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Why don't electrons leave a negatively charged metal in air?

  1. Feb 3, 2008 #1
    I'm familiar with lightning rods taking advantage of the mutual repulsion of charges to shoot off a corona discharge off the sharp end and start a thunder, but why doesn't corona discharge happen to all charged metals? What makes air such a good insulator, when it's just gases, relatively few molecules moving all over the place bouncing on each other, how can this be a good insulator?
  2. jcsd
  3. Feb 3, 2008 #2
    To answer you question : think of the essential difference between a conductor and an insulator. What makes a metal a good conductor ? This property is not inherent to air !
    wink wink, enjoy

  4. Feb 3, 2008 #3
    > What makes a metal a good conductor ?

    Dunno. If you were an electron in free space, would there be anything to stop you from moving? Space should be a great conductor, better than an electron-crowded metal. E/M waves move faster too.

    You see what I'm saying? Air is closer to free space than it is to metal, is it not?
    Last edited: Feb 3, 2008
  5. Feb 4, 2008 #4
    I think you're confusing electric current in a conductor with the movement of a free electron through space or through air. They are not the same.
  6. Feb 4, 2008 #5
    What about electrons jumping from one metal to another separated by space, like in an old amplifier lamp. It doesn't matter what we call current, I just want to know what's stopping electrons from leaving a metal based on first principles, quantum principles, wavefunctions and all that. That's why I'm asking you guys in this section.

    From what I know there is nothing stopping an electron from jumping from a negatively charged metal to another in free space, and very little in air.
    Last edited: Feb 4, 2008
  7. Feb 4, 2008 #6
    In other words, how do we account for the insulating properties of air from first principles.
  8. Feb 4, 2008 #7
    Air consists mainly of N2 and O2, both of which are highly stable molecules with the valence electron shell completely filled (triple bond for N2, double bond for O2). It takes a lot of energy to put an electron into the next shell, far more than would be given up when an electron leaves a negatively charged metal.

    Metals by contrast do not have their valence shells filled. In fact, electrons in a metal can move fairly freely like water in an ocean because of this, jumping from atom to atom. For metals, even a small electric potential imbalance will cause current flow because the electrons, being totally free to move, will gravitate (no pun intented) toward the positive end of the imbalance. But, for air, because it requires so much energy for an electron to jump to the next shell level, this requires a huge potential difference, along the scale of lightning.

    In a vacuum, electrons will only flow toward a positive charge, which is then simply governed by F = kq1q2/r^2.
  9. Feb 4, 2008 #8
    Yes but electrons don't have to jump from air molecule to air molecule, they can just fly between molecules, right?
  10. Feb 4, 2008 #9
    No, with a lightning bolt, there is a current flowing through the air, making the air itself a conductor - i.e., jumping from atom to atom. It's because the air is ionized that a conductive path is formed. No one is certain, however, _why_ an ionized path is created in the air.

    Again, remember that electrons "want" to be bound to atoms. They ideally want to be in a full valence shell, but they'd much rather attach themselves to a higher orbital spot in an atom than be free. Free electrons are usually only seen in labs or electronics like vacuum tubes or CRT televisions.
    Last edited: Feb 4, 2008
  11. Feb 4, 2008 #10


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    Cathode rays are electrons travelling in space. But you have to make a pretty hard vacuum for them to travel a foot between emitter and target. Old vacuum tubes use a stream of electrons - note the name. The point is that at atmospheric pressure, electrons can't fly between the gas molecules very easily.
  12. Feb 4, 2008 #11

    Ben Niehoff

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    This is a good question; one way to answer it is to use the Method of Images.

    Consider an infinite conducting, charged plane (which is what any conductor looks like up close), and a point charge of the same sign, near the plane. The point charge induces an additional, opposite charge on the conducting plane, which can be modeled by considering a fictional "mirror-image" of the charge, sitting the same distance away, but on the other side of the conductor.

