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!
 A: 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 $q_b$.
A: Came across this problem in Smythe, Static and Dynamic Electricity, phrased slightly differently but with the same idea:

Problem 1.14. If two charged concentric shells are connected by a wire, the inner one is wholly discharged. If the force law were $r^{-(2+p)}$, prove that there would be a charge $B$ on the inner shell such that if $A$ were the charge on the outer shell and $f,g$ the sum and difference of the radii $2gB=-Ap[(f-g)\,\ln(f+g)-f\ln f + g\ln g]$, approximately.

This problem took me several days to figure out, and I couldn't find any readable solutions online, so I wanted to share mine.
There are different approaches, including a modification of Gauss's law (determine enclosed charge by electric flux, with corrections) and direct calculation by using the electric field (which must be zero between the shells) or the electric potential (which must have zero potential difference from one shell to the other). I think the easiest approach is to calculate the electric potential, but it's also useful to review why Gauss's law requires modification, as you did in your answer.
Modified Electric Potential. First, we need to find the expression for electric potential by the modified Coulomb's law. For this, we calculate the work performed bringing a test charge from infinitely far away; the modified electric field for a central charge $q_0$ is
$$
4\pi\epsilon E(r) = \frac{q_0}{r^{2+\delta}}
$$
and the work done per unit charge becomes
$$
4\pi\epsilon V(r) = 4\pi\epsilon \int_r^\infty E(r)\, dr
$$
thus
$$
4\pi\epsilon V(r) = \frac{q_0}{\delta+1} \frac{1}{r^{1+p}}
$$
so that the modified potential of an infinitesimal charge density $\rho\,dV$ is
$$
4\pi\epsilon V(\mathbf r) = \frac{1}{\delta+1} \int \frac{\rho(\mathbf r')}{|\mathbf r - \mathbf r'|^{1+\delta}}\,d^3 r'.
$$
Nothing is lost if we ignore the constants (we're only going to be looking at potential differences), so for the rest of the problem, I'll write this in the form
$$
V(\mathbf r) = \int \frac{\rho(\mathbf r')}{|\mathbf r - \mathbf r'|^{1+\delta}}\,d^3 r'.
$$
Modified Potential of a Charged Spherical Shell. Now that we've verified the expression for modified electric potential in general, we can calculate the potential of a single charged spherical shell of radius $r_1$ and charge $q_b$. By taking the surface integral over the shell for a given observation point at radial distance $r$, we find
$$
V(r) = \frac{q_b}{2} \left[ \frac{(r^{2}+r_{1}^{2}-2rr_{1}u)^{(1-\delta)/2}}{rr_{1}(\delta-1)} \right|^{1}_{-1}=\frac{q_b}{2rr_{1}(\delta-1)} \left[  [(r-r_{1})^{2}]^{(1-\delta)/2}-[(r+r_{1})^{2}]^{(1-\delta)/2} \right]
$$
which comes to
$$
V(r) = \begin{cases}
\frac{1}{\delta-1} \frac{q_b}{2rr_{1}} \left( \frac{r-r_{1}}{(r-r_{1})^{\delta}} - \frac{r+r_{1}}{(r+r_{1})^{\delta}} \right)  & r>r_{1} \\
\frac{1}{\delta-1} \frac{q_b}{2rr_{1}} \left( \frac{r_{1}-r}{(r_{1}-r)^{\delta}} - \frac{r_{1}+r}{(r_{1}+r)^{\delta}} \right)  & r<r_{1}.
\end{cases}
$$
Comparing this with the expression for the inverse-square law form, which would be
$$
\begin{cases}
\frac{q_r}{r} & r>r_{1} \\
\frac{q_r}{r_{1}} & r<r_{1},
\end{cases}
$$
we see that whereas with an inverse-square law the potential within a spherical shell is constant, with the modified law this is no longer the case due to the $(r_1 \pm r)^{-\delta}$ terms. When $\delta=0$, we recover the inverse-square expressions.
