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I Conductor in an External Electric Field

  1. Mar 1, 2016 #1


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    I just had a test question in my E&M class that asked what would happen to a conductor when placed in an external electric field. One of the "correct" answers was that all of the free electrons would move to the surface of the conductor. I understand that the free electrons would rearrange themselves to cancel out the external field, but would all of the free electrons move to the surface? Wouldn't that build up an electric field inside the conductor?
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  3. Mar 1, 2016 #2
    Wouldn't that be just enough to counteract the external field's "penetration"?
  4. Mar 1, 2016 #3


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    That's what I thought, but if ALL of the free electrons moved to the surface wouldn't that build up a substantial electric field inside the conductor that is much more than required to counteract the external field?
  5. Mar 1, 2016 #4
    I just picture it like a fluid fluctuating over the surface bouncing every which way. Way inside is like opposite to counter the external magnetic field and electrons are nudged more strongly to the surface to blend in where they like to all balance each other and they tend to stay near the lines is more realistic I think.
  6. Mar 1, 2016 #5
    Although the answer does appear to be wrong, an explanation for it does come to mind.
    First the number of free charges required to neutralize the external field go to the surface ( electrons on one side and +ve charges on the other). Then the remaining free charges arrange themselves all over the surface ( not on just one side) such that the net field inside would still be zero ( there will of course be some neutralization of charges)
    I don't know this surely, just seems like a plausible idea.
  7. Mar 2, 2016 #6


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    Why would the net field inside be zero if all of the negative charges are on the outside of the conductor and all of the positive charges are on the inside?
  8. Mar 2, 2016 #7


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    I believe not all the electrons will move to the surface. Once the surface electrons have cancelled the external field, their movement will stop. So, number of electrons just enough to cancel the external electric field will move on the surface. If the conductor is a thin metal plate, -ve charges will be on one of its two surfaces. If the conductor is cylindrical and the external field is along its axis , rings of charge will form on it's surface.
    Last edited: Mar 2, 2016
  9. Mar 2, 2016 #8
    You can see that just some of the free electrons will come to the surface and not all of them from Gauss' law: [itex]\nabla E=\frac{\rho}{\epsilon}[/itex] . We know that electric field is zero in a perfect conductor, so the right hand side of that equation also has to be zero, hence, [itex]\rho = 0[/itex] . Now, if all of the free electrons moved to the surface, [itex]\rho[/itex] certaintly wouldn't be zero. The simplest way you can imagine that is very thin cylindrical metal stick in a homogenus electric field pointing down its axis. Lets say field points in x direction, and stick is also in x direction. Since it's very thin, it can be imagined from left to right as surface-inside-surface. So the outer field moves every free electron to the left, and that one from the right surface comes a little bit left (leaving the right surface negative), the next one travels the same distance (but he came near the stationary positive ion situated inside, and now you have electron and positive ion=neutral in the inside just as we said from Gauss' law) and so on, and when we come to left surface,it is negatively charged.
  10. Mar 2, 2016 #9
    The answer should depend on how many free electrons are, to start with. And how strong is the field. You need less electrons to compensate a very weak field. It is not a general, yes/no answer.

    If it is a very poor conductor, with very low density of free carriers, it may not be possible the compensate the external field, even if they all move to the best positions on the surface.

    It may be a situation with the right electric field and the right electron density where all free electrons are on the surface but this would be very special case.
    But in general there should be just a gradient of electron density, with higher density near the surface.

    Assume a copper plate with 1 mm thickness. If all the free electrons would be crowded (uniformly) on one of the two surfaces, the charge density will be

    σ=e n h

    where e is the electron charge, n is the number density of electrons and h the thickness of the plate.

    n is the order of 1028 m-3 for copper and other good conductors.

    So σ would be of the order of 10-19 x1028 10-3 = 106 Q/m2
    The field produced by this charge density alone will be of the order
    σ/εo or 106/10-11 =1017 V/m.

    The dielectric strength (dielectric breakdown) of most materials is of the order 106-107 V/m.
  11. Mar 2, 2016 #10
    Ugh! I made the stupid assumption that the +ve charges also move ( I'm embarrassed). Sorry
    So what did you conclude? The given "correct" answer must be wrong?
  12. Mar 2, 2016 #11


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    Given my own knowledge and that of the posters in this thread, yes.
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