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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 $0$ ?


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 $$ \lim_{r \rightarrow \infty} \lim_{l \rightarrow 0} R(l,r,d) $$

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$r \gg d \gg l$ means that it's a really short cylinder? –  genneth Feb 15 '11 at 10:01
    
Yes, like a surface –  user1708 Feb 15 '11 at 10:03
    
In the other limit of small $d$, it's never going to be zero; in that limit you could approximate things by assuming an infinitely large conductor, in which case the conductance drops as something like $1/d^2$ (I think... no calculations!) –  genneth Feb 15 '11 at 10:05
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If the two points are close enough to the boundary of the disk (at radius $r$), you surely need to know how close both of them are. If both of them are much further from the boundary of the disk than $d$ as well as $l$, then the boundary is irrelevant and you approximate the situation by an infinite material. For $d\gg l$, it is a 2D material; for $d\ll l$, it is a 3D material. –  Luboš Motl Feb 15 '11 at 10:39
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Hi @kakemonsteret, good. For $d$ much smaller than $l$, the resistance between the points is of course much smaller than the whole $R$ but it is not zero. To derive genneth's scalings, imagine that you cut a plane in between the two points, at distance $d/2$ from each point. The relevant radius of the area where the current is flowing goes like $d$. So in the 2D or 3D material, you have about $O(d)$ or $O(d^2)$ parallel branches, and each of them has resistance going like $d$, I think, i.e. conductance $1/d$. This fast guess: the total conductance goes like $\ln(d/l)$ or $d$ for 2D, 3D case. –  Luboš Motl Feb 15 '11 at 11:26
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4 Answers

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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.

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I think the answer is infinite because of the singular nature of a point. If we assume a steady state situation then the divergence of the potential is zero, by symmetry in a full 2D model the current density J will scale like 1/r. This implies the voltage scales like log(r), which diverges as r goes to zero. In the 3D case its even worse, as J scales as 1/r**2, and thus voltage scales as 1/r, which diverges even faster as r goes to zero. Note that we can compute restance, by fixing the current, and computing the voltage difference. The problem is that the voltage difference doesn't converge in the vicinity of a point current source/sink.

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see 4th comment to the answer (by me) –  Georg Feb 15 '11 at 16:52
    
Georg, yes but your emamples have finite current sources not points. The problem with the initial question is with the point surfaces. You are correct that he needs to create current contact over a finite area in order for the question to have meaning. –  Omega Centauri Feb 15 '11 at 17:22
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I'll tackle the 3D case. I am using the SI system. It should be noted that the electrical resistance of an electrical element (in given case a homogeneous medium) measures its opposition to the passage of an electric current (in given case direct current). The resistance of a homogeneous medium between two electrodes is defined as $$R=\frac{U}{I}=\frac{\rho\epsilon_0}{C}$$ where C is a capacitance between two electrodes.

Let's start with the assumption that instead of two points we have two conducting tiny spheres with a radius $r_0$. The distance between the centers of the spheres is $d$ and $r_0<<d$. To simplify the calculation let us assume that the charge on the spheres is distributed spherically symmetric. Then $$C=2\pi\epsilon_0r_0$$ Finally, required resistance:

$$R=\frac{\rho}{2\pi r_0}$$

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"Resistance between two points" is not a well-defined concept. If c is a curve connnecting point A to point B, the electromotive force V along c is a well-defined thing (it's the circulation of the electric vector field along c), and resistance would be the factor R such that V = RI when I is the intensity flowing through a thin tube borne by c. But it's clear that intensity depends on the cross section of this virtual tube, a non-defined entity here. So we have no proper definition of "resistance R between A and B" that would allow us to compute it.

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Maybe I'm just being stupid here. If you connect a thin wire to point A and another at point B, and apply a potential difference between the two, a certain amount of current will flow (spread out through the entire object). The ratio is the resistance. –  Ted Bunn Feb 15 '11 at 16:29
    
It's not "a certain amount of current" but an uncertain one. How much current flows in the situation you describe depends on unspecified parameters, mainly, the contact resistance, and/or the area of effective contact between the wire and the object. –  Bossavit Feb 16 '11 at 13:10
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