- #1

Laudator

- 18

- 1

My thought:

When ##r>b##, the potential behaves like the potential of a point charge, so ##V=\frac{1}{4\pi\epsilon_0}\frac{q}{b}##.

When ##a<r<b##, a conductor is an equapotential, so again, ##V=\frac{1}{4\pi\epsilon_0}\frac{q}{b}##.

When ##r<a##, I'm a bit confused, but I think it's ##V=\frac{1}{4\pi\epsilon_0}(\frac{q}{r}+\frac{-q}{a}+\frac{q}{b})##.

Is this correct? Second, what is the energy stored in this configuration?

My thought:

I learned that the energy stored in a surface charge distribution is ##E=\frac{1}{2}\int\sigma Vda##. The surface charge densities on the cavity wall and the outer shell should all be uniform and the potentials on them are the same constant, but the surface charge densities have opposite sign, does this mean the energy is 0? This doesn't seem right... Also I don't know how to take the energy of the point charge at the origin into account, the integral seems only applies to distributions, not an isolated charge.

This is, in fact, a problem in David J. Griffiths'

*Introduction to Electrodynamics*, 4th edition, problem 2.60. I checked the solution manual, the answer is ##-\frac{q^2}{8\pi\epsilon_0}(\frac{1}{a}-\frac{1}{b})##. I brainstormed for days, still don't get it...