electromagnetism - Resistance between two points on a conducting surface

Suppose we have a cylindrical resistor, with resistance given by $R=\rho\cdot l/(\pi r^2)$

Let $d$ be the distance between two points in the interior of the resistor and let $r\gg d\gg l$. Ie. it is approximately a 2D-surface (a rather thin disk).

What is the resistance between these two points?

Let $r,l\gg d$, (ie a 3D volume), is the resistance $0$ ?

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Clarification: A voltage difference is applied between two points a distance $d$ apart, inside a material with resistivity $\rho$, and the current is measured, the proportionality constant $V/I$ is called $R$. The material is a cylinder of height $l$ and radius $r$, and the two points are situated close to the center, we can write $R$ as a function of $l$, $r$ and $d$, $R(l,r,d)$, for small $d$.

The questions are then:
What is $$ \lim_{r \rightarrow \infty} \lim_{l \rightarrow \infty} R(l,r,d) $$
What is $$ \lim_{r \rightarrow \infty} \lim_{l \rightarrow 0} R(l,r,d) $$

Answer

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:

1. Nonzero current leads to infinite voltage: $I\neq0\,\Rightarrow\,\Delta U \to \infty$.


2. If $A\Rightarrow B$, then $!B\Rightarrow !A$.


3. $\Delta U\mbox{-finite}\,\Rightarrow\,I=0$

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:

1. Radius of the disk.


2. Thickness of the disk.


3. Distance between contacts $|s-d|$


4. 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.