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Electron distribution in a metal under the influence of a field.

  1. May 11, 2009 #1
    Under an influence of a field, the electron distribution can be -

    Theory A -

    http://img147.imageshack.us/img147/1690/quantumelectrondistribu.png [Broken]


    Theory B -

    [PLAIN]http://img115.imageshack.us/img115/1690/quantumelectrondistribu.png [Broken]

    Theory B can be state wrong cause cause if this was so, the charge density made would have been extremely high.

    But using the same principle (that at one place here are tons of electrons, while at the other there are none) we can also model a lower charge density...when only a few positive ions will be exposed, i.e at the place where the electrons reside, the density of electrons will not be that high as compared to the natural charge density of the metal.

    For example -
    In this case, using the same principle (theory B), the charge gained is low -

    [PLAIN]http://img115.imageshack.us/img115/1690/quantumelectrondistribu.png [Broken]

    While with this is high -

    http://img152.imageshack.us/img152/1690/quantumelectrondistribu.png [Broken]

    It might also be stated by someone that the cause the minimum number of electrons that can be withdrawn from a positive ion in a metal (considering the temperature is feasible) is equal to the number of its valence electrons, there should be a minimum amount charge that the metal will gain if charged, and higher charges will be a multiple of this minimum charge, which does not happen practically -

    Minimum amount of charge -

    http://img164.imageshack.us/img164/1690/quantumelectrondistribu.png [Broken]

    Other charges will be a multiple of this -

    http://img4.imageshack.us/img4/1690/quantumelectrondistribu.png [Broken]

    i.e we have twice the charge now.

    However, this is not true considering this can also happen -

    http://img246.imageshack.us/img246/1690/quantumelectrondistribu.png [Broken]

    Or the valence electrons have been just displaced by a bit, resulting in partial polarisation.

    So theory B is explaining all possible scenarios, with no glitches........at least that's what it appears.

    The problem that arises from theory A is that -- how can some electrons not move (when the field falls), while the others start moving...the field doesn't know how to discriminate among electrons, its gonna influence all of the electrons, so all the electrons will move together, leaving a few places in the metal completely electron-less...while at other places, even electron density.

    So by the above point, theory B is showing a weak sign............so which one is correct?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 11, 2009 #2


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    Gold Member

    A snip of the relevant part of my post in the other thread.

    Again, this is a simple energy minimization problem. The applied electric fields will move the charges freely around the conductor. They will naturally find the state that will cancel out any fields on the interior because they always move in a manner that counteracts the applied field. They will only reach equilibrium once there is no longer any field inside the conductor.

    The problem with your diagrams is that they are too extreme. I tried to show you that even realistic conductors have more than enough electrons just on the surface to counteract any realistic applied fields. This is why we make the assumption that they are perfect conductors with an infinite number of electrons. The difference this assumption makes is minimal. So we do not have a situation where you will strip every single electron and thrust it to one side of the conductor. In reality, you will strip a few electrons per unit area on the surface and move them from one side to the next. This may cause an imbalance because if we perfectly cancel out just the applied field, we still have a charge distribution that is not what we would like at equilibrium without any applied fields. So the bulk needs to do a shift, but this shift, due to the very few charges moved per volume, will be very slight and it will probably be a very very very small gradient. It wouldn't surprise me if the shift in the charge densities would be overpowered by the changes in time due to thermal variations. And because the extra shift to make everything happy again is probably very very small, the area between the surfaces will more or less remain neutral.
  4. May 11, 2009 #3
    Lets keep the E.F inside the conductor question away for a while, and................lets just see if theory A or B is true.

    Lets assume an unrealistic scenario...lets talk E.Fs of stars, nurseries etc...

    You mean if we use either of theory A or B...it does not matter?

    I think you mean something else.

    Again, lets take an extreme scenario, and in that extreme scenario, how will you justify theory B is wrong?

    The imbalance is justified by the interfering field...so the electrons will stay at a place while other places with remain departed from electrons.

    Of course we are assuming we have applied a field.

    Considering everything here is symmetrical, why will only 'few' of the many electrons move?

    Actually the question here is, how did this scenario even occur?...after the bulk electrons shifting to one side of the conductor, why will they move back considering no change has occured?

    How did the temperature come into the picture? This manifests the fact that I'm thinking something very different from what you are.
  5. May 13, 2009 #4
    I'm trying to sort things out............there seems to be around 35 assertions and 7 major doubts that I've made in my notes.....................and none of them are even close to getting cleared/confirmed.

    Overall it seems that I'm in a state of 'chaos' as one might say after understanding what I'm going through.
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