3 ms·

Your objection applies equally to the 2D crystal. As far as I can tell the fact that charges reside on the boundary only really applies to induced charges. Whe

Your objection applies equally to the 2D crystal. As far as I can tell the fact that charges reside on the boundary only really applies to induced charges. When the conductor itself has a net charge then you'd normally expect these to configure themselves so they're as far apart from each other as possible. If the conductor doesn't have any kind of weird shape this should normally result in a roughly uniform distribution of charge (if it has spikes then most charges will move there to be away from the bulk).

c1ccccc1 5y ago
I think for my objection to apply in the 2D case, you'd need for the force law to be 1/r instead of 1/r^2. That would happen in a truly 2D world, but not in this case where we have charges confined to a 2D plane, but with 3D electric fields extending above and below that plane. I think charge still goes to the surface, even if not induced. A spherical conductor with a net charge would have a uniform distribution of charge on its surface, and 0 charge density inside the bulk. Any net charge density inside the conductor would create a diverging electric field around itself, which would cause a current to flow, dissipating that net charge in the process. The same argument should apply to a bunch of free electrons, shouldn't it?
- contravariant 5y ago
  Ah you're right by definition a conductor does have a current provided there's any electrical field inside it (divergent or not). So indeed they can't have charges inside them in equilibrium. However if you induce an electrical field in a vacuum then no charges at all will flow because a vacuum is the perfect insulator. And even if there is nonzero electric field in and around an individual electron they won't move as long as things cancel out at their exact position. Really the theory of conductors and charge densities seems to break down somewhat once you get to the point of individual electrons near absolute zero in a vacuum. One way to put this to the test would be to charge a conductor to its absolute limit, basically removing all free charges from it and seeing if the rest will crystalize. This might require an impractical amount of energy.

c1ccccc1 5y ago

I think for my objection to apply in the 2D case, you'd need for the force law to be 1/r instead of 1/r^2. That would happen in a truly 2D world, but not in this case where we have charges confined to a 2D plane, but with 3D electric fields extending above and below that plane. I think charge still goes to the surface, even if not induced. A spherical conductor with a net charge would have a uniform distribution of charge on its surface, and 0 charge density inside the bulk. Any net charge density inside the conductor would create a diverging electric field around itself, which would cause a current to flow, dissipating that net charge in the process. The same argument should apply to a bunch of free electrons, shouldn't it?

contravariant 5y ago
Ah you're right by definition a conductor does have a current provided there's any electrical field inside it (divergent or not). So indeed they can't have charges inside them in equilibrium. However if you induce an electrical field in a vacuum then no charges at all will flow because a vacuum is the perfect insulator. And even if there is nonzero electric field in and around an individual electron they won't move as long as things cancel out at their exact position. Really the theory of conductors and charge densities seems to break down somewhat once you get to the point of individual electrons near absolute zero in a vacuum. One way to put this to the test would be to charge a conductor to its absolute limit, basically removing all free charges from it and seeing if the rest will crystalize. This might require an impractical amount of energy.