How Does Charge Distribution Affect Electric Fields in Concentric Spheres?

[Arisa]

Homework Statement

A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b and outer radius c. The hollow sphere has not net chare. a) Derive expressions for the electric field magnitude in terms of the distance r from the center for the regions r<a, a<r<b, b<r<c, r>c.

Homework Equations

E = [1/(4*pi*epsilon_0)](q/r^2)

The Attempt at a Solution

I've managed to correctly answer the first two parts of the problem, however when it comes to b<r<c and r>c, I do not get the answers I should.
Apparently, for b<r<c, E = 0 since a -q cancels the inner +q. Then, for r>c, E = [1/(4*pi*epsilon_0)](q/r^2) since the total charge enclosed is +q again.

I think my problem lies in the fact that I don't fully comprehend what a concentric sphere is or how charge distribution on a concentric sphere works. Based on the solution, I feel I should intrepret that the neutral concentric sphere is neutral because it contain an equal number of positive and negative charges that have all collected on opposite surfaces - the negative charges on the inner surface of the concentric sphere (radius b) and the positive charges on its outer surface (radius c.) Otherwise, I don't quite understand how the -q and overall +q come into play...

Thank you very much for taking the time to read this!

 

[Astronuc]

This might help - http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

The inner sphere has charge +q. Being a conductor, the charge resides at the surface.

It would have a surface charge density of +q/4$\pi$ a².

This positive charge induces a corresponding -q charge on the inner surface of the hollow conductor (r=b), and consequently there is a +q charge on the outer surface r = c.

The electric flux is based on the enclosed charge, and the +q at r=a cancels the -q charge at r=b.

Then there is a charge +q at r=c.

The surface density of the charge at r = b is -q / 4$\pi$ b², and the surface charge density at r=c is +q / 4$\pi$ c².