How would charge be distributed in charged conductors if the Coulomb law was not ${1}/{r^2}$?

Question

Would the excess charge on a conductor move to surface until the electric field inside become zero if the Coulomb law was for example $\frac{1}{r^3}$? If yes, would the distribution $\sigma(x,y)$ be different from when it is $\frac{1}{r^2}$?

Answer 1

James Clerk Maxwell thought about this one and showed the following. Suppose we have two concentric conducting spheres and we charge one up to a potential $\Phi$ relative to some grounding plane. Then the voltage of the inner sphere relative to the same ground is:

$$\Phi_{inner} = \Phi \,q\, \left(\frac{\rho}{2}\log\left(\frac{\rho+1}{\rho-1}\right)-\frac{1}{2}\log\left(\frac{4\,\rho^2}{\rho^2-1}\right)\right)\quad\quad\quad(1)$$

where $\rho = r_{outer} / r_{inner}$ is the ratio of the radiuses of the outer to inner spheres and $q$ is the deviation between the power of $r$ in the Coulomb law and 2. Thus, the radial dependence in Coulomb's law is $r^{-(2\pm q)}$; if there is exactly inverse square dependence, then $q=0$. This fact has been used to test the Coulomb law to high precision, see:

Plimpton, S. J.; Lawton, W. E., "A Very Accurate Test of Coulomb's Law of Force Between Charges", Physical Review, vol. 50 (1936), Issue 11, pp. 1066-1071

If the photon has a mass $m$, the Coulomb $1/r$ potential generalises to a Yukawa potential:

$$\Phi = -\frac{q}{4\,\pi\,\epsilon_0} \frac{\exp\left(-\frac{m\,c}{\hbar}\,r\right)}{r}\quad\quad\quad(2)$$

and so the experiment described can be used to bound the photon mass. According to Wikipedia (see the "Experimental checks on photon mass" on the "Photon" page), this bound is $10^{-14}\mathrm{eV}/c^2$, or about $1.6\times10^{-50}\mathrm{kg}$, i.e. about $10^{-20}$ electron masses. So now I'd like to show how to relate the Coulomb-Yukawa potential and the photon mass, and show how to interpret the experimental null result. A good review paper (at least it was clear to me) here is:

Liang-Cheng Tu and Jun Luo, "Experimental tests of Coulomb’s Law and the photon rest mass", Metrologia 41 (2004) pp136–146

It is much easier, and equivalent, to talk about this kind of thing in terms of potentials rather than forces (assuming we have irrotational forces). Also, the following discussion in terms of photon mass is actually a much simpler framework to talk about straight deviations from the postulated $1/r$ Coulomb potential than Maxwell's (incidentally, the Maxwell expression (1) is derived in the review paper as well). Instead of talking about a deviation $q$ of the power $1/r^{1\pm q}$ in the Coulomb potential law from its postulated power as Maxwell does, we speak of a multiplicative error factor $f(r) \approx 1+\epsilon_1\,r\approx e^{\epsilon\,r}$ (the approximation holding for $r \ll 1/\epsilon$) so that we assume our actual potential law is $e^{\epsilon\,r}/r$ rather than $1/r$.

The photon mass would make itself felt by changing the propagation equation for the electromagnetic potentials from the massless wave equation to the Maxwell-Proca equations (see the Wikipedia page for "Proca Action" ):

$$\nabla^2 A_\mu - \frac{1}{c^2}\,\partial_t^2 A_\mu - \left(\frac{m\,c}{\hbar}\right)^2\,A_\mu = -\mu_0 J_\mu\quad\quad\quad(3)$$

where $J_\mu$ is the four-current source for the field. To understand that the scaling constant $m^2\,c^2/\hbar^2$ in the new term $m^2\,c^2\,A_\mu /\hbar^2$ has the interpretation of being a mass, we can:

1. Make the observation that in freespace $\hbar^2 \nabla^2 - \hbar^2 \partial_t^2/c^2$ is the operator (quantum observable) equivalent to the squared length of the four-momentum $E^2 /c^2 - |\vec{p}|^2$, which is the proper (rest mass) term $m^2 c^2$; or

2. Think of the solution to freespace version of (3) ($J_\mu = 0$) as a Fourier decomposition into plane waves: plane wave (wavenumber $k$), time harmonic (frequency $\omega$) solutions of (3) are defined by

$$\omega = \pm \sqrt{k^2 +\frac{m^2\,c^2}{\hbar^2}}\,c\quad\quad\quad(4)$$

• which is, of course $\omega = c\,|k|$ when the mass is nought, whence the group speed equals the phase speed equals $c$. But with nonzero $m$, the group speed for low frequencies $k \ll m\,c/\hbar$ is nought, and a wavepacket can stand roughly still for a time that is proportional to the $m$ term.

