Charge distribution within a solid conductive sphere

[m.e.t.a.]

I have been attempting to deduce the arrangement of electrons within a solid spherical conductor in different states of charge. I'd like to know which, if any, of my conclusions are correct; and of the incorrect, I'd like to understand where I went wrong.

A cool conductive (metal) sphere is placed in vacuum. The sphere is initially neutrally charged and is then given a net-positive charge followed by a net-negative charge. Between these changes in charge, time is allowed to allow the system to come to equilibrium. I have modeled the free electrons as free-moving particles whose movement is uninhibited by the temperature or structure of the metal lattice.

Rather than explain my reasoning, which would be tiresome for people to read, I will just list my conclusions for the different states of charge.

1. Neutrally Charged
Electrons are arranged homogeneously about the whole volume of the sphere.

2. Positively Charged
The sphere is divided into two distinct regions. All free electrons locate themselves in a central “core” of radius $$R_C$$. The electrons in this core are arranged homogeneously. The remainder of the sphere (between radii $$R_C$$ and $$R$$) contains zero free electrons and is therefore composed of positively charged metal lattice. There is a sharp boundary between the two regions. (http://img182.imageshack.us/img182/9082/positivelychargedconduchy1.png" .)

3. Negatively Charged
The whole sphere carries a homogeneous neutral charge with the exception of a thin spherical shell at radius $$R$$. In this thin surface-shell, all free electrons are located.

It is point 3 that I find most difficult to accept, partly because I could only be vague (how thin is "thin"?) and partly because I don't know how electrons behave in close proximity.

Are any of my conclusions correct?

- m.e.t.a.

 

[James R]

Regardless of whether the sphere is positively or negatively charged, all the excess charge is on the surface. This follows from Gauss's law. So, if the sphere is negatively charged, all the excess electrons lie on the surface of the sphere. If it is positively charged, then the "missing" electrons are missing at the surface.

The simple application of Gauss's law treats a macroscopic charge distribution as continuous, however, and not made up of individual quanta (electrons, protons etc.). To get down to the nitty gritty of where the excess electrons are in the solid metal lattice (rather than just saying they are at the surface, for example), you need to look at the microscopic energy-band structure of the metal itself. But then the electrons don't look like point particles anymore - they are "smeared out" as quantised waves in the conduction band of the metal.

 

[m.e.t.a.]

If it is positively charged, then the "missing" electrons are missing at the surface.

Could you elaborate? It sounds as if you agree with me, though I have a problem understanding how all of the sphere's positive charge could be located "at" the surface.

Negative surface charge can be very dense, because free electrons are both mobile and small. But positive surface charge cannot become similarly concentrated because the metal lattice -- which carries the positive charge -- is immovable. Furthermore, inter-atomic distance is large compared to the small minimum separation of free electrons. Also, a typical conductor holds only one or two free electrons per atom. These three factors together (stationary lattice, large inter-atomic distance, low maximum electron liberation per atom) place a very low limit on the maximum density of positive charge.

The density of negative charge, however, suffers from no such limitation. This is why I do not understand the assertion that positive charge locates "at" the surface.

 

[Domenicaccio]

I think meta's conclusions make all sense.

The simple application of Gauss's law treats a macroscopic charge distribution as continuous, however, and not made up of individual quanta (electrons, protons etc.)

Not only that, but also the normal model of a conductor (by which at equilibrium the inner E field is zero and all the excess charge is on the surface) tends to assume that the object has as much movable charge as needed.

But clearly, if you imagine a conductor being charged by only removing electrons, unless something else occurs in the process, clearly the area of "uncovered positive charged" extends more and more inwards and cannot be an arbitrarily thin layer.

If you were charging it positively by adding protons instead, assuming those protons would be free to move (I have no idea), then they would occupy a very thin layer on the surface.

 

[Domenicaccio]

The density of negative charge, however, suffers from no such limitation. This is why I do not understand the assertion that positive charge locates "at" the surface.

My conclusion, to make it clearer, is that this is true indeed only if you assume that for the problem's purposes the conductor has enough atoms on the surface (not surface in a geometrical sense, but you may afford to approximate it that way if that's ok for your purpose) to "provide" all the + charges needed.

 

[m.e.t.a.]

I think I understand what you mean now, Domenicaccio. You are saying that, under everyday conditions, the electron deficit of a positively charged conductive sphere (or other object) is normally so small that the layer of charge can be modeled as infinitely thin.

Of course, an equal quantity of negative charge would occupy a much thinner layer still. But, as you point out, under ordinarily achievable conditions it doesn't matter what polarity your charge is. Both layers are thin, and you're unlikely to encounter a positive charge so vast that the shell containing that charge need be modeled as anything but "thin".

I attempted to calculate how thin this layer might typically be. By taking the plausible example of a solid copper sphere charged to 20kV, I got a figure for the "virtual" thickness of the charge-carrying shell at around 2.30 × 10$$^{-16}$$ m. I'd happily bet that my answer is likely to be wrong, but it certainly is "thin" indeed; in fact, it is a good deal thinner than the diameter of an atomic nucleus. This suggests that, unless I were to massively increase the charge on the sphere (or apply the same charge to a much smaller sphere) I would only have ionised a tiny fraction of a single one-atom-deep layer. (In trying to roughly estimate this fraction I arrived at a figure of between 1 and 10 +1-ionised atoms per million neutrally charged ones. It seems like an impossibly small figure.)

But again, exact figures aren't important. Domenicaccio and James R are both correct; thank you for your answers!

 

[James R]

m.e.t.a:

One more thing: Most of the time in the real world, when we talk about a "positively charged metal sphere" or something like that, we mean a sphere which has had some electrons removed, so that the number of protons remaining is greater than the number of electrons, thus giving the sphere a net positive charge. We can still refer to this situation as a "positively-charged metal sphere, with all the positive charge at the surface". This is essentially a shorthand way of saying that the "missing" electrons are all missing from the surface layers of the metal.