Potential for 2D problem
Let's start with a 2D disk and try to solve the general problem for infinitesimally flat disk. I will change notations a bit -- the surface resistance will be $\sigma$ and the radius of the disk will be $a$.
Starting with basic electrodynamics:
$\vec{j} = -\sigma\frac{\partial u}{\partial \vec{r}},\, div\vec{j}=0\,\Rightarrow\,\Delta u = 0$ with the boundary condition: $\vec{n}\cdot\vec{j} = 0 \Rightarrow \vec{n}\frac{\partial u}{\partial \vec{r}} = 0$
Let's first consider the current $I$ flowing into the surface in the centre and uniformly flowing away from the edges. solution for potential is well known:
$U(r,\phi) = -\frac{I}{2\pi\sigma}\, \ln r$
I use the conformal map $z\to a\frac{z-s}{a^2-s*z}$ to "shift the centre" into the point $s=x_{source}+iy_{source}$. The potential is then:
$U(r,\phi) = -\frac{I}{2\pi\sigma}\, \ln\left|a\frac{re^{i\phi}-s}{a^2-s^*re^{i\phi}}\right|$
Now I substract the similar potential, with different parameter $d=x_{drain}+iy_{drain}$ to compensate the outgoing flow. Obtaining:
$U(r,\phi) = -\frac{I}{2\pi\sigma}\, \ln\left|\frac{re^{i\phi}-s}{re^{i\phi}-d}
\cdot\frac{a^2-d^*re^{i\phi}}{a^2-s^*re^{i\phi}}\right|$ or $U(z) = -\frac{I}{2\pi\sigma}\, \ln\left|\frac{z-s}{z-d}
\cdot\frac{a^2-d^*z}{a^2-s^*z}\right|$
This is the harmonic function, satisfying the boundary conditions. You can play here with it.
Interpretation of the solution
The potential is divergent in points $s$ and $d$. This happens because the resistance is strongly dependend on the microscopic details of the problem. Indeed -- as you get closer to the source -- all your current have to pass through smaller and smaller amount of conductor. And in the limit of infinitely small source you get infinite resistance.
Formulation issue
I admit that while solving I first fixed the current and then found the potential, while you formulated the problem differently -- "set the potential here and there and find the current". But let us use logic:
- Nonzero current leads to infinite voltage: $I\neq0\,\Rightarrow\,\Delta U \to \infty$.
- If $A\Rightarrow B$, then $!B\Rightarrow !A$.
- $\Delta U\mbox{-finite}\,\Rightarrow\,I=0$
At finite voltage you'll get zero current or, equivalently, infinite resistance.
What happens in 3D case?
Same thing. Just consider single pointlike source -- and the potential $U\sim\frac{1}{r}$ is divergent. Don't need to go into further details.
"Cutoffs"
In order to move on I introduce the "cutoffs" -- new small (real) quantities $\epsilon_{s,d}$ which denoting "sizes" of the source and the drain. Using them I obtain the voltage:
$U(d+\epsilon_d)-U(s+\epsilon_s)=\frac{I}{2\pi\sigma}\left[\ln\frac{\epsilon_s}{|s-d|}+\ln\frac{\epsilon_d}{|s-d|} +\ln\left|\frac{a^2-s^*d}{a^2-|s|^2}\cdot\frac{a^2-sd^*}{a^2-|d|^2}\right|\right]$
Scales
Putting together everything above. One can say that in the problem there are four (or, even five) scales:
- Radius of the disk.
- Thickness of the disk.
- Distance between contacts $|s-d|$
- Sizes of those contacts $\epsilon_{s,d}$
Since you are talking about "points" -- then first we have to take $\epsilon_{s,d}\to0$, right? But if $\epsilon_{s,d}$ is much smaller that any other scale then they introduce divergent contribution into the resistance. And any other detail of the problem becomes irrelevant.
Therefore, the answer to your question is: The resistance between two points is infinite, whatever the geometry of the problem is.
Best Answer
Nonlocal resistance is the ratio of the current in a material to the voltage between some other two points. It is a much less useful quantity, because this depends on the details of induced changes on the conductor and elsewhere to determine, it isn't something that is determined only by local material quantities.
The only advantage is that you can measure it away from the material, if you can't stick a probe in. It can be measured using the electric field far away from the material. The disadvantage is that you then have to break your head to figure out what is going on inside the conducting material itself, which is what you usually care about.
The ratio of voltage to current, where voltage is between to other points.
In the linear regime, with only materials with linear response and conductors around, the answer is always yes, with one important caveat--- if the resistance is negative (meaning you measured the voltage between two points which for some reason have the opposite voltage than two points at successive position in the wire), it gets more negative as you increase the current, so it technically goes down. The precise statement is that if you multiply the current by a factor of k, you also multiply the additional voltage elsewhere by a factor of k, so it's a linear relationship.