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?

1. Uniformly throughout the shell
2. -Q on the inner surface, -Q on the outer surface, and the rest uniformly throughout
3. -Q on the inner surface, -4Q on the outer surface
4. -Q on the inner surface, -6Q on the outer surface
5. +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 r2

and points toward the center

For 1.5R < r < 2R, E = 0

For R < r < 1.5R, E

=

kQ r2

, directed away from the center

For r < R, E

=

kQr R3

, 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 – r2 R2 )	–kQ 1.5R	–2kQ R	–kQ 6R ( 7 + 3r2 R2 )

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.