Consider a thin, hollow, fixed and overall uncharged metallic sphere of radius r, with centre O. Now suppose a point of charge Q is held at A. OA = d (d > r). Let N(x) be the charge : surface area ratio at any point P on the surface of the sphere such that the angle between the lines OA and OP is x. What possible function N(x) minimises the total electrical potential energy of the sphere?

Thanks,

Michael

disclaimer: this may be a load of rubbish

I am not particularly good at this type of thing, (so feel free to point it out) but if you hold your charge outside the sphere, then there will appear a section of opposite charge on the surface of the sphere. Electrical energy is E: E = QV (I hope this is the thing you mean) so lets say you have a +charge (unfeasible assumptions coming up:)

then there will be a zone of negative charge (possibly an equal amount?) on the surface of the sphere around the bit where the line OA cuts it, and a zone of +charge diametrically opposite that part. I'm not sure how you get the energy potential, but the test charge will want to go towards the sphere, and vice-versa, so a force is acting:

F=[1/4pe₀][(Q₁ Q₂/r²] If the charges are equal, then it will be Q²

now E(or strictly work)=Fd

so E=[1/4pe₀][(Q₁ Q₂/r]
and then V=E/Q

so V is E but cancel one or other of the Qs (?)

er..

your use of the words charge:surface area ratio confuses me, but it seems to be a similar idea to the patch of opposite charges. This is total guesswork, but this ratio as you go through an angle x (through the plane with P and the place where OA cuts the surface on it, now called Z) may be described by -cos x, which is just an idea, since the max -charge will appear at Z (ie x=0), and at x=pi, will be the focus of +charge and at x=pi/2, you will be halfway between the two focuses, so there will be no charge here. The bits in between will have smaller amounts of charge because not all the electrons will want to focus too near each other, so they will have to space out, becoming more spaced out as you go back to the x=pi/2

I don't see that you have mnuch choice of functions for N(x) if any of this either A: helps (ha!) or B: makes you laugh then I have at least tried to answer a fairly alien question. Was this in a physics paper?

Hi! Thanks for the reply.

I think that perhaps I was a bit ambiguous when posing the question. You said you don't understand what N(x) actually represents. Well I'll try to explain.

Imagine you take any section on the surface of area A. Let the total net charge on this section be Q. Let's describe Q/A as the "average charge density" or "average charge:surface area ratio" in this region. But what we actually want is the charge density at a point. To do this, we take a sequence of areas converging on a point and look at the limit of the average charge densities. (This is analogous to the instantaneous speed of an object being the limit of the average speed over continually shorter time intervals.) So now we know what the charge density at any point is.

N(x) simply denotes this charge density at any point at an angle x. (Between OA and OP.) For any angle x there is a circle of points, but it doesn't matter which point we take. The charge density will be unaltered. By symmetry we can be sure of this.

So now we've got the definition of N(x). There are an infinite number of possible functions N(x) and we want the one that minimises the total potential energy. The only thing we know for certain is that the total charge is zero. (Thus we could form an integral equation for N(x).) I want to minimise the potential energy because this will show the charge density at each point when the electrons on the surface are in equilibrium. I have managed to find the potential energy as double integral (neither of which I can evaluate let alone minimise!)

When you say the ratio is -cos(x) then are you referring to N(x)? I'm not too sure this would work as the minimum function should not really be symmetric about the plane through the diameter perpendicular to OA. But it's not a bad guess; it may be possible to modify it (make it lopsided) and get what we require.

By the way, this was not in a physics paper. (The mathematics in the sort of physics papers we do rarely goes beyond re-arranging F = ma for m and a...) The reason I wanted to answer it was to find out how large a dipole is induced on a sphere when a charge is brought near it. This will always be done to minimise the electrical potential energy. I admit it is an exceedingly hard question. (It's my current favourite for your page on impossible questions, although I doubt that this one would win the special prize!) I'm wondering if there is any way of showing that the answer I'm looking for cannot be written in closed form.

Many thanks,

Michael

Hi,

I'm still not sure what you mean by electrical potential energy, and do any of the formulae for F, E, V etc that I gave serve any purpose in this question?

Regards

Neil

Yes, they were what I was after.

The potential energy between two particles is kQ₁ Q₂ /r like you said. The total potential energy of any particle is the sum of the potential energy between this particle and all the others in the system (both on and off the sphere). Then the PE of the entire sphere is the sum of this over all particles on the surface of the sphere! Not easy.

However, saying I wanted the potential energy minimised was just a convenient way of saying that I wanted the function N(x) that would leave the electrons in equilibrium on the sphere (given that the sphere is fixed). Clearly the sphere will perceive an overall attraction to the point charge, but if it is fixed then the individual charge carriers are in equilibrium. (Because in a metal they cannot leave the surface.) What I mean by equilibrium is that they don't want to move along the surface of the sphere. It is possibly best not to bring energy into it at all. This doesn't mean that I have made any progress any other way... I just think it is quite interesting that physicists use the idea of dipoles all the time. (Moving a point charge close to a collection of charged objects distorts the charge density.) Yet when it comes to actually calculating the extent of the distortion, it seems almost impossible.

Many thanks for your trouble,

Michael