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Charge in a Conductor

  1. Sep 23, 2004 #1
    Hello,

    Previously, I asked about a single charge inside a solid metal sphere. And I said that it would not be possible to represent the charge at the surface, because there is no driving force to cause migration. I talked with a professor and he told me that in that case it is called ("metastable?"). I forget the exact word. He said what I posted in here, but could not get an agreeing post on. That point charge would just migrate around freely within the sphere. So as viewed from outside, you would see a charge on the sphere, but it would be constantly changing as the charge moves. He told me that if you knew the position of the particle as a function of time, you could determine the charge at some location. I asked this question before, but everyone kept telling me it would be represented at the surface. His anwser seems to make more sense. Maybe my question was not clear at first, or maybe you still disagree with me. Id like to know what you think thought.


    Edit: No, no im sorry. I think I asked him about having a point charge directly in the center of a metal solid sphere, and having a perfectly uniform charge around the sphere. In that case, it would attract the particle in all directions equally, and he called that metastable. Im sory if its not called metastable, the word started with an m, i think that was it though. He said that it would be impossible for the charge to exist in the very center though, because the atom vibrates, and so it could not remain still, once it got displaced a small amount, it would then be attracted to the surface. ( Tide, i believe this is the thermal energy you were talking about, but I thought of what you told me the moment vibration came out of his mouth, as vibration is a thermal property of molecules, they all vibrate, faster or slower depending on the amount of heat into or out of the system.) I guess only in absolute zero would the charge remain in place, but even that is a stretch!
     
    Last edited: Sep 23, 2004
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  3. Sep 24, 2004 #2

    Tide

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    Cyrus,

    It looks like you finally got it! Yes, classically you would call it a metastable state when the test charge is exactly at the center. Thermal motion will eventually knock it off center but invoking absolute zero won't really help either since, by Heisenberg, you cannot simultaneously position the charge exactly at the center with zero momentum.
     
  4. Sep 24, 2004 #3

    ehild

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    Tide, do you mean that positioning a charged body at the exact centre of a solid metal sphere would not change the surface charge unless this very same charged body moves there? And its charge gets to the surface only because the body can not stay at the exact center?

    Imagine that an oxygen ion is placed at the exact center of the solid metal sphere. It would not give the extra electrons to the metal, it stays charged. On the other hand, the oxygen ion itself can migrate in principle but it is a very slow process at room temperature so we can assume that it stays in the center. What will happen? Does the surface charge change or it stays the same as before? How fast the new equilibrium charge distribution is set up?

    Otherwise you are right that it is impossible to place an extra free electron at the very centre of a metal sphere. Or if that electron stayed there at a certain instant of time it would not stay there and would migrate out to the surface. But anyway, if the test charge is an electron, how do you know the same electron would go out? Can you distinguish the added electron from those belonging to the metal? As far as I know electrons are indistinguishable particles.

    ehild
     
  5. Sep 24, 2004 #4

    Tide

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    Ehild,

    Notice we haven't addressed the problem of how to get that test charge to the center in the first place! Setting that aside, the problem originally posed involved a conducting sphere with excess charge on its surface. In that situation, the excess charge will eventually distribute itself uniformly on the surface of the sphere. Clearly, there is no electric field inside the sphere by symmetry or Gauss' Law.

    The problem was what happens to a new additional charge (the test charge) placed at the center of the sphere. Again, by symmetry, it should stay there provided it is not subject to being displaced by thermal motion. However, the test charge will produce a net electric field inside the conductor! The field lines will be radially symmetric and will tend to pull the surface charge inward or push them outward depending on the relative sign of the charges. Of course, they won't continue moving outward or inward. Rather, they (along with the net neutral charges in the conductor) will redistribute themselves radially creating a dipole layer over a very small region at the surface (the Debye length) until their mutual repulsion brings them into equilibrium. In this ideal case, the internal charges will simply rearrange themselves and effectively shield the electric field of the test charge so that the only electric fields will be in a region very close to the test charge and in a diple layer at the surface.

    My original point was that when the test charge is displaced even slightly from the center (and it WILL be displaced either by thermal motion or from the uncertainty principle!) the field lines produced by the test charge will no longer be radially symmetric. Since the sphere is a conductor, it cannot sustain electric fields at the surface that are not normal to the surface. Therefore, they will redistribute themselves and become asymmetrically arranged on the surface.

    With the symmetry broken, the surface charge will provide an electric field inside the sphere. It is that field that will ultimately pull or push the test charge to the surface. In simple terms, the conductor (including all charges) will do whatever it can to eliminate electric fields internally and nonradial fields at the surface.

    The process is similar to placing a charge near a conducting sheet. The charges in the sheet redistribute themselves in such a way that the sheet remains an equipotential. Sometimes people visualize (analyze!) this using an "image charge" in which case the test charge is attracted to the sheet - whether positively or negatively charged.

    As to how long it takes, that depends on the conductivity of the material but it would be of the same order as the time it took for the original excess charge to redistribute itself after it was placed onto the surface!
     
  6. Sep 24, 2004 #5
    So is the situation about a single charge correct also? Will the charge just wander around inside the sphere if it is a neutral sphere with exactly one elemental charge placed inside it.
     
  7. Sep 24, 2004 #6

    Tide

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    Basically, yes! There will be some redistribution of charges to maintain or achieve zero net electric field inside the conductor and the test charge will wander about.
     
