# Two concentric conducting shells connected by wire with modified Coulomb law?

This question is related to this one, but is a bit different. Consider two concentric and conducting spherical shells connected by a wire. Inner shell has radius $b$ and (unknown) charge $q_b$ while outer shell has radius $a$ and (known) charge $q_a$. Imagine that the electric field of a point charge is given by

$$\vec E=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^{2+\delta}}{\hat r}$$

Find $q_b$ neglecting terms of order $O(\delta^2)$ and higher.

One can show that $\nabla\times\vec E=0$, so that we can imagine $\vec E$ to come from a potential $\vec E=-\nabla\phi$. Considering that the wire is connecting the two shells, the charge will balance in some way such that no current will be flowing. This means that the potential at $a$ has to be equal to the potential at $b$. Integrating from infinity:

$$\int_\infty^a E~dr=\int_\infty^b E~dr$$

$$\int_\infty^a \frac{q_a+q_b}{r^{2+\delta}} dr=\int_\infty^a \frac{q_a}{r^{2+\delta}} dr+\int_\infty^b \frac{q_b}{r^{2+\delta}} dr$$

$$0=q_b\underbrace{\int_a^b \frac{1}{r^{2+\delta}} dr}_{\neq0}$$

Therefore, we clearly get $q_b=0$, which is consistent with the familiar result when $\delta=0$ exactly. However, since we are asked to compute corrections up to linear order in $\delta$, I wonder if I missed some subtlety, which would produce a non-zero answer. Can someone clarify? Thanks for any suggestion!

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 $q_b$.