How Do Charges Distribute Between Concentric Spheres When Connected by a Wire?

[GwtBc]

Homework Statement

A metal sphere with charge q=3.7 μC and radius r=3.1 cm is concentric with a larger metal sphere with charge Q=13 μC and radius R=5.6 cm. (a) What is the magnitude of the potential difference between the spheres? If we connect the spheres with a wire, what then is the charge on (b) the smaller sphere and (c) the larger sphere?

Homework Equations

[/B]

The Attempt at a Solution

I don't know what the answer is, but if I find an expression for the electric field outside the larger conductor using Gauss' law and then integrate that from infinity to 0.056, then find the potential at the surface of the second conductor by simply using q/4*pi*epsilonnaught*r I should have the correct potentials at both surfaces, and for the second part the constraints are that the electric field between the two conductors is zero and q+Q= 16.7e-6 C?

edit: I just realized that according to Gauss' law, all the charge must move on to the larger sphere once they two are connected (since E_inside = 0). Not sure if I can make sense of this result, could someone please explain?

 

[Simon Bridge]

The potential at the surface of a charged conducting sphere is the same as if all the charge were confined at the center.
When the spheres are connected, remember that like charges repel... so they will try to get as far apart as possible.

 

[GwtBc]

The potential at the surface of a charged conducting sphere is the same as if all the charge were confined at the center.
When the spheres are connected, remember that like charges repel... so they will try to get as far apart as possible.

Not sure where you're going with the second point.

 

[Simon Bridge]

The second point is in response to this:

I just realized that according to Gauss' law, all the charge must move on to the larger sphere once they two are connected (since E_inside = 0). Not sure if I can make sense of this result, could someone please explain?