|Dec28-12, 10:26 PM||#18|
Charge Between Two Conducting, Connected Shells
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.
|Dec29-12, 05:45 PM||#19|
I found this thread rather helpful in addressing a similar difficulty I had. Can someone verify I understand this problem:
Given a system of two concentric spherical shells, these are the areas I can divide it into:
1) Outside both shells.
2) The outer layer of the outer shell
3) The inner layer of the outer shell (or equivalently, the thickness below the outer layer if thinking microscopically)
4) The void between the shells.
5) The outer layer of the inner shell
6) The inner layer of the inner shell
7) The area below the inner layer of the inner shell.
There exists a point charge Q at 1.5R.
So, this is what happens:
1) For r > 2R, the electric field behaves as a point charge Q at the center. This is because the electric field from the point charge, even though its spherically asymmetric, is canceled out by the inner surface of the outer layer. Therefore, the Gaussian surface I construct has to only worry about the surface charge and my E dot dA integral reduces to a constant.
2) The outer surface of outer layer has a charge Q distributed over it, spherically symmetric.
3) ...because the inner surface of the outer layer has charge -Q distributed so as to cancel out the electric field lines (within the conductor itself) from point charge at 1.5R as this is the behavior of any conductor.
4) In the void, I'm not sure exactly what's happening because the point charge isn't exactly spherically symmetric but I know there's some sort of electric field. In fact, if I were to just consider equipotential surfaces for the point charge without the influence of anything else, there are some surfaces that actually intersect the outside conductor at some points but don't at others. Seems really weird but is my interpretation correct?
5) So at this point... does the outer surface of the inner shell also have a charge of -Q distributed over it to cancel out the electric field lines such that the conductor's thickness has an electric field of zero (and the inner surface gains charge +Q)? It seems strange that there is -Q distributed in two places to cancel out the electric field from the point charge.
6) Applying Gauss law now on the inside, the Electric field is zero.
However, I'm slightly confused as to the case in which the a cavity DOES have an electric field of zero. So electrostatic shielding only applies when:
1) There are no point charges in conductor cavities
2) Your within the conductor itself when point charges are present in cavities.
Also, given the spherical asymmetric of the point charge, why can I assume the charge Q is distributed uniformly? Again, is this because the electric field is cancelled out by the inner surface so the outer surface uniformly distributes itself?
|Dec30-12, 12:10 AM||#20|
If the shells are insulated I think, there would be -Q charge induced on the inner surface of the outer shell and no induced charge on the inner shell.
You get the electric field by solving Laplace equation and appliying the boundary conditions. But sometimes you can calculate it by assuming charged surfaces substituting metal plates/shells. Do not forget that the electric field lines can not cross really the metal layer. The electrons on the outer surface do not feel the inner charge.
|Dec30-12, 03:37 AM||#21|
Thank you very much, ehild
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