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Experimental proof of electrostatic equilibrium

  1. Mar 2, 2008 #1

    In my physics we derived certain properties of conductors in electrostatic equilibrium using Gauss' law. One property seemed to "mathematical" to me. It says that the whole charge of a conductor in electrostatic equilibrium is located at the surface.
    The proof is at follows. Since the electrical field inside the conductor is zero, then one can put a Gaussian surface inside the conductor. The net flux through the surface is given by
    [tex]\Phi = \oint \bar{E}d\bar{A} = 0[/tex]
    Then, according to Gauss' law,
    [tex]\Phi = \frac{Q_{in}}{\epsilon_0} = 0[/tex]
    [tex]Q_{in} = 0[/tex]
    One can make a Gaussian surface that gets bigger and bigger but stays inside the conductor. The charge inside the surface is then 0. From that follows that the charge is located at the surface.
    What bothers me in this proof is that, what stops you from making a Gaussian surface that is exactly a copy of the conductor? That means that the total charge in the conductor is 0 and that is a contradiction. Why do you have to stop "growing" your Gaussian surface just before you reach the surface of the conductor? Furthermore, if that's true, then that means that the whole charge is a located in the infinitesimal thin outer layer of the conductor.
    How is the charge really distributed in a conductor in electrostatic equilibrium? I ask that question because the physical world doesn't know infinitesimal thin layers. Or does it?

  2. jcsd
  3. Mar 2, 2008 #2
    The point is that the E field must be zero everywhere inside the conductor, or else the charges will move until it is. That's what lets you set the surface integral to zero. If you pick the actual surface of the conductor as your Gaussian surface, then you won't have a zero-valued integral any more, since there is a net field; it's just normal to the surface at all points (or else the charges move until it is).

    I don't know the answer to your second question about the microscopic charge distribution. There is such a thing as "skin depth" for currents, but I'm not sure if something like that applies to static charge distributions. Also, if you pump enough charges onto a conductor, eventually the charge density reaches the point where charges start to leak off the surface, hence the operation of lightning rods.
  4. Mar 2, 2008 #3
    Thanks for your answer. Of course, I forgot that the electric field at the field is normal to the surface and that it isn't zero. Thank you for your argument.
    Has anyone got any ideas about the charge distribution? How is it really distributed? And is there actual experimental proof about the charge being distributed at the surface?
  5. Mar 2, 2008 #4
    I'm quite sure people know to a fair accuracy what the charge density would be for different conductive materials. I'd guess you have to use QM to do it right, which is a calculation I've never seen.
  6. Mar 3, 2008 #5
    Ok, that goes a bit too far :-). In fact, I'm just a computer science student following some physics courses. Don't ask me anything about QM :-).
    Anyway, thanks for your answer.
  7. Mar 3, 2008 #6


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    It has nothing to do with QM or conductivity material, once equilibrilum is reached.
    Newton performed the first experiment in what is called the "Newton Ice Pail" experiment.
    It is demonstrated daily in devices, like the Van de Graaff generator that use the neutrality of the inner surface of a conducting shell. Very accurate measurement have been made testing the neutrality of the inner surface, and are used to give the most accurate determination of the 1/r^2 dependence of Coulomb's law.
  8. Mar 3, 2008 #7
    I don't understand. Are you saying that the free charge density goes to zero for any depth greater than an electron radius below the surface? The OP was asking what the real charge density looks like, given that "infinitesimally thin" is usually an approximation of the truth - albeit an approximation that is more than adequate for all practical purposes. Do you mean that there is no measurable depth for the charge distribution?
  9. Mar 3, 2008 #8


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    Science Advisor

    The usual equations of electromagnetism apply in the abstraction of a continuous material.
    To describe the thickness of static surface charge on a conductor, the treatment would have to consider the individual atoms near the surface. There probably are "surface science" calculations of this, but they are not in my field.
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