So now we look at the static situation ($\partial_t = 0$) for the electrostatic charge so that (3) becomes

$$\left(\nabla^2 - \left(\frac{m\,c}{\hbar}\right)^2\right)\,\Phi = \frac{\rho}{\epsilon_0}\quad\quad\quad(5)$$

and the Yukawa potential (2) is the relevant Green's function for this equation, i.e. the solution to

$$\left(\nabla^2 - \left(\frac{m\,c}{\hbar}\right)^2\right)\,\Phi = \delta(\vec{r})\quad\quad\quad(6)$$

whence we can build up fields arising from general charge distributions $\rho$ by linear superposition:

$$\Phi(\vec{r}) = -\frac{1}{4\,\pi\,\epsilon_0} \int_V \rho(\vec{r}^\prime)\frac{\exp\left(-\frac{m\,c}{\hbar}\,|\vec{r} -\vec{r}^\prime|\right)}{|\vec{r} -\vec{r}^\prime|}\,\mathrm{d}V^\prime\quad\quad\quad(7)$$

Note that the static field in freespace away from the charge for any distrubution of charges each having the Yukawa potential (2) still fulfills the freespace equation

$$\left(\nabla^2 - \left(\frac{m\,c}{\hbar}\right)^2\right)\,\Phi = 0\quad\quad\quad(8)$$

by linear superposition: $\nabla^2$ is unaffected by either a shift in the origin or a rotation of the relevant co-ordinate system. Notice that we could not say the same thing if, say, we had $\Phi\propto 1/r^n$ for $n\neq 1$, because then the relevant differential equation would be:

$$\nabla^2 \Phi -\frac{n\,(n-1)}{r^2} \Phi = 0\quad\quad\quad(9)$$

and the factor $n\,(n-1)/r^2$ most certainly changes its form in response to shifts in the origin. The Coulomb and Yukawa potentials are special insofar that they are the Green's function of constant co-efficient, linear partial differential equations.

Now we look at a hollow conductor. The first thing to note here is that the uniqueness theorems for Laplace's equaton and the static Maxwell-Proca potential equation work in exactly the same way. If we know the potential on the boundary $\partial V$ of a volume $V$, then if there are no singlarities in $V$, we suppose there were two real valued solutions $\phi_1$ and $\phi_2$ with the same behaviour on $\partial V$ and we apply Gauss's divergence theorem to $\psi\,\nabla \phi$ where $\phi = \phi_1-\phi_2$ (noting $\phi$ on $\partial V$ is nought by assumption):

$$0 = \oint_{\partial V} \phi\,\nabla\phi\cdot \hat{\vec{n}} \mathrm{d}S = \int_V\left( |\nabla \phi|^2 + \phi \nabla^2 \phi\right)\,\mathrm{d} V = \int_V \left(|\nabla \phi|^2 + \frac{m^2\,c^2}{\hbar^2}|\phi|^2\right)\,\mathrm{d} V\quad\quad\quad(10)$$

so that $\phi$ must be nought throughout $V$ since the integrand on the RHS is positive or nought, i.e. we have proven uniqueness given we can find a solution in the first place. For a perfect conductor, any charges inside will shift until there is no force tangential to the conductor's surface ($i.e.$ the charges move feely until tethered by the surface), for otherwise they could further re-arrange themselves (by moving along the surface). So the electric field is always orthogonal to a conductor's surface - this fact is independent of the form of the Coulomb law. So the inner surface of any hollow conductor is always an equipotential surface, independent of the form of the elctrostatic force law (as long as the force from a lone charge is radially directed towards or away from the charge). Now, in the case of the $\Phi = 1/r$ potential, if the potential on the hollow inside surface is $\Phi_0$, then a constant potential of $\Phi_0$ is a solution of Laplace's equation and, by the foregoing discussion, it is the only solution fulfilling our boundary conditions. So $\nabla \Phi = 0$ and there is no electric field inside the conductor.

