Potential of Concentric Shells Say I have a metal sphere of radius $R$, with charge $q$, and it is surrounded by a concentric metal shell, with no net charge, of inner radius $a$ and outer radius $b$.  I ground the outer surface of the concentric shell and lower its potential to 0.  
I don't understand what the grounding has done.  Will the inner region of the shell have a $-q$ amount of charge as induced by the center sphere of charge $q$?
Just to be clear, I understand this problem if the outer surface of the concentric shell is not grounded.  Then the inner region of the shell has $-q$ amount of charge and the outer region has a $q$ amount of charge.  This way they cancel out giving no net charge.  I am having trouble understanding what the grounding does. 
 A: It is sometimes helpful to think of the ground as both


*

*an additional boundary condition (bringing r = infinity to r = a finite value)

*an infinite supply of charges


Uncharged, isolated shell

In the original case, where the shell is uncharged and isolated, you can solve for the charge on the inner surface (-q) and outer surface (+q) using Gauss' law. You don't need to ask what the value of the potential is anywhere in  order to solve this problem.
The electric field is the physical observable in the problem, and in a physical situation the electric field is uniquely defined. It is the electric field that shows up in the equation of motion, in the Lorentz force 
$$m\vec{a} = \vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$$
However, the electric field is a vector field and doing calculations with vectors can sometimes be cumbersome. So, we often instead use the made up quantity potential, which is a scalar field, and it is often easier to manipulate scalar functions. We can construct this made up scalar field by saying that our electric field is just the slope (gradient) of our potential field
$$\vec{E}(\vec{r}) = -\vec{\nabla}V(\vec{r})$$
the minus sign meaning our E vector points 'downhill'. But you can see that our potential is not unique, in fact we can often define V(r) in an infinite number of ways. For example, if we add a constant to V(r), our electric field is unchanged
$$\vec{E}(\vec{r}) = -\vec{\nabla}(V(\vec{r}) + C)$$
because the gradient of a constant is zero. Therefore, we see that the value of the potential is not physical, only gradients of potential are physical.
In this uncharged case, when solving for the electric potential outside of the shell, we can write
$$V(r) = -\int dr\frac{q}{4\pi\epsilon_o r^2} = \frac{q}{4\pi\epsilon_or} + C$$
Here, the value of this constant is the value of the potential at infinity
$$V(\infty) = C$$
Often in these problems, it is our convention to say that that this constant is zero for convenience, so we don't have to remeber whatever arbitrary value we assign it. This is our conventional boundary condition
$$V(\infty) = C = 0$$
Grounded shell

In the grounded shell case, the charge on the inner surface (-q) can be determined from Gauss' law, just like above. It is the the charge on the outer shell that is different.
I think the trick to solving the grounded shell case is to assume that our boundary condition for the potential at infinity is still
$$V(\infty) = 0$$
The role of the ground, then, is to say that we have an additional boundary condition, that the potential in the shell is equal to the potential at infinity, which we arbitrarily assigned the value of
$$V(\infty) = V(shell) = 0$$
And, as we now see, there is no potential difference between the shell and infinity, meaning there is no gradient in potential outside the shell. Therefore, the electric field outside the shell is zero.
Now, the charge on the outer surface (q=0) can be determined from Gauss' law.
Role of the ground
So, you can think of the ground as


*

*Bringing the arbitrary boundary condition at infinity to a finite position is space (this case the location of the shell).

*An infinite source of charge that remains neutral no matter how many charges you give to it or take from it. If the shell was grounded, but there was no charge in the middle, the shell would be uncharged. If you then moved a +q to the center of the shell, negative charges from the ground will feel that positive charge and move to the inner surface of the shell from ground.

A: Consider first the outer shell ungrounded. Then, as you correctly state, a charge of -q is induced on the inner surface of the shell and a compensating +q is induced on the outer surface so that in the end you have an outer electric field corresponding to the inner +q charge. When you ground the outer shell, this means that the shell is at the same potential as the environment, so there can be no outer surface charge and electric field outside the shell. The grounding removes the outer shell surface +q charge to the ground. The inner -q charge of the shell has to remain there in order to prevent the electric field of the +q charged metal sphere to penetrate the outer shell metal. 
