse grained scale it would be as if the material was electrically neutral, with a very small pressure gradient which immeasurably increases its density closer to the bottom, but no electric field.
 
  • #22
Count Iblis
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Are you looking for an answer like F_g = -F_e => mg = eE, so E = -mg/e everywhere inside the bar? But that's wrong. Any internal electric field caused by a displacement of electrons would affect the ionic cores. Gravity would also affect the ionic cores. Both these effects would attract the ionic cores back to the electrons. When the dust settles (equilibrium conditions), on a coarse grained scale it would be as if the material was electrically neutral, with a very small pressure gradient which immeasurably increases its density closer to the bottom, but no electric field.

What happened to the metal? If we swich on a weak electric field parallel to the metal, then the free electrons do respond cancelling out the applied electric field. Now, in that case there is also a force exerted on the ions in the lattice, but the bottom line is that the free electrons respond to make sure that there is no potential energy difference. The ions do not play a significant role in this process. It doesn't matter one iota that the potential difference they are eliminating is electrical in nature or not.

The conclusion that there exists a net electric field inside the metal is, however you look at it, unavoidable. This then leads to an electric field paralllel to the metal's surface outside the metal bar. One can try to measure such effect. There have been contradictory experimental results, http://prola.aps.org/abstract/PRB/v1/i12/p4649_1" [Broken]
 
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  • #23
Ulysees
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Here's a more fun problem for students:

If you spin the rod fast enough, ie create artificial gravity that is strong enough, the electrons will end up at the ends with more concentration there, giving rise to an electric field. The rotation would cause a magnetic field too, that can be calculated.

So a spinning rod is a magnet and a transmitter of radio waves. :tongue2:
 
  • #24
Redbelly98
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Just to put some numbers to the problem:

The field required to balance an electron's weight is a miniscule mg/e = 5.6e-11 V/m

A single static charge of 1 electron (or proton) on the insulating surface will produce a field greater than that for up to 5 meters away.
 
  • #25
Count Iblis
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Just to put some numbers to the problem:

The field required to balance an electron's weight is a miniscule mg/e = 5.6e-11 V/m

A single static charge of 1 electron (or proton) on the insulating surface will produce a field greater than that for up to 5 meters away.


That's right, but what makes this effect detectable is the fact that the electric field has a component parallel to the surface of the conductor.
 
  • #26
Count Iblis
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Here's a more fun problem for students:

If you spin the rod fast enough, ie create artificial gravity that is strong enough, the electrons will end up at the ends with more concentration there, giving rise to an electric field. The rotation would cause a magnetic field too, that can be calculated.

So a spinning rod is a magnet and a transmitter of radio waves. :tongue2:

Indeed! And even without this effect a rotating conductor would emit electromagnetic waves via the so-called dynamical Casimir effect.
 
  • #27
Ulysees
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On the other hand, if a rotating rod emits electromagnetic waves, one has to ask where that energy is removed from. Maybe it slows down a little.

Check your messages Count Iblis. Can't believe they even removed a post pointing to the scepticism and debunking section! Hey, admins! Ever heard of a thing called epistemology?
 
  • #28
Count Iblis
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On the other hand, if a rotating rod emits electromagnetic waves, one has to ask where that energy is removed from. Maybe it slows down a little.

Check your messages Count Iblis. Can't believe they even removed a post pointing to the scepticism and debunking section! Hey, admins! Ever heard of a thing called epistemology?

Hmmm, haven't received any messages...

The rotating bar will lose energy and thus slow down. Compare this with neutron stars that emit radio pulses and slow down as a result. Accelerated charges always emit radiation.
 
  • #29
Ulysees
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That was an email I sent you. It's about the "dynamical Casimir effect" (oopsa, I said a bad word :smile: ). I've opened a thread that this effect seems to be an example of. It's in the scepticism and debunking section.
 
  • #30
Phlogistonian
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Kanato is right.

Gedanken experiment: Suppose, for the sake of argument, that gravity induced a polarization in an object, with one charge on the bottom and the opposite on top. Then the charges on the bottom would feel a feel a force downward and a force upward. No problem there. But the charges on top would feel two forces downward. They would fall. And the end result would be no polarization and no net electric field. This applies to any substance, including aluminium. Therefore, gravity does not induce electric fields.
 
  • #31
Count Iblis
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That was an email I sent you. It's about the "dynamical Casimir effect" (oopsa, I said a bad word :smile: ). I've opened a thread that this effect seems to be an example of. It's in the scepticism and debunking section.

I received your mail, I'll visit that thread later today. :smile:
 
  • #32
Count Iblis
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Kanato is right.

Gedanken experiment: Suppose, for the sake of argument, that gravity induced a polarization in an object, with one charge on the bottom and the opposite on top. Then the charges on the bottom would feel a feel a force downward and a force upward. No problem there. But the charges on top would feel two forces downward. They would fall. And the end result would be no polarization and no net electric field. This applies to any substance, including aluminium. Therefore, gravity does not induce electric fields.


This argument doesn't make any sense whatsoever. The ions are not very mobile. If the electrons sink downward, there is no way the aluminium ions will follow suit.
 
  • #33
Phlogistonian
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This argument doesn't make any sense whatsoever. The ions are not very mobile. If the electrons sink downward, there is no way the aluminium ions will follow suit.

The ions don't fall from top to bottom. Each ion moves a little bit.

And what makes you think the electrons fall without the ions?
 
  • #34
Count Iblis
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The ions don't fall from top to bottom. Each ion moves a little bit.

And what makes you think the electrons fall without the ions?

In a conductor the electrons are far more mobile than the ions. If you switch on a weak electric field, then the electrons will move and neutralize the electric field. The ions are not relevant here. But the electrons do that because a force acts on them, they don't care that it is an electric force. The reason why (in the absense of gravity) there is no component of the electric field along the surface of the conductor is because the free elecrons respond to any applied electric field in this direction. The sum of the field produced by the electrons and the applied field is zero.

So, when we apply an electric field we do get polarization. There is a net electric field generated by the metal which cancels the applied electric field, so the total electric field is exactly zero in the metal and just outside the metal he component of the total electric field along the metal's surface is zero.

If we now replace the applied external electric field by the gravitational field, the results will be similar. The metal will become polarized and an electric field will be generated that so that the electron's potential energy as a function of height becomes constant.

If someone disputes this then they should also dispute the standard textbook explanation of why the electric field inside conductors is exactly zero.
 
  • #35
Phlogistonian
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...

If we now replace the applied external electric field by the gravitational field, the results will be similar. The metal will become polarized and an electric field will be generated that so that the electron's potential energy as a function of height becomes constant.

Gravity affects the ions too. They don't move much relative to each other, but they do move. The whole lattice will move, like an iron cage sinking in water.
 

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