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Problem with the physics behind charged, isolated conductors

  1. May 6, 2010 #1
    There is something that I do not understand completely with regards to charged, isolated conductors.

    Gauss' law implies the following:
    "If an excess charge is placed on an isolated conductor, that amount of charge will move entirely to the surface of the conductor. None of the excess charge will be found within the body of the conductor."

    So let us suppose that we have a solid, spherical conductor made up of copper.

    The previous statement says that if you were to add any quantity of negative charge, say n electrons, that they will disperse to the surface of the material in an attempt to be as distant from one another due to electrostatic repulsion. This makes sense to me, because in a conductive metal the free electrons are allowed to move about. Throughout the volume of the metal there will be the normal amount of electrons associated with the protons in the copper molecules, and the n added electrons will be dispersed across the surface area of the sphere.

    However, I don't see how the converse is true: suppose that we begin with a neutral sphere of copper and then remove n electrons. Gauss' law says that there should be no excess charge will be within the conductor. The protons are fixed in place, though, there are no free protons to move about to the surface; the net charge is induced by our removal of mobile electrons, rather than the addition of protons. Therefore, in order for our theory via Gauss' law to hold, that means that there are n less electrons on the surface and still the required amount of electrons within the sphere to make the inner neutral.

    My problem with this is that the electrons within the metal, the conductive electrons, are still free to move about and because of their repulsion should still want to be as far apart as possible. How/why do the electrons not act like this; why, in this scenario of electrons removed, does the entire volume of the sphere stay uniformly charged?
  2. jcsd
  3. May 7, 2010 #2
    The electrons are also attracted to the protons. If there are more protons than electrons, the attractive force will be bigger
  4. May 7, 2010 #3
    I think this may occur because: If you consider a local positive charge in the center of the volume, it has free electrons around it in every direction that will try to move to nullify the charge. Whereas if you have a local positive charge on the boundary of the volume, there are less free electrons available to move in and nullify that charge. Hence the electrons tend to be drawn inwards and the charge imbalance exists concentrated on the surface.
  5. May 7, 2010 #4
    When electrons are removed from the conducting sphere, they are nominally all removed from surface atoms, leaving positively charged copper ions on the surface . When this supply of electrons has been exhausted, no further "positive charging" of the sphere is possible. The best way to understand the behavior of conductors, more generally, is that Ohm's law ensures that the conduction electrons/positive ions are always arranged in such a way that the electric field at interior points is zero. It isn't true that conduction electrons move to the sphere's surface in order to get as far away from one another as possible. Consider an uncharged sphere. The conduction electrons are evenly distributed within the sphere. If you could add extra electrons to a point inside the sphere (say via an insulated wire), they would engender an electric field at points within the sphere, and conduction electrons at such points will move such as to drive the electric field at interior points to zero. By Gauss' law, this indicates that they must all reside on the surface (of any conductor) once electrostatic conditions have been attained.
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