How do charges behave in a spherical conducting shell?

[FrogPad]

I never really understood this...

Lets say we have a positive point charge Q at the center of a spherical conducting shell of an inner radius Ri and an outer radius R0.

Now my books says that an amount of negative charge equal to -Q must be induced on the inner shell surface (R=Ri), and that an amount of positve .
charge +Q is induced on the outer shell surface (R=R0).

Why is this?

What would happen if a negative point charge was placed in the center?
I would assume that the inner shell would have an induced positve charge, and the outer shell would have an induced negative charge. Again, why is this?

Also what would happen if there was no charge at the center, and instead a charge was placed on the outside of the shell near the conducting surface?

 

[cyberfrenzy]

The main reason for this is because an isolated conductor cannot cannot have electric field lines within them. A simple application of Gauss' law ( reread that should give you all the answers.

Cheers
Nithin

 

[FrogPad]

What I wasn't understanding is why there is an induced charge on the surface of the shells. After thinking about it some more, I think I have I remember why it happens. Since the outer electrons are free to dissociate themselves they are effected by an electric field. So when we place the +Q point charge in the center, all the free negative charge is pulled towards the +Q. On the other hand the positve charge in the shell moves as far away from the +Q as possible, this happens to be the bounds of the shell.

A similar argument can be made for placing a negative charge inside the shell, or any charge outside of the shell.

I guess my problem is visualizing the symmetry. Yes, I can visualize the charge arrangment in this case. But, what happens if we place the near a conducting wall that is not centered (R \new 0). From my argument above I would conclude that negative charge would get as close as possible to the +Q point charge, and positve charge would form on the surface. However this would produce a non-uniform electric field around the conducting shell, and we would need some more information, because the charges on the conducting shell are not cancelling nicely.

Is this logic correct?