Resistance between two point voltages on infinite plane

[Baluncore]

Your original question hypothesised an infinite 2D plane with point charges. But current cannot flow in 2D, it requires 3D, so your model becomes a section through two circular conductors.

A current filament is a very thin thread of current. It is sometimes round with a finite radius, at other times it can be treated as being infinitesimally thin. Your simplest model requires conductors to be of circular section that can map to equipotentials.

Take any two parallel cylindrical conductors, of any radius ratio, and you can map them onto the equipotential circles of the field diagram. That makes much of the maths easier when it comes to fields, such as transmission line analysis.

The same happens in your model when two conductive discs or rings are placed on a plane in 2D. They make it trivial to plot the surrounding field as circles.

Where an electric field exists, a current will flow, the magnitude of the current is determined by the voltage divided by resistance. The equipotentials have the same pattern no matter what the resistivity of the infinite medium surrounding the conductors. Only the current flow is different.

 

[Roger44]

Hi

I have a result for this problem. :

We can see that as the electrodes become further apart, the non-straight paths to the opposite electrode become "worthwhile".

The Shape factor formula is quoted in thermal conductiviy documents as the Shape Factor for heat conduction between two tubes in an infinite medium. I couldn't find on the Net the walk through to get to this formula, if someone knows it or gan give us a link, thanks in advance. I don't know if it's an empirical or exact mathematical solution, but I can say that their Shape Factor looks exact for one tube inside a larger tube, off-axis by Z. When we put Z=0, it becomes the the simple case of the conductivity of the arc of a circle which is an easy Ln function.

You had showed me

- that the current spreads out on the surface initially uniformally in all directions;

-the paths followed by the current are at right-angles to equipotential lines which are circles whose formula for diameter and displacement along the axis for any voltage are given in xxx's enclosed document.

Thanks, I didn't know all that. Then we can easily show geometrically that the current paths are also circular because they cross at right angles circles which obey the conditions of diameter and displacement. From there I wanted to find an expression for the conductivity of the small shaded d surface in the image below.

Its four sides are arcs of circles whose radii and lengths can be expressed in terms of alpha, theta, d alpha d theta. We can integrate first with respect to beta, and then to alpha. However I could't find the expression for its conductivity.. If you assimilate it to a rectangle the result is not exact, I suppose it would be the same thing if you assimilate it to an arc.

That's how far I could get. My contribution is really rather negligible. Maybe someone can find the last jump to get to the above S formula. But it might be that the S formula comes from a completely different approach.

 

[Roger44]

Hi
I have the complete solution to this problem in the enclosed image. It comes from a book on thermal conductivity, so replace "source" and "drain" by .electrodes, and "isotherm" by "equipotential line". I've re-written it with S the half-distance between the electrodes and r0 their radius to conform with Baluncore's enclosed PJ. Those who have read Baluncore's PJ can skip all what's above the thick black line.

There's no integration and you don't need to know anything about the current paths, everything is deduced from the equipotential lines.