  8. Sep 24, 2004 #7
    So the test charge wont reside at the surface, it could be moving about inside the surface. The only time it will reside in the surface is when there are two or more test charges. Two would mean a force is now present to drive them as far appart from eachother as they possibly can, and so they will be radially opposed from one another at all time at the surface.
     
  9. Sep 25, 2004 #8
    tide, can you reply to post in the question above plz? I just want to double check and make sure you agree with that statement too.
     
  10. Sep 25, 2004 #9

    Tide

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    If you have a single test charge, it can reside anywhere including at or near the surface. Reside is somewhat misleading in this case since it won't stay in one place for any appreciable time due to thermal motion. If you have two like test charges in an otherwise neutral conductor then, yes, they will tend to migrate as far from each other as possible and reside there!
     
  11. Sep 25, 2004 #10
    Just want to put in my two cents...

    There can be *no* field inside a conductor by definition. The central (positive) test charge will be neutralised by the free electrons in the conductors, giving zero charge at the centre. This state is not even metastable.
     
  12. Sep 25, 2004 #11
    Sure it is, why not wong? it be metastable equilibrium. The slightest offset will cause it to move to its natural equilibrium. Were talking about a dt in time, and in this dt of time, then yes i think im correct. But in the long run, you are right. It wont remain in the center, and so later in time it will do as you say and neutralize. I was just hypothesizing about an ideal situation. In the case of the single charge, I dont see how that could reside in the surface as you say though. That would not be possible, but again, were talking about a single charge, and that too is ideal. So you would never see that happen in real life. Even if such a case did exist, it would be so small a field that is constantly changing, I would highly doubt any instrument would detect something on the order of 1.6x10-19 C!
     
    Last edited: Sep 25, 2004
  13. Sep 25, 2004 #12

    Tide

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    It is metastable. As the test charge is displaced from the center its shielding will lag (however briefly!) and communicate the presence of the polarized charge (including shielding) to the surface charges which in turn will redistribute themselves. The field they produce will add impetus to the (polararized) test charge and will continue moving it to the surface. The final state will be no electric fields internally.

    If you don't accept that then consider that the shielding is, for the most part local with an e-folding distance equal to the Debye length. However, even with the exponential falloff with distance from the test charge the outer surface will "know" of the presence of the test charge within and if it's offcenter they will eventually pull it up!

    According to your argument there would not even be a uniform distribution of the original excess charge on the surface since each individual charge would effectively have been cut off from the rest of the conductor by the shielding effect.
     
  14. Sep 26, 2004 #13

    ehild

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    And what happens if there are about 30 electrons already at the surface, and you place a new one in the centre of the sphere? Those electrons at the surface would be closer to it than R, the radius, if it goes out. Why don't it stay in the middle keeping the longest distance possible from the others? :rofl:

    ehild
     
  15. Sep 26, 2004 #14

    ehild

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    I do not understand why does that test charge communicate with the far-away surface charges sooner than with the free electrons in the vicinity?

    The other question: Consider a metal sphere with a hole in the center. There is vacuum in the hole. We place a little uniformly charged isolating sphere at the middle of this hole, its radius being considerably smaller than that of the hole and it is fixed to the center so it can not move, and it is isolated from the metal bulk, so no charge transfer is possible between this charged body and the metal. The charge of the metal sphere was previously Q, all this charge at the surface already (or you can say inside the surface layer of Debye length which is a negligible distance with respect to the radius of the sphere). The little sphere in the center carries charge q. How much charge will be at the surface of the big sphere when equilibrium is established? According to your arguments, it should stay Q as the charge of the little sphere can not move outward.
     
  16. Sep 26, 2004 #15

    ehild

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    There is no static field inside the conductor. But it is an electric field inside a wire that drives the current when a potential difference is maintained between the two end of the wire.

    Otherwise you are right. If the equilibrium charge distribution is disturbed inside the conductor, it tends to equilibrium by the redistribution of the free electrons and this happens in a very short time. The equilibrium charge density is zero inside a conductor at macroscopic scale.

    ehild
     
  17. Sep 26, 2004 #16
    Two things I hate so far, (1) linear algebra, (2) calculating those damm integrals for coulombs law in terms of electric fields of wires, planes, and any other thing that does not integrate easily! Oh yeah, and I REALLY HATE binomial expanisons arg, why dont they just let me use mablab to solve it all!!!! One problem in the solutions manual was solved using a variable that was in the picture, but they never bothered to mention it was not the same variable. So I spent an hour trying to figure it out and finally realized what they ment ARGG. Its 3am, sorry i had to vent. :-p. Everyone write complaints to the university physics publisher for sears and zemanski for that one problem LOL. I need someones reasurance that it will get easier, please for the love of god! :rofl: :cry: :cry: :cry:
     
    Last edited: Sep 26, 2004
  18. Sep 26, 2004 #17

    Tide

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    cyrus,

    If you relied on matlab or mathematica you would never know if you got the right answer and you certainly wouldn't learn the physics as well!
     
  19. Sep 26, 2004 #18
    True....what a silly mistake...I'm talking about static field.

    But the main point is that the state is not even metastable.
     
  20. Sep 26, 2004 #19

    Tide

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    But it is metastable! Perturb the test charge from the central location and Gauss' Law clearly shows a net charge inside the sphere even with shielding and the surface charges will end up pulling the test charge to the surface.
     
  21. Sep 26, 2004 #20
    A conductor contains many free charges. One should not think that the *excess* charges are all free charges that are present. There are also free electrons. It is because of this that in *static* cases there can be no electric field inside a conductor.
     
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