    The repulsive force of the infinite plane is constant, but the attraction between the point charge and its image goes as 1/r^2. Therefore, no matter how strong the repulsion is between the charge and the plane, there is some distance r within which the attraction between the charge and its image is stronger. Call this distance D.

    Therefore, to fully remove an electron from a conducting surface, it must be removed at least a distance D, or it will actually be attracted back to the surface! To remove the charge a distance D against this attractive force requires some energy; calculation will show that this energy is equal to the "work function" of the material in question (the same work function used to calculate the photoelectric effect, where electrons are knocked out of a metal by energetic photons).
  13. Feb 9, 2008 #12
    Metals are good conductors because they exhibit the property of conduction band electrons. Such electrons are not bound to one atom but can move from one atom to another in the socalled conduction band. Such bands exist thanks to :

    1) putting many atoms together (you need a many particle system, single atoms do not have electronic bands)

    2) the specific crystal structure, ie its symmetry, of that material, which makes up the electronic energy structure

    Last edited: Feb 10, 2008
  14. Feb 9, 2008 #13
    They can but they can also interact with air-molecules. Besides, electrons loose kinetic energy when passing through air ! This happens much less in a metal ! Why ? because electrons passing through a metal are essentially helping each other to pass through the conductor, a bit like "domino's". When you place a potential across a wire, the electric field at the end of the wire changes, this causes the electrons at the end of the wire to move. Once these electrons have moved, they change the electric field experienced by the neighbouring electrons, which as a result feel a net force causing them to move, thus again perturbing the field for their neighbouring electrons. This is repeated down the whole length of the wire, each electron perturbing the field of its neighbours causing it to move,

    Last edited: Feb 9, 2008
  15. Feb 9, 2008 #14


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    I think you meant to say "conduction" instead of "valence."
  16. Feb 9, 2008 #15
    When you have a lighting rod, the tip is usually very pointy. At these sharp edges, you can build up an extremely large electric field compared to that along the length of the rod. The stronger electric field allows the air to become ionized much easier, dissipating much of the charge on the rod. If your rod is too pointy, though, the efield drops off too rapidly, and you do not get as much ionization.

    Any charged piece of metal could do this, it would just have to be fairly pointy.
  17. Feb 10, 2008 #16
    Opps, indeed, my mistake. Post corrected.


  18. Feb 10, 2008 #17
    The air is not insulator for the same reason as the solid insulators are, for sure! Solid insulators are insulators because the available momentum eigenstates form a filled ball, and the expectation value of the momentum of all particles remains zero. Such effect does not occur in gas, because there is no periodic potential on the background.

    I don't think that considering the bound states of electrons on the gas molecules was helpful either. If the electrons cannot get onto the bound states of molecules, shouldn't the free electrons then just behave as a gas as if the electrons were like other molecules?

    A great post! I think this answers the original question, and is also related to one other problem I was thinking about some time earlier. Doesn't this also explain why electrons don't form gas in normal circumstances? Free electrons always get attracted to the solids for this reason? I might guess that this effect is present also with insulators, because you don't need macroscopic currents for this.
    Last edited: Feb 10, 2008
  19. Feb 11, 2008 #18
    But it's not obvious why the redistribution of charge is equivalent to a mirror image of equal and opposite charge appearing on the opposite side of the mirror, where the mirror is the metal surface. What is the proof?
  20. Feb 11, 2008 #19


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    Have you ever looked at the Method of Images before? It is a standard technique in solving for the fields in electrostatic problems, i.e. it is in textbooks on classical E&M.

  21. Feb 11, 2008 #20
    Thanks, I'm sure it is a standard method, but does that satisfy your curiosity? It is so much more fun to look for the fundamentals of things, otherwise we'd just say:

    1. metal, good conductor
    2. air, bad conductor
    3. therefore current does not flow
    4. end of story

    Isn't the root of things in terms of basic, basic elements, so much more satisfactory and profound understanding?
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