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In a conductor, any excess charge will distribute itself evenly over the surface of the conductor. Because of quantum mechanics, this is possible with small charges (i.e. 1e).

But if electrons were balls, they wouldn't always be able to distribute themselves evenly over the surface of the conductor. For example, if the excess charge was 2e, there would be two points with charge 1e each and the rest would be neutral.

So, what would happen in conductors with excess charge if electrons were like balls that couldn't necessarily evenly distribute themselves over the shell (on account of them being discrete)?

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    $\begingroup$ @John I think the point of this question is to ask the counterfactual in which you can't have delocalized electrons. I'm always a little edgy about counterfactuals on Physics SE, because they are a legitimate tool at times, but could be easily abused on the site. $\endgroup$ – dmckee --- ex-moderator kitten Feb 23 '14 at 18:27
  • $\begingroup$ Comment to the question (v2): It would be good if OP (or somebody else?) would clarify/simplify/stress what is really being asked so that it doesn't superficially look like a duplicate of this by OP. $\endgroup$ – Qmechanic Mar 5 '14 at 9:54
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if electrons were balls, there wouldn't be conductors. Coulomb systems are unstable. you wouldn't be able to have free electrons and other stuff, they'd all lump together with positive nuclei. the remaining balls would run away as far as the Coulomb force pushes them.

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As you say, the electric charge is quantized and the charge on a charged body is always an integral multiple of the charge on a single electron or a proton. In most practical situations, charge on a charged body is large as compared to the magnitude of charge on an electron or a proton that the quantization of charge may be ignored. In other words, we can assume that on a charged body of reasonable size, charge has continuous distribution or in your words charge can be assumed to be distributed evenly. For small charges like 2e, we can't assume even distribution or continuous distribution of charge, no matter what you consider to electron to be a ball or anything.

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  • $\begingroup$ But then what would the equilibrium state look like with 2e? $\endgroup$ – dfg Feb 23 '14 at 18:39
  • $\begingroup$ With 2e electron charge, it would be negligible that we can't assume even charge distribution and charge will not be distributed evenly. As you say, there will be complexity in distributing it evenly. We can assume linear, surface or volume charge densities only when magnitude of charge is large. $\endgroup$ – Immortal Player Feb 23 '14 at 18:47
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    $\begingroup$ I think you have underestimated the sophistication of the question. See the comments on physics.stackexchange.com/q/95291. $\endgroup$ – dmckee --- ex-moderator kitten Feb 23 '14 at 22:14
  • $\begingroup$ @dmckee: Thank you for the comment. "He who seeks for methods without having a definite problem in mind seeks for the most part in vain"-D.Hilbert I think I am going to get most part in vain now. If where the problem is going on, directly mentioned. I will for sure make corrections or quit by deleting my post. $\endgroup$ – Immortal Player Feb 24 '14 at 2:09
  • $\begingroup$ The answer can explore that in the conductor there are free electrons, even if not charged, and all conductors have defects with potentials different from the mean where... I remember a game of N seat chairs and N+1 persons. Up N, move around N+1, seat N, and repeat...without leaving the game. Randomly 1 will be up. Which one is not relevant because the game is endless. $\endgroup$ – Helder Velez Feb 28 '14 at 2:54
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We have to make some assumptions. Let's first consider what we mean by a neutral conductor with small balls of negative charge representing electrons. For the positive charge, let's assume a continuous uniform background ("jellium", if you will) whose total charge is equal to the total charge of the electrons. Let's also assume that the electron balls are small enough so that they don't fill the volume of the conductor, and that they can move freely except for when they contact another electron. Also assume that the charge is concentrated at the center of the ball, and that there is some kind of shell that forms the ball. I make this assumption so that we can allow the balls to touch each other. (uniformly charged balls might be able to touch each other, but I'm not going to take the time to figure out if they can or can't) [Notice already how many strange assumptions one needs to make progress with the question.] The electrons will move to a configuration of lowest potential energy (or close to the lowest; It might not find the absolute minimum). They will be distributed relatively uniformly in a layer of non-zero thickness at the surface, but not completely uniformly because they have finite size. The field inside the conductor will not be identically zero. The graininess of the positions of the electrons prevent them from achieving the distribution that produces zero field. Near the surface, within the layer of electrons, the field will vary wildly. The field will get smaller and smaller the further you probe from the surface.

Now add two more electrons. Again, the electrons will redistribute to "try" to minimize potential energy. If the conductor is a solid sphere (a "ball"), then you will end up with two areas where there is a higher concentration of negative charge, and these areas will be diametrically opposite each other across the sphere. The field within the layer of electrons will again vary wildly from place to place, but differently than for the neutral conductor. The field inside the sphere will become smaller the closer to the center that you probe, but the "healing" of the field will occur over a longer distance than that for the neutral conductor.

I hope this helps, but it's rather non-physical. I suppose it might model some system, maybe charged plastic microscopic balls in a fluid. Not sure.

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You are basically confining two negative charged small spheres in a bigger, neutral sphere (this image is better since the you don't need to worry about quantum effects, and there's no need to consider unphysical situations). Since both spheres would repel each other, they'd try to maximize the distance between them, so they would set up in a configuration on the surface of the sphere in which the equilibrium position is with both spheres in diametrically oposite points. Depending on the initial conditions of the problem, the spheres could move around the equilibrium positions. Notice that ANY pair of diametrically opposite points are an equilibrium position.

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electrons will only distribute evenly if the conductor is spherical. if it is not, coronal discharge will occur at sharp points until the voltage is too low to sustain it. of course, breakdown will occur even with a spherical conductor, e.g., van de graff generator, but much more charge will accumulate before any one point has sufficient potential to cause dielectric breakdown. when you refer to a conductor, the fact is that there are so many electrons in even the thinnest wires that they are easily able to distribute the electrical potential amongst themselves. the only place you really see effects from electrons behaving as (more) discrete particles is at the atomic level, where it is the dominant factor in determining the shapes of molecules. it's called VSEPR theory, but it's more of a chemistry topic (at least, AP chem's where i learned about it).

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  • $\begingroup$ I think in the realm of ideal E&M, which I think the OP is all about, discharge can be left out of the discussion. I also think he's talking about an imaginary world where charge carriers are large. I may be wrong. $\endgroup$ – garyp Mar 5 '14 at 3:37
  • $\begingroup$ @garyp large as in massive, or in volume? or even more imaginary than that? $\endgroup$ – user40753 Mar 5 '14 at 3:48
  • $\begingroup$ The OP would have to answer, but I thought volume. $\endgroup$ – garyp Mar 5 '14 at 12:11

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