# 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: 9,675 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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 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: 9,675 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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 Quote by Contingency 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?
They can't make the field in the void zero if there are point charges in it. Imagine getting really close to one of those point charges. As you get closer, the potential due to the point charge increases without limit, while that due to the relatively distant conductor is barely changing. So net potential changes, so there is a field.
 P: 41 Gotcha, that makes sense. So there is field. Can you please elaborate more as to why exactly the solution remains the same? Why does this field not affect anything outside? What exactly is going on in this problem?
 P: 41 I just know for certain that the field outside of the conductor behaves like that of a point charge at the origin. I know this must be true irrespective of what's going on in the conductor by guessing a solution to Laplace's Equation with the boundary conditions of a shell-like non-zero equipotential surface, and a zero equipotential surface at infinity. I can't even justify why the point charge at the at the origin should be of same charge q, because I don't know the potential on the conductor.. I'd really appreciate a detailed explanation
 P: 41 The problem is phrased with shells that have zero width, which really made things difficult to understand. I'm picturing them as having width, since zero width doesn't exist in nature. Then I looked at the system without the connection nor the inner conducting shell. In that instance it was clear to me why the same charge of opposite sign moves to the inner boundary of the outer shell - applying gauss's law and requiring there be no field inside the outer shell. Conservation of charge means the exact same charge as the point charge is spread across the outmost boundary of the outer shell. That helped, because the null field inside the outer shell is exactly what prevents the very outmost charges from feeling and force/being aware of electric field from inside. This is why they distribute uniformally and therefore field outside is that of a point charge in the center. Would still very much like to know whether the inner conducting shell affects the problem in any way, and also, from guessing a solution to Laplace's Equation, how could one determine that the magnitude of the virtual point charge creating the field is the same of the original point charge in the problem.. Thanks
 HW Helper Thanks P: 9,675 Maybe we do not understand the same thing on the "inside of the outer shell". The shall has a wall of some thickness, although it is negligible compared to the radius. The electric field is zero inside the wall, but there is an electric field around the charge in the void between the shells. Think to make a Gaussian surface around the charge Q. The surface integral for this surface can not be zero as it encloses charge. At the same time, the electric field is zero inside the inner shell, as it does not contain any charge. The electric field lines emerge from charges and end in charges or at infinity. There must be some surface charge on the inner surface of the big shell and also on the outer surface of the small shell. The shells are connected, they are at the same potential, and the charges can move from one shell to the other. The charge Q induces -Q surface charge on the conductors at the interface between conductor and void. The conductors are electrically neutral, so Q charge has to appear at the outer surface. (As the electric field is zero inside the small shell, there is no surface charge on its inner surface ) ehild Attached Thumbnails
 P: 41 Thank you very much! That was really clear. I came pretty close in my last post as you see.. Definitely not as clear cut though. A question remains - if I were just guessing a solution to Laplace's Equation - how would I know what magnitude to assign the charge at the origin? I'm asking because whenever there's a problem area whose outer boundary is a spherically symmetric conductor, a point charge in its center solves Laplace's Equation for the area outside the conductor.. I just need to know what magnitude of charge to assign it..
 HW Helper Thanks P: 9,675 You want the solution of the Laplace equation in the space between a spherical surface and infinity, with the boundary condition that the potential is constant on the spherical surface and zero at infinity. The solution of the Laplace equation with given boundary condition is unique. If you find a solution, it is the solution. The potential around a single conducting sphere is spherically symmetric. If the spherical surface encloses Q charge, the electric field is kQ/r^2 according to Gauss' Law, and the potential is kQ/r at distance r from the centre if r>R. In your problem, the outer spherical surface encloses the charge Q placed into the void. So the potential is kQ/r for r>R. To find the electric field and potential inside the cavity would be more difficult, but it was not the question. ehild

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