Ground "draining off" charge - Griffiths Problem 2.38

Question

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!

acarturk · Answer

How can you define the ground if you don't have any extraneous information? Just as it happens, in finite charge distributions, selecting $|mathbf r|toinfty$ as $V=0$ has its perks, such as it being the canonical ground while using $V(mathbf r)=sum_{i}frac{q_i}{4piepsilon_0|mathbf r-mathbf r_i|}$, 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?

Answered by acarturk on December 25, 2020

Ground "draining off" charge - Griffiths Problem 2.38

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