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.
It depends on the internal resistance of the source.
Fist consider a "voltage supply".
What does "voltage supply" even mean?
A voltage supply is supposed to output a fixed voltage no matter what we connect it to.
Is this even possible?
Suppose we connect the two terminals of the voltage supply together through a piece of wire, i.e. a really low resistance load resistor $R_L$.
The current output should be $V/R_L$.
If this is a 9 Volt battery and my resistor is 0.1 $\Omega$ then I'd have a current of 90 Amps.
A 9 Volt battery most certainly cannot output 90 Amps.
The way to model the limitation in the battery's maximum output current is to imagine that it is an ideal voltage supply in series with an internal resistance $R_i$.
Now when we connect it to a load resistor $R_L$, the total current is $I=V/(R_i + R_L)$.
If $R_L\rightarrow 0$ then the current goes to $I_{\text{max}}=V/R_i$, a finite value.
In other words, the internal resistance sets a maximum output current.
Since the total current is $I=V/(R_i + R_L)$, the total power dissipated in the load resistor is
$$
P_L = I^2 R_L = V^2 \frac{R_L}{\left(R_i + R_L\right)^2}.
$$
In order to find the value $R_L^*$ for which the power is maximized, differentiate with respect to $R_L$ and set that derivative equal to 0:
$$
\begin{align}
\frac{dP_L}{dR_L} &= V^2 \frac{(R_i + R_L)^2 - 2 R_L(R_i + R_L)}{(R_i + R_L)^4}\\
0 &= V^2 \frac{(R_i + R_L^*)^2 - 2 R_L^*(R_i + R_L^*)}{(R_i + R_L^*)^4} \\
2 R_L^* &= R_i + R_L^* \\
R_L^* &= R_i \, .
\end{align}
$$
This is the result to remember: the power dissipated in the load is maximized when the load resistance is matched to the source's own internal resistance.
Now, any circuit you would reasonably call a "voltage source" must have a low internal resistance compared to typical load resistances.
If it didn't then the voltage accross the load would depend on the load resistance, which would mean your source isn't doing a good job of being a fixed voltage source.
So, because "voltage sources" have low output resistance, and because we showed that the power is maximized when the load resistance matches the source resistance, you will observe that the load gets hotter if it's low resistance.
This is why you found that with batteries, which are designed to be voltage sources, the lower resistance loads got hotter.
Current sources are the other way around.
They're designed for high internal resistance so you get a hotter load for a higher load resistance.
But in general, you get more power in the load if it's matched to the source.
Best Answer
For a homogeneous material characterized by a resistivity $\rho$ (in $\Omega m$) the resistance between any two points of contact is unbounded. Such "infinite" resistance even applies if one point of contact is replaced by a spherical contact area centered around the point. Just check for yourself and calculate the resistance for this latter configuration by integrating $ \rho /(4 \pi r^2)$ from zero to any finite radial distance.
Another way of recognizing this divergence is by dimensional analysis. To get from a resistivity $\rho$ measured in $\Omega m$ to a resistance $R$ measured in $\Omega$, one has to divide $\rho$ by a length scale. This length scale can not be the distance between the contacts, as this would lead to the unphysical behavior of the resistance between two points decreasing with increasing distance. It turns our that the relevant length scale is the linear size $r$ of the electrical contacts: $R \approx \rho / r$.
Physically what happens is that the electrical field strength diverges towards a current injection point. You have to assume finite contact areas to obtain a meaningful answer.