chargeelectricityelectrostaticsgauss-law
A solid metal sphere of radius $R$ has charge $+2Q$. A hollow spherical shell of radius $3R$, concentric with the first sphere, has net charge $-Q$.
What would be the final distribution of the charge if the spheres were joined by a conducting wire?
Answer: $+Q$ on outer shell, $0$ on inner sphere.
Why is that? I would assume that the total charge should just evenly spread between the spheres.
Ok, I think I figured it out. We are so used to assuming that a shell can be collapsed to a point charge, that it is easy to miss the subtlety that this is only true for the regular Coulomb law. As soon as we deform the Coulomb law as given above, we get corrections. If you do all the integrals over the charge distribution you will get the following fields outside and inside of a uniformly charged (total charge $Q$) spherical shell of radius $R$:
$$\vec E=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\left(1-\ln(\sqrt{r^2-R^2})\delta\right)\hat r+O(\delta^2)~~~~~~~~~~~~~~~~~\text{for}~~~r>R$$ $$\vec E=\frac{1}{4\pi\epsilon_0}\frac{-Q~\delta}{r}\left(\frac{1}{R}+\frac{1}{r}\ln\left(\sqrt{\frac{R-r}{R+r}}\right)\right)\hat r+O(\delta^2)~~~~~~\text{for}~~~r<R$$
So not only does the correction depend on the radius of the shell $R$, but there is even a non-vanishing electric field inside the shell. Sure, the dimensionful quantity inside the first ln is weird, but we made no effort to balance units while doing the deformation in the first place, so thats fine. With this the integrals as done in the question above will yield a nontrivial relation between $q_a$ and $q_b$.
As you know by Gauss Law, electric field inside a conductor is zero.Thus in what ever orientation the negative charge is induced in the inner surface of the conductor it creates no electric field through the conductor thus it cannot affect the orientation of the positive charge on the outer surface. This is because a charge can only be moved or displaced by electric field so no inner charge can manipulate the charge distribution on the outer surface. See the image attached below,
Best Answer
Using Gauss' law and the condition that there can be no electric field inside a conductor the initial charge distribution is as follows:
$+2Q$ charge on the outside of the inner sphere.
$-2Q$ charge on the inside of the outer sphere ($-Q$ original charge and $-Q$ induced charge)
There is an electric field between the two sphere and so there must be a potential difference between the two spheres.
Joining the two spheres with a conducting wire means that charges will flow until the potential difference between the two spheres becomes zero ie there is no electric field between the two conductors and inside the conducting wire.
So no charge can reside on the outside of the inner sphere and the inside of the outer sphere and this is achieved by the $-2Q$ on the inside of the outer sphere and the $+2Q$ on the outside of the inner sphere neutralising one another leaving charge $+Q$ on the outside of the outer sphere.