Concentric Spheres
Now we'll put the two cases together. The insulating sphere at the center has a charge +Q uniformly distributed over it, and has a radius R. The concentric conducting shell has inner radius 1.5R and outer radius 2R. It has a net charge of 5Q.
What is the electric field as a function of r?
What is the electric potential as a function of r?
To answer the above, first answer the following: How is the 5Q of charge distributed over the conducting shell?
 Uniformly throughout the shell
 Q on the inner surface, Q on the outer surface, and the rest uniformly throughout
 Q on the inner surface, 4Q on the outer surface
 Q on the inner surface, 6Q on the outer surface
 +Q on the inner surface, 6Q on the outer surface
There is Q on the inner surface and 4Q on the outer surface.
Let's try figuring out the field first:
For r > 2R, E 
= 
4kQ
 
r^{2}


and points toward the center 
For R < r < 1.5R, E 
= 
kQ
 
r^{2}


, directed away from the center 
For r < R, E 
= 
kQr
 
R^{3}


, directed away from the center 
To find the total potential we can simply use superposition. The net potential is the potential from the different pieces, the insulator, the Q charge on the inner surface of the conductor, and the 4Q on the outer surface of the conductor. The table below shows the potential from each one of these three pieces in the different regions.
Region  Insulator  Conductor Inner surface  Conductor Outer surface  Net Potential 
r > 2R  +kQ
 
r

 –kQ
 
r

 –4kQ
 
r

 –4kQ
 
r


1.5R < r < 2R  +kQ
 
r

 –kQ
 
r

 –2kQ
 
R

 –2kQ
 
R


R < r < 1.5R  +kQ
 
r

 –kQ
 
1.5R

 –2kQ
 
R

 +kQ
 
r


–  8kQ
 
3R

 
r < R  kQ
 
2R


( 3 –  r^{2}
 
R^{2}

 )   –kQ
 
1.5R

 –2kQ
 
R

 –kQ
 
6R


( 7 +  3r^{2}
 
R^{2}

 )  
While the electric field can be discontinuous (the graph has jumps in it) the potential is continuous. This means the graph of potential is continuous, and that the solutions to the different regions above must agree at the boundaries. For example, if you plug in r = R to the r < R and the R < r < 1.5R solutions, they give the same answer, 5kQ/3R. That's a great way to check your answer.