Ground "draining off" charge - Griffiths Problem 2.38 I'm currently reading through Griffiths Electrodynamics 4ed and I'm struggling with problem 2.38c (see attached picture of question and solution).
I get that, using his statement that the potential goes to 0 after grounding, I can show that there must be a charge of -q on the outer shell (using that information) since potential is now 0 all the way to radius a from infinity, requiring E=0 everywhere up to a. Generally speaking, it makes sense that potential goes to 0, but I'm curious as to how we can show or know this. That is, if I had not known a priori that potential was 0 over the surface after grounding, I could not have solved the problem, and yet I'm certain that there must be a way of solving it without using that (extraneous) piece of info.
I'm therefore wondering if anyone can provide me with the intuition for why the surface charge on the shell at b drains off as opposed to any other possibility. It is not immediately clear to me that +q should drain off (though I generally get that this is charging by induction).
Thanks!


 A: In the first case, the shell has no charge.
There is a charge $q$ on the sphere. Hence the flux of the electric field one every closed surface enclosing this sphere, in particular any spherical surface with the same center, will be $q/\epsilon_0$, as long as there is no possibility to have any more charge inside the sphere, that is, in the vacuum between the sphere and the inner surface of the shell.
Within the shell, which is a conductor there can be no  electric field. Thus the charge within any sphere included in the shell must be zero. Therefore there must be a charge $-q$ on this inner surface, so that the global charge included in any sphere within the shell be zero.
But in the first case the shell is neutral, no net charge. If a charge $-q$ is located on its inner surface, a balancing charge $q$ must be located in the only other possible place, its outer surface.
Any sphere in the vacuum outside the shell therefore encloses the $q$ charge on the sphere, $-q$ on the inner surface of the shell and $q$ on the outer surface, (a global zero charge coming from the entire shell which is neutral). Hence the field in the vacuum between the outer surface of the shell and infinity, the same field as if there were no shell, but only the charge $q$ of the sphere.
Now in the second case, you touch the shell with a "grounding wire". For people who work with electricity (like my father, who had a degree an Electrical Engineering and taught me all about it) the word "grounding" is a technical term meaning bringing perforce the potential of the "grounded object" to that of the reference point which, in the case of this problem is infinity. Sinking a wire into very dry earth, as you suggest in your comment to R.W. Bird's answer, is not "grounding". If in your bathroom you try to "ground" your water pipes by connecting them to a metal post just randomly sunk into the earth, you can easily get electrocuted. "Grounding" is serious business.
Thus the shell being "grounded" means that it has the same potential as infinity. Hence there can be no electric field outside it, no electric flux on any sphere enclosing it, and thus the net charge inside any such sphere must be zero. Since there is already a charge $q$ on the sphere and a charge $-q$ on the inner surface of the shell, this shows that there must be no more charge on its outer surface. The charge that was there in the first case has entirely been drained by the "grounding" wire.
Otherwise, the wire would just not deserve the name "grounding".
I hope this is now intuitive enough.
A: How can you define the ground if you don't have any extraneous information? Just as it happens, in finite charge distributions, selecting $|\mathbf r|\to\infty$ as $V=0$ has its perks, such as it being the canonical ground while using $V(\mathbf r)=\sum_{i}\frac{q_i}{4\pi\epsilon_0|\mathbf r-\mathbf r_i|}$, corresponding to there being no charges at all. Any other attempt to define a ground will ignore this fact. In the end calculations would work out by why not just select the standard one?
A: In the absence of a power source, there can be no E field inside of a conductor. (If there were such a field, free charges would move until there was no field.)  This means that every point in the conductor must be at the same potential. The “ground” in an electrostatics problem is defined as having a potential of zero.  Any conductor connected to the “ground” (as by a wire) must come to that zero potential.
