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A metal sphere of radius R, carrying charge q, is surrounded by a thick concentric metal shell (inner radius a, outer radius b). The shell carries no net charge.

(A) Find the surface charge density R, at a, and at b.

(B) Find the potential at the center, using infinity as the reference point.

(C) Now the outer surface is touched to a grounding wire, which lowers its potential to zero (same as infinity). How do your answers (A) and (B) change?

2007-10-20 14:52:45 · 2 answers · asked by Nate-dawg 2 in Science & Mathematics Physics

2 answers

Stephie K did a pretty good job. But let me add a little more

For the metal shell, all of the charge q, at least to an outside observer, will appear as if it were on the outer surface of the shell. If you think about it, the strength of the field at this point is exactly what you would get if the shell was not there, basically q divided by the square of the distance or q/r^2 (where r is the radius of the shell)

What this means is that to balance the charge q on the outer surface, you need a -q charge on the inner surface of the shell. (This is because the shell, en total, carries no charge).

The potential at the center, visa vie infinite, remains the same before it had the shell

When you ground the shell, to the observer outside the shell, there is no charge, which means that the outside surface has no charge density.

But the shell has now aquired a charge of -q to balance the charge of q in the center. This charge is distributed across a, the inner surface, exactly like the example above.

2007-10-20 17:54:05 · answer #1 · answered by Frst Grade Rocks! Ω 7 · 1 0

I am no physicist but I wanted to tackle this anyway. So dont anybody embarass me if what I am saying is stupid Besides, I think that a non-physicist perspective might be more helpful anyhow since a conceptual hurdle... So I looked up some physics online and this is my interpretation...The "surface charge density" is a formula that simplifies the energy being emitted by the sphere down to a unit area. So the total charge 'q' is divided by the total area (A= pi Rsquared) This is actually an average charge expected at any single point on the sphere. That makes this value q a hypothetical "point charge" which you must use in order to calculate the potential. The potential is basically the amount of energy you would predict over a set of points. In order to calculate potential you must first calculate the "electric field" E. The electric field is the way the charge is distributed among points. Its formula is a ratio of avg charge q and the distance between two points. Which for some reason the distance influence is weakened, it has to be divided into a fraction. For this case the points are infinity and center, which is where I get to a conceptual problem myself. I think the center can be considered as -R, or -a, or -b, if you imagine the surface as the starting line. Maybe infinity cancels and you just get radius by radius cubed (look up the formula for electric field at the source I noted below) Question (C) I really dont know except it seems like you should set potential to zero in a formula and then see if the electric field E changes but it feels to me that only the surface loses charge and the energy has to go somewhere else so it could increase elsewhere. OKAY perhaps this is an exercise in futility but I'll be the first to admit I have nothing better to do right now. Good Luck. I am really curious just how wrong I am.

2007-10-20 16:54:36 · answer #2 · answered by stephiek 2 · 1 0

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