So now we do the same for the static Maxwell-Proca potential. We consider a hollow sphere of radius $R$ and we charge it up to a monstrous voltage $\Phi_0$. Then a nonsingular, axissymetric solution to (8) inside the hollow is:

$$\Phi(r) = \Phi_0\,\frac{R_0\,\sinh\left(\frac{m\,c}{\hbar} r\right)}{r\,\sinh\left(\frac{m\,c}{\hbar} R_0\right)}\quad\quad\quad(11)$$

and, by the foregoing, this must be the only solution. Notice that, as an aside, the solutions to this problem are the Helmholtz equation solutions, to wit, spherical Bessel functions, but for imaginary wavenumbers, since the static Maxwell-Proca potential equation is the Helmholtz equation with an imaginary $k$. The electric field inside our sphere is:

$$\vec{E}= R_0\,\Phi_0\,\frac{\sinh\left(\frac{m\,c}{\hbar} r\right) - \frac{m\,c\,r}{\hbar} \cosh\left(\frac{m\,c}{\hbar} r\right)}{r^3\,\sinh\left(\frac{m\,c}{\hbar} R_0\right)}\vec{r}\approx \frac{m^2 c^2\,\Phi_0}{3\,\hbar^2}\,\vec{r}\quad\quad\quad(12)$$

so suppose we charge a one metre radius sphere to a million volts and measure no electric field with a probe just inside the sphere, accurate to within, say, 100 volts per metre. Then the experiment has yielded an upper bound on the photon mass of:

$$m < \sqrt{\frac{3\times 100\mathrm{V\,m^{-1}}}{10^6\mathrm{V}\times1\mathrm{m}}} \times \frac{\hbar}{c} = 6\times 10^{-45}\mathrm{kg}$$

Notice also that, by the uniqueness theorem we looked at above, the experimental result does not hinge on the sphere's being exactly spherical. We can numerically solve the Maxwell-Proca potential equation for distorted spheres and thus test the sensitivity of our experiment to such distortions.

The above figures represent a very crude and easy experiment in a modern high voltage laboratory. As noted in Wikipedia, the actual photon mass bound acheived by this experiment is about six orders of magnitude smaller than this ($1.6\times10^{-50}\mathrm{kg}$), the photon mass bound achievable by any current method (observation of the galactic plasma) is about thirteen orders of magnitude smaller again ($10^{-63}\mathrm{kg}$) and lastly, as noted in the Liang-Cheng Tu and Jun Luo paper, for the present universe the maximum achievable accuracy for measuring the energy (mass) of something can be reckonned with the Heisenberg inequality $\Delta E \Delta t \geq \frac{\hbar}{2}$ with $\Delta t$ set to the age of the universe ($4\times10^{17}$ seconds), so the minumum achievable mass bound is $\hbar/(2\,c^2\,\Delta t)\approx 10^{-69}\mathrm{kg}$.

Answer 2

Would the excess charge on a conductor move to surface untill the electric field inside become zero [...]?

Think about the mechanism of that motion for a moment. The charges moves because

1. they are free (not bound)
2. there exists a non-zero fields, so from $\vec{F}_E = q \vec{E}$ a force on them

Those two facts are independent of the exact form of the Coulomb interactions. So the short answer is "Yes."

If yes will the distribution $\sigma(x,y)$ be different [...] ?

Sure. WetSavannaAnimal pointed you in the direction of the closed form solution, but it should be intuitively clear that to get the same condition (no field interior to the conductor) with a different form for the field the charge distribution must be different.

Not that the charge distribution may not be strictly a surface distribution at all, and should be written $\rho(\vec{r})$.

Answer 3

Suggestion to the question (v3): Generalize the question to a $1/r^s$ potential law in $n$ spatial dimensions! Then according to Henry Cohn's mathoverflow answer here, the charges rush to the boundary iff $s\leq n-2$. So in OP's example $(s=2,n=3)$, the charges don't rush to the boundary, in contrast to the real world $(s=1,n=3)$.