hyleave?
Charge will indeed flow off the inner shell out to infinity, up to the (very short) time when the charge on the shell equals the value calculated in Problem 3.9. The error in the opposing logic is the following. If we consider one small point charge, then this charge would be happy to hang out on the wire right at the hole in the outer shell. However, if we try to do the same thing with another small point charge, then we can't put both of them in the same place, because they will repel each other. So we certainly can't pile up charge right at the hole in the outer shell. What if we stretch out the charge and create a linear density along the wire? Could the charges stabilize in a linear distribution, partially between the shells and partially outside the outer shell?
To answer this, we must use the fact that a very thin wire has essentially zero capacitance (see Exercise 3.59; the outer radius in that setup can be assigned some arbitrary fixed value). So charge can't pile up on the wire. If we give the wire a tiny radius, then in a static situation only a tiny bit of charge can pile up. The very strong forces (which go like 1/d², from Coulomb's law) between nearby charges on the wire will make the linear charge density λ nearly uniform (see Problem 3.5). So in a sense we have an essentially rigid stick of (a tiny bit of) charge extending from the inner shell out to a very large radius. The question is: what is the direction of the net force on this stick? Will it be drawn inward or be pushed outward?
There are competing effects, because the field E₁(r) between the shells points outward, and the field E₂(r) outside the outer shell points inward (at least once some charge has left the inner shell, leaving behind a net negative charge on the two shells). We are ignoring the forces between the charges in the stick here, because they are internal.
Assuming constant density λ, the outward force on the stick equals ∫ᴿ¹ᴿ² E₁ λ dr, and the inward force equals ∫ᴿ²∞ E₂ λ dr (this is negative since E₂ is negative). The stick won't move if these two forces add up to zero:
∫ᴿ¹ᴿ² E₁ λ dr + ∫ᴿ²∞ E₂ λ dr = 0
⇒ ∫ᴿ¹ᴿ² E₁ dr = -∫ᴿ²∞ E₂ dr
⇒ φ(R₁) - φ(R₂) = φ(∞) - φ(R₂)
⇒ φ(R₁) = φ(∞). (12.149)
We see that the stick won't move if the potential at the inner shell equals the potential at infinity. So charge will flow off the inner shell until the time when this condition is met, at which point the charge equals the value calculated in Problem 3.9.
If the wire does have a nonzero capacitance, then things are different. If we have, say, a little spherical bulge in the wire somewhere outside, then charge will build up there, but only until the potential equals the potential everywhere else along the wire.
As an exercise, you can think about an analogous setup involving two large capacitor sheets separated by a small distance, with charges ±Q, and with one sheet grounded by connecting it to infinity via a very thin wire that passes through a very small hole in the other sheet.
Note that the above reasoning took advantage of the fact that ∫E dr has two interpretations. It equals the potential difference φ, of course. But if we multiply it by a constant λ, it also equals the total force on a uniform stick. Basically, multiply ∫E dr by q and you get the total work done on a charged particle moving between two given points. Multiply it by λ and you get the total force on a charged stick lying between the two points.
644 Solutions to the problems