# Charge Between Two Conducting, Connected Shells

by Contingency
Tags: conductors, field
 P: 41 1. The problem statement, all variables and given/known data On the XY plane there exist two concentric, thin, conducting shells, centered at the origin; one is of radius R, the other of 2R. They are connected. There exists a point charge at (0, 1.5R). What is the potential at some point outside the outer ring? 2. Relevant equations I don't really understand what's going on here 3. The attempt at a solution Tried method of image charges for outer sphere, and inverse for inner sphere... didn't get anywhere. I'm trying to understand the general behavior of charges inside a void inside a conductor... I was told that given any conductor, with a void inside of it containing charge, the electric field outside of it will be the same irrespective of how you move the charge around.. I'd really appreciate an explanation!
 HW Helper Thanks P: 9,313 The question is about the electric field outside the rings. The charge in the void attracts opposite charges to the inner surfaces of the conducting shells, which result in the same surface charge on the outer surface. But the electric field in the void does not penetrate through the conducting wall of the outer shell. The charges on the outer surface "do not know" about the inner electric field. The outer electric field is the same as that of a charged sphere of radius 2R. There is also a law of Electrostatics that you can fill a closed equipotential surface with metal, without changing the field outside. The two shells are connected, so they are at the same potential. You can fill the void with metal. What happens with the charge inside? ehild
 P: 41 How is it that the charge attracts opposite charges to the inner surfaces of the conductors, if it is outside of one of them?
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## Charge Between Two Conducting, Connected Shells

Just for clarification, are we dealing with "rings" or "spheres" (or perhaps even cylinders perpendicular to the xy plane)?
 P: 41 Spherical shells, or, just for 2D, rings. Both very thin
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P: 9,313
 Quote by Contingency How is it that the charge attracts opposite charges to the inner surfaces of the conductors, if it is outside of one of them?
I meant "inner" with respect to the void.

ehild
 P: 41 I got it. It's essentially no different from a charge in a void of a conductor. There can be no field between the shells, so there is a total charge distribution equal to the point charge inside on the outer shell. Using the '4πσ argument', the distribution on the outer shell must be uniform since there are no charges outside. Therefore the field outside is like that of a point charge in the center of the shells.
 HW Helper Thanks P: 9,313 There is field between the shells in the void, but that does not influence the outer field. Yes, the field outside is the same as of a point charge in the centre. ehild
 P: 41 There is field? I don't understand why..
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 Quote by Contingency There is field? I don't understand why..
The charge on the point charge will be opposite in sign to that on the nearest parts of the conductor, so not only is there a field in the cavity, it's stronger than with the point charge alone.
 P: 41 But the shells are connected, don't they become one conductor and make the field in the void between them zero? Could you explain this in more depth please?
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