Question involving potentials and spherical conductors/capacitors/things

[schattenjaeger]

Say you have a spherical conductor of radius a located inside the cavity of a larger spherical conductor of radius b, and that the larger sphere's outer radius is c

If you charge the inner conductor at a to a charge Q, the inside part of the outer conductor becomes charged -Q and the outer outer part with Q again, ok, I got that

then it wanted the potential anywhere. I understand the potential outside the whole thing, the potential between b and c, but when r is from a to b I had trouble. I had the answer beforehand, which is kQ/(rbc)*[bc-r(b-c)] or I may've mixed that up

I see that if you unsimplify that it's like a sum of potentials. It's the potential between a and b, kQ/r, minus the potential AT the surface b(well, the Q is negative there so I guess you can just say plus the potential at b)plus the potential at c

Simple question: Why is it like that? Is there a more systematic way of reaching that answer? oh and of course the answer from a to b is like that one with r=a

 

[OlderDan]

Say you have a spherical conductor of radius a located inside the cavity of a larger spherical conductor of radius b, and that the larger sphere's outer radius is c

If you charge the inner conductor at a to a charge Q, the inside part of the outer conductor becomes charged -Q and the outer outer part with Q again, ok, I got that

then it wanted the potential anywhere. I understand the potential outside the whole thing, the potential between b and c, but when r is from a to b I had trouble. I had the answer beforehand, which is kQ/(rbc)*[bc-r(b-c)] or I may've mixed that up

I see that if you unsimplify that it's like a sum of potentials. It's the potential between a and b, kQ/r, minus the potential AT the surface b(well, the Q is negative there so I guess you can just say plus the potential at b)plus the potential at c

Simple question: Why is it like that? Is there a more systematic way of reaching that answer? oh and of course the answer from a to b is like that one with r=a

It is the magic of 1/r^2 and Gauss and all that. Inside any spherically symmetric charge distribution the electric field is zero. That can be shown by direct integration, or by Gauss's Law. That means it takes zero work to move a test charge from one place to another inside a cavity surrounded by such a charge distributiuon, which means the potential due to that charge distribution is constant in that region.

The potential is additive (superposition). It's usually easier to work from the outside into add things up because the outer charges contribute only a constant potential to the inner regions.