Three Spheres: Newsletter Challenge (December 2015)

Q...

Waay too long since this was done as assignment.

Similar charges repel, and charge accumulates at the surface, so Q/2 on the extreme surface of each of the outer spheres with a bias to the outer "ends", no charge on the centre sphere.

Though I'm certain one of my electromechanical energy conversion profs would be asking whether the spheres are hollow (Just to stir) and which way is magnetic North (for interaction) and whether they are moving in relation to the observer or any field.

My physics prof would then interject that the mere action of trying to measure the result would cause disturbance. (Something he learned from some Heisenberg guy.)

1/2 Q on A, 1/2 Q on C, zero on B. By symmetry, A and C have to be equal. Like charges repel, so half go to A, half go to C.

I'm having second thoughts about my answer.

Since the charge density is all on the surface and is proportional to the surface curvature...

All the charge will be in the wires. None will be on the spheres.

My analysis was the same as your first answer, and I was going to give you a GA for it.

If Q was one charge, it could be anywhere on any sphere or wire. If Q was 2 charges, they would be as far from each other as possible. As more charges are added, the mutual repulsion keeps them as far apart as geometrically possible. If Q is an odd number, there could be at most one charge on the center sphere; if Q is an even number there would be no charges on the center sphere.

For a very large value of Q, the average would be 1/2 Q on the left sphere and 1/2 Q on the right sphere.

Here's a quickie sketch of what I think the charge distribution would cause, in terms of field lines. (The field lines in the center woldn't actually touch; they'd be asymptotic to infinity.)

The charges cannot move to the far ends, because then there would be an electric field caused by the potential difference between spheres A and C and sphere B. Since the spheres and wires are perfectly conducting, the charges have to be uniform. So it's not as simple as the charges just moving as far away as possible. That solution seemed too easy! (A good puzzle has to have some irony in it.)

All of the charge is on the surface. There is no potential difference inside a conductor. The electric field is everywhere perpendicular to the surface.

The charge density (electric field) is inversely proportional to the radius of curvature. That's how lightning rods work. Consider a very long thin charged conducting cylinder. There is no voltage parallel to the surface, so the entire cylinder is at the same potential, a given charge per unit length. If you increase the diameter, the surface area increases proportionally, but the charge density drops by the same factor. So the charge density per unit length remains constant when you change the wire size (within reason, until leakage out the ends is significant).

So if the diameter of the wire were increased to 20 cm, it would still be similar to a long thin cylinder (20 cm/240 cm), and the amount of charge per unit length would be about the same.

So my final answer (I hope ) is that about 50 percent of the charge is in the connecting wires 120 cm/240 cm), and the remaining 50 percent equally shared by the three spheres.

Ok, Yes. Enough of that makes sense to merit a GA.

Charges on conducting spherical surfaces are not uniform except in the spacial case where the sphere is isolated.

Obviously, a positive charge will attract negative charges, etc. This is not an "easy" problem.

P.S. think of the field strength with all that charge in a very thin wire. This bit of the question is sensible: the potential at the surface of the thin wire is the same as on the spheres, so although the charge-density may be high the total charge will be small (and the wire's contribution to any potential will decay quite rapidly away from the wire).

Metal spheres. Q/3.

All spheres including the wire has the same potential since connected by a conducting wire. It makes the potential difference on any point on any of the surfaces zero. The three spheres and the wire obviously share the same charge Q - homogeneously (cumulative sum of all). Thus, cutting the wires with perfectly insulated tool or even not would make the charge of each sphere a 1/3 of Q, considering wire is so thin so that q in wire is negligible.

each sphere has a charge of Q/3

Voltage being common to all three, charge will be distributed according ONLY to each capacitor's "isolated sphere capacitance" which is ALSO the same for all three, since they have the same dimensions. For the record, Isolated sphere capacitance is air or vacuum is in the order of r/(9*10^9) Farads, r of the sphere in meters. Since spheres are electrically connected, their relative capacitances (due to their surfaces and proximity to each other) will have no significant effect in charge distribution. So I will stick to Q/3. S.M.

Something must be done with the short time you have to correct typing errors like mine ... "each capacitor's "isolated sphere capacitance"... above which should read: ..."each sphere's "isolated sphere capacitance"... to make some sense. This is not tweeter, we try to write meaningful things on short notice and I'm getting fed-up with the lack of correction possibility (within reasonable time). Or after admin approval for that matter. S.M.

Just a very occasional dip, as I'm usually up to my eyes...

This is not an easy problem! I suspect that the setter assumed that you could calculate the fields from each sphere independently and simply add the potentials*. Such simple superposition appears to be incorrect, because the superposed fields would be non-zero inside all of the spheres.

Feinman http://www.feynmanlectures.caltech.edu/II_06.html (section 6-9) tells you how to handle the two-sphere case, and once you take advantage of symmetry this is a reasonably straightforward extension; but there's still more manipulation involved than insight, so it's not the sort of thing that I would expect to see as a CR4 challenge question.

Best to all

Fyz

*We would then simply solve Qa=Qc, Qa+10Qb+2Qc=10Qa+Qb+Qc/2 and Qa+Qb+Qc=Q), but this is not worth doing as it does not give a precise answer (though it's not all that far out because the spheres are small compared with their separation).

Well, "The Answer" is an approximation: charges treated as being uniform on the surface of each sphere (so outside the sphere this is equivalent to the charge being concentrated at the centre of each sphere). That would be fine if the question had specified such an approximation.