Boundary conditions in Electrostatics If I have a grounded conducting material, then I know that $\phi=0$ inside this material, no matter what the electric configuration in the surrounding will be. 
Now I have a conducting material that is not grounded, then there will be (as long as I am dealing with static problems) no electric field inside this material. Therefore the potential will be constant inside this material, right?


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*Question 1: Therefore, is there any difference in the boundary conditions if I am dealing with a grounded conducting material and an insulator around or a non-charged insulated conducting material and an insulator around? 

*Question 2: Is it possible to get a non-zero potential inside an uncharged insulated conducting material? Especially, would you get a non-zero potential inside a conducting insulated material due to image charges?

*Question 3: Of course, I read a few pages in Jackson's book about this and saw that he substituted the problem of a charged insulated conducting sphere in an external field with the one of having a grounded conducting sphere in the external field that has a charge sitting in the center of the sphere. Then, the magnitude of the extra charge was given by the difference of the initial charge of the sphere minus the induced image charge on the grounded conducting sphere. 
Is it possible to make a general substitution like this: Thereby I mean, that we substitute a charged insulated conducting material carrying a charge by a grounded conducting material that has an additional charge(magnitude given by the difference of total charge-image charge) sitting on its surface? So, I would solve the grounded problem and would add the difference of the total charge-image charge to the surface of the material and add this field to the field calculated for the problem of the grounded material. 
 A: Question 1:
Yes. If the PEC (perfect electric conductor) is grounded it has a fixed potential of zero volts on all points of the boundary. That's a standard Dirichlet boundary conditions in differential equation terminology. The charge $Q$ on the conductor will generally not be zero, but will be given by the Gauss equation:
$$\oint_\Gamma\vec{E}\cdot\vec{\mathrm{d}A}=\frac{Q}{\varepsilon_0}$$
In the other scenario, you do not know the potential on the boundary, and IMHO you can not call it a Dirichlet condition. PEC's are still equipotential, you just don't know what this potential is. Instead, you are given the charge $Q$ (zero in the case of neutrality) and the Gauss equation must be solved as part of the problem to obtain the potential on the boundary.
It's worth noting that since $\vec E = -\nabla\phi$ any constant offset in $\phi$ is irrelevant. It's the electric field that eventually leads to motion of particles in the physical world (through the Lorentz' force), so in the end only relative potentials matter. This means that you can usually set the potential to zero at an arbitrary point of your choice, i.e. you get to choose the ground/reference node.
The insulator is irrelevant. It only changes the surrounding field, not the boundary conditions of the PEC.
Question 2:
Yes. It's the electric field that must be zero within a PEC, which according to $\vec E=-\nabla\phi$ means that the potential must be constant, not necessarily zero. Any practical circuit will have conductors at different voltages. Again, you get to choose ground, but the relative potentials must remain.
I can't quite make sense of how you would involve the image charges in this but I think the answer to that is: no, image charges do not necessarily make the potential non-zero. Since you get to choose the ground you could choose the object under consideration to be ground. In that case it would be zero, but it could still have image charges according to Gauss equation. 
As an end-note: For non-ideal conductors you may have a slight electric field within it, and therefore a varying potential, but I take it that we're discussing ideal conductors here. Most metals have a conductivity in the order of $10^7\,\mathrm{S/m}$ so it's usually a valid approximation.
Question 3:
Without having seen the entire problem myself I assume Jackson decomposed the problem into two simpler problems.
First problem: It's likely simpler to solve the problem with the external charges by fixing the potential of the sphere to for instance zero volts, since this means you have Dirichlet boundary condition. However, by doing so the sphere will have a charge $Q'$ following from the Gauss equation for the determined solution and not what was specified ($Q$).
Second problem: The difference between the original problem and the "first subproblem" is that of a sphere with charge $Q-Q'$ with no external charges. We know that the fields produced by a sphere and point charge of equal charge in the middle of it is the same (this is only valid outside the sphere). This problem has a well known solution.
By linearity of the Poisson equation, the two solutions can be added to give the solution of the original problem (superposition).
A: Question 1:
In general, the stipulation that something is "grounded" does change boundary conditions. In electrostatics, however, I do not think there are major differences between a grounded but still insulated wire and a not-grounded but still electrically neutral and insulated wire.
This can be further compounded by the fact that "grounded" items are sometimes just a matter of perspective. You can choose any item (or place) in a system to be your "ground," but then everything else in the system needs to be adjusted for that.
Question 2
Yes, you can have a non-zero potential inside an uncharged and insulated wire. If you expose a piece of wire (even an insulated one) to an external electric field, you can get charges to move within the wire to create an internal potential difference within the wire. This potential can vary, but it depends on the strength of the electric field. If the electric field is constant, then the potential within the wire will also be constant.
This is due to the electrons moving from one end or side of the wire to the other end or side of the wire, as they follow electric field lines. You should never have to worry about a wire's internal potential varying in electrostatics, but that potential can exist. (A varying potential within a wire would be fair game in electrodynamics, though!)
Is this caused by "image" charges? No, not really. It's just electrons following electric field lines. The method of images is helpful to solving electrostatic problems, but this effect isn't caused by it.
Question 3
Sometimes. Jackson was apparently able to do so in that situation. You can only do that substitution when the surface charge density does not matter. For instance, if you dropped a charge inside a conducting sphere with a charge density on its surface, this substitution does not work.
