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Charge inside a conductor

  1. Aug 15, 2009 #1
    friends,
    there is no charge inside a conductor or to put in a better way " why electric charge resides only on the outer surface of the conductor"
    thanks in advance
     
  2. jcsd
  3. Aug 15, 2009 #2

    negitron

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    The only mobile charge carriers in a conductor are electrons. Electrons have a negative charge. Like charges repel.

    QED
     
  4. Aug 15, 2009 #3

    Dale

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    If there were a free charge inside the conductor then by Gauss' law there would be an e-field in the conductor. Since you cannot have an e-field inside a conductor then you cannot have a charge inside the conductor, so it has to reside on the outer surface.
     
  5. Aug 15, 2009 #4

    Born2bwire

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    A conductor has, for the purposes of most real world situations, a near infinite amount of free electrons that move around with little interference. Anytime an electric field is applied in a conductor, the electrons will move in response to that field, leaving behind positive ions in the conductor's lattice. The positive ions and the moving electrons will produce their own electric field. They will move about until they create a secondary electric field that perfectly cancels out the applied electric field. Since the electrons can move about freely, they will be pushed to the outer edges of the conductor where they will collect due to the fact that the conductor has ended.

    So, like DaleSpam stated, the charges move freely about to cancel out any electric fields. Since there is no net field inside, there can be no net charge.
     
  6. Aug 18, 2009 #5
    As DaleSpam wrote, it is because of Gauss' law. But Gauss' law is valid because of Coulomb's inverse square law. If the photon were to have a small mass, Coulomb's law would not be exactly valid and then you would have charges inside a conductor, because to cancel all the electric fields inside the conductor when an external electric field is applied, would require some nonzero electric charge density inside the metal. In case of a hollow uncharged sphere, you would get an electric field in the interior cavity.
     
  7. Aug 18, 2009 #6

    Vanadium 50

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    I don't think Count Iblis' explanation is entirely correct.

    Reason 1: the charges want to minimize their potential energy, which they do by moving as far away from each other as possible. This does not require that the force between them be inverse square, merely that it be monotonically decreasing with distance.

    Reason 2: a photon in a superconductor acquires a mass, but the charge still goes to the outside.
     
  8. Aug 18, 2009 #7
    If you start with an uncharged conductor, then you would get a volume charge density in the bulk as a response to an externally applied electric field. In case of a hollow sphere, it is well known that you get an electric field inside hollow space. In fact that latter effect has been used to obtain an upper bound on the photon mass.

    Reason 2 is wrong, because the effective photon mass in that case is ultimately due to interaction with the electrons and the ions. So, you've obtained an effective theory by integrating out some effects and you now want to apply that effective theory to those things that you've already integrated out.
     
  9. Aug 18, 2009 #8

    Dale

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    I don't know about the "if a photon had mass" part, but this part is certainly correct:
    Gauss' law and Coulomb's law are mathematically equivalent. You can derive either one from the other.
     
  10. Aug 18, 2009 #9
    About the massive photon, the limit on the photon mass via the experiment I was referring to is relevant again because of the point made in this recent article:

    http://arxiv.org/abs/hep-ph/0306245

    Anyway, it shouldn't be too difficult to work out a few examples of conductors in electric fields and charged conductors within the framework of Proca theory.
     
  11. Aug 18, 2009 #10
    If there exists a free charge inside a cavity in a conductor, which does not touch the coundoctor itself, there will be a field inside, but otherwise there will not be. (If the charge touches the conductor, it flows into it and goes to the outside.)
     
  12. Aug 18, 2009 #11

    Vanadium 50

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    The paper was interesting - but I don't quite see it's relevance to this discussion.

    If a hollow charged sphere had an electric field inside, a free charge on the inside would migrate to the outside. So the charge still ends up on the surface. Like I said before, it's easiest to think of in terms of energy. The lowest energy configuration has the charges as far as possible away from each other. This relies only on the force on like charges always being repulsive, not on the exact functional form of the force.
     
  13. Aug 19, 2009 #12
    I don't buy this energy argument. I give you a cube and a trillion charges. Are you saying that to minimize the electrostatic energy you would place them all on the surface because then the distance between the charges is smaller? That doesn't seem to be plausible at all.

    Let me give an argument without any complicated math. In a metal, you have a reservoir of free electrons causing any added charge (that could be immobile) to be screened. You can simply treat the free electrons as a plasma and then consider this as Debye screening with a very short Debye length. The inverse Debye length can be considered to be the effective mass of the photon, which is thus quite large. You get a charge on the surface because the screening charges have to come from somewhere.


    First consider very simple model for a massive photon. Suppose that neutrinos are millicharged particles and then the neutrino backround (which is neutral on average) will lead to a Debye screening of charges so we have an effective photon mass.

    If we consider again the charge in the conductor, then the neutrinos do part of the screening, so the free electrons will screen the charge slightly less. Of course, this leads to the same result: The net charge in the bulk is zero and you only have a surface charge.

    Now, since the neutrino background surrounding the charge has a net charge (note that the Debye screening length due to the neutrinos is huge), the metal will have the opposite charge distribution due to the electrons so that the total charge will be zero in the metal.

    But now we have to consider that the charge distribution in the metal due to the free electrons is the way it is, simply because the way the neutrino screened electric field of the charge depends on the distance from the charge, not because of whether or not the charge is really screened by neutrinos or not.

    So, if the photon has a mass which is not due to a millicharged background, then you will inevitably get a net charge in the bulk of the metal.
     
  14. Aug 19, 2009 #13

    Born2bwire

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    There isn't any complicated math needed. It is a simple energy optimization problem, just like any molecular dynamic problem. There is the lattice potentials that bind the conductor's atoms together. Then there are the potentials that hold the electrons to the ions and the Coulombic-like potentials that cause the repulsion between electrons. When you apply a field, the electrons that move leave behind ions in the lattice. The resulting net positive and net negative regions create an opposing field. As long as any net field exists inside a conductor, the electrons will move in response to it, creating an opposing field. The only equilibrium state that will arise due to the free movement of the electrons is one where there no longer exists any field inside the conductor, the only barrier to the movement of the electrons is the boundaries of the conductor. So the electrons will collect on the surface. Due to the extremely high density of electrons in a good conductor, you only need to strip the electrons off of the atoms on the exterior boundary of the conductor to get enough charges migrated to produce an opposing electric field that can negate any reasonable applied electric field. So even the immobile positive ions will only be created on the surface.
     
  15. Aug 19, 2009 #14
    I disagree. As explained above, Coulomb's law is essential. Energy minimization is not the reason why you don't have a charge in the bulk.
     
    Last edited: Aug 19, 2009
  16. Aug 19, 2009 #15

    Born2bwire

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    Of course the Coulombic fields are important, they are the ones that counteract he applied field inside the conductor.
     
  17. Aug 19, 2009 #16

    Vanadium 50

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    Born2bwire, that's exactly right.
     
  18. Aug 19, 2009 #17

    It doesn't follow that the charge density inside the conductor is zero from your argument.
     
  19. Aug 21, 2009 #18
    Anyway, I've found an exact expression for a charged spherical conductor, I'll start a new thread challenging others to derive this result.
     
  20. Sep 2, 2009 #19
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