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If we have a configuration of metal concentric spheres (each of negligible thickness) of radii $r_1,r_2,r_3$ respectively and $r_1<r_2<r_3$, and we are given the potentials of the spheres to be $0, constant,0$ respectively. How might we find the capacitance of the configuration?

EDIT: My question: How is capacitance defined for such a system?

Added: I think I have managed to find the charge on each shell :) What is missing now is just a definition of capacitance for such a system.

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Here, they are looking for self capacitance. Self capacitance is basically the charge divided by potential of a conductor. One way to look at is is that the conductor is one plate of a capacitor with the second plate at infinity(or wherever your datum for 0 potential is) . From this, we can easily see that capacitance of a sphere is $\frac{kq}{r}$.

Now, the question could be asking one out of many things. This is because a standing condition (what is fixed) is not given.

Firstly, we can consider it to be a pair of capacitors, one between the first and second shell, and the second between the second and third shells. Assuming that potential of inner and outer plate remains constantly 0, we can take these as a pair of parallel capacitors.

Secondly, we can look at only the outer sphere. We can calculate the change in potential $\Delta V$ of the outer shell when we add a charge $q$ to it, and calculate \frac{q}{\Delta V}$. I personally think that this is the correct interpretation, though this will require you to calculate charge distribution of the sphere twice.

Thirdly, we use the same approach for the second and inner shells.

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this is a good answer... it's tricky to define the capacitance because voltage is relative to some arbitrary zero point. – Timtam Feb 13 '12 at 3:15
Usually we assume the arbitrary 0 point to be infinity. If we have infinite charged sheets, we assume the point as one of the plates. If there is grounding, we take the grounded part to be 0 (we can usually take this simulataneously with potential at infinity as 0, but the earth does have a charge, so there is a small p.d.) – Manishearth Feb 13 '12 at 3:29

Although there is a theoretical ambiguity because there are 3 conductors, the particular specification of the potentials here implies that the desired capacitance $C$ is the ratio of the charge $Q$ on the middle sphere to its voltage $V$: $C=Q/V$.

Note that the inner and outer spheres are grounded so $V$ is the (only) potential difference in the problem.

Note also that the total charge on all the spheres must add to zero: otherwise there would be electric fields external to the sphere-assembly, contrary to the assumption that the outer sphere is grounded.

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