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If the charge is large, considering charge density is usually meaningful despite the discrete nature of electrical charge. The following sentence is part of a problem in a textbook on electromagnetics:

Imagine a sphere of radius a filled with negative charge of uniform density, the total charge being equivalent to that of two electrons.

The charge is only 2e, and so it seems that the model wouldn't work at all. Is there any real-world application of such a calculation?

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Buckyball with shared orbitals? Maybe QM will allow some sort of psuedo-distributed charge? –  AlanSE Jun 14 '11 at 19:27
    
Why not? Electrons in metal are not "point-like" and the metal is polarizable. –  Vladimir Kalitvianski Jun 14 '11 at 19:35
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As the Fermi velocity of electrons in such a metal is generally quite high (~10^6 m/s), I presume two extra electrons could bounce around in the sphere at such high speeds that it might seem as a uniform charge density. Of course, if we can just say that it is in an eigenfunction of the Hamiltonian, the electrons will be spread about the entire and with it their charge. –  Kasper Meerts Jun 14 '11 at 19:52
    
The Fermi velocity is a quantum mechanical notion and it does not provide a classical averaging. In metals the extra electrons are not normally localized so with a metallic ball the charge distribution can by spherical. As well as with an axial symmetry (a circular current). –  Vladimir Kalitvianski Jun 14 '11 at 21:21
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2 Answers 2

No, I would suspect there is no particle with such a uniform charge density. As you state, at this level, charges are discrete with quarks having multiples of $$\pm\frac{1}{3}e$$ and leptons having either +/-e or zero charge. The charge density of a proton falls off nearly exponentially, see

http://prl.aps.org/abstract/PRL/v99/i11/e112001

All I found was the charge density of Helium ions which seems to be very roughly constant. But the total charge would be +1. see

http://www.ae.utexas.edu/~lraja/APL_80(4)_Kothnur_et_al.pdf

This is probably not intended to be a real world problem but just a homework problem which keeps the numbers simple. At this scale you won't be able to use classical electrodynamics anyways.

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Solid conductive metals have a pool of electrons that are naturally delocalized, i.e., it is energetically favorable for these electrons to spread out somewhat, so they don't have a well-defined location. This may be difficult to grasp if you think of electrons as objects having well-defined locations at all times, but such an intuition is imported from experiences in the macroscopic classical world, and does not approximate quantum mechanical behavior well. That is, while electrical charge takes on a discrete set of values, the position of electrons does not. There is a brief description in the Wikipedia article on conduction.

The delocalized electrons are free to move about when acted on by potentials, and this is what allows such metals to conduct electricity. If you add two electrons to a neutral conductive metallic sphere with no potentials applied, it will be energetically favorable for the charge density to be uniform, and there are no obstructions to realizing such a configuration. In summary, the answer to your title question is "yes".

Regarding real-world applications, if the radius of the sphere is small enough (read: nanoscale), the charge density will be high enough to affect the outcome of chemical reactions. Even if the radius is larger, small static charges may contribute noise to the readings of sensitive instruments.

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The question is not dealing with a solid metal sphere, and charge in a metal sphere will only have a surface density. It talks of uniform density in three dimensions. Even assuming that the electrons are shared all over the surface, why would the collective wavefunction be necessarily uniform? there could be dipole and higher pole terms in general. –  anna v Jun 15 '11 at 10:56
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