Concentric Spherical Shells Connected by a Wire. The condition for equilibrium when the two spherical shells are connected is that the potential difference from one shell to another is zero. Let the inner sphere have radius $r_1$ and charge $q_b$, and the outer sphere have radius $r_2$ and charge $q_a$. To calculate the total potential difference, we will first determine the potential due to the inner sphere and then the contribution of the outer sphere. For the first, we have
$$
V_{b}(r_{2})-V_{b}(r_{1}) = \frac{q_b}{2r_{1}r_{2}}\left( \frac{r_{2}-r_{1}}{(r_{2}-r_{1})^{\delta}} - \frac{r_{2}+r_{1}}{(r_{2}+r_{1})^{\delta}} \right) - \frac{q_b}{2r_{1}^{2}}\left( \frac{r_{1}-r_{1}}{(r_{1}-r_{1})^{\delta}} - \frac{2r_{1}}{(2r_{1})^{\delta}} \right) 
$$
where the right-most term has an indeterminate of the form $\frac{0}{0}$, but we can show that this has a limit of $0$. Thus we find
$$
\Delta V_{b} = \frac{q_b}{2r_{1}r_{2}}\left( \frac{r_{2}-r_{1}}{(r_{2}-r_{1})^{\delta}} - \frac{r_{2}+r_{1}}{(r_{2}+r_{1})^{\delta}} \right) + \frac{q_b}{r_{1}} \frac{1}{(2r_{1})^{\delta}}
$$
or with $f=r_{1}+r_{2}$ and $g=r_{2}-r_{1}$,
$$
\Delta V_{b} = \frac{q_b}{2r_{1}r_{2}}\left( \frac{g}{g^{\delta}} - \frac{f}{f^{\delta}} \right) + \frac{2q_b}{f-g} \frac{1}{(f-g)^{\delta}}.
$$
We can rewrite this in a more convenient form by making further use of the relationships $2r_{1} = f+g$ and $2r_{2} = f-g$ giving
$$
\frac{q_b}{(f+g)r_{2}} \left( g g^{-\delta} - f f^{-\delta} + (f+g) (f-g)^{-\delta} \right) .
$$
By a similar procedure, we arrive at the potential difference due to the outer sphere,
$$
\Delta V_{a} = - \frac{q_a}{2r_{1}r_{2}} \left( \frac{g}{g^{\delta}}-\frac{f}{f^{\delta}} \right)-\frac{2q_a}{f+g} \frac{1}{(f+g)^{\delta}} 
$$
or rearranged,
$$
\Delta V_{a} = - \frac{q_a}{(f+g)r_{2}} \left( g g^{-\delta}-f f^{-\delta}+(f-g) (f+g)^{-\delta} \right).
$$
For equilibrium, these must be summed and set equal to zero, then solved for $B$ in terms of $A$.
Approximation to $O(\delta^2)$. To this point the analysis has been exact. To continue, we will use approximate methods by using the expansion with respect to $\delta$ of
$$
x^{-\delta} = 1- \delta \ln x + O(\delta^2).
$$
This can be derived from the Taylor series for the exponential $e^\delta$.
Substituting this in for the potential difference due to the outer sphere, we have
$$
\begin{align}
-\frac{q_a\delta }{(f+g)r_{2}}&\left[ \frac{g-f+f-g}{\delta}-g\ln g+f\ln f-(f-g) \ln(f+g) \right] \\
= & -\frac{q_a\delta}{(f+g)r_{2}} [f \ln f - g \ln g - (f-g) \ln (f+g)]
\end{align}
$$
and we can perform a similar substitution for the inner sphere. However, for the problem as written in e.g. Smythe, we will actually approximate the inner sphere's contribution to $O(\delta)$ by setting $\delta=0$, to give
$$
\frac{q_b}{(f+g)r_{2}}(g-f+f+g) = 2g \frac{q_b}{(f+g)r_{2}}
$$
which allows us to solve the equilibrium condition,
$$
2g \frac{q_b}{(f+g)r_{2}} - \frac{q_a\delta}{(f+g)r_{2}}[f \ln f - g \ln g - (f-g) \ln (f+g)]=0
$$
for an approximate residual charge on the inner sphere of
$$
q_b = -\frac{q_a\delta}{2g}[(f-g) \ln (f+g)-f \ln f+g \ln g].
$$
This form agrees with Smythe's expression, but other orders of approximation are easy to obtain.
Conclusion. I left out a lot of the grunt work in the above derivation. There are many decisions necessary throughout in order to get the right result, e.g. how to write an expression, where to introduce approximation, what approach to take in general. I think the problem itself is very difficult, but it's also very practical (testing Coulomb's law experimentally by Cavendish's method would require something like the above analysis before results could be interpreted) and it forces you to think carefully about how to approach a problem, what assumptions you're making when solving problems, and how you can introduce approximate methods.
