Resistance between two point voltages on infinite plane

[Roger44]

Hi
What would be the mathematical approach for calculating the resistance between two close points on an infinite plane of resistive material? And if the plane was circular/square/rectangular with the two points at the cente?

 

[Babadag]

In my opinion, the resistance of one vertical rod electrode does not depend on the surrounding space limits, but only on the length and rod diameter [or radius].

The distance between rod locations has to be at least as the rod length [or height].See for instance IEEE 81/1983 or BS7430 ch.10.2 Rods or pipes.

R=ro/2/pi()/L*[ln(8*L/d)-1] where:

L is the length of the electrode, in meters (m);

d is the diameter, in meters (m);

ro is the resistivity of the soil, in ohm meters[ohm.m].

The resistance between 2 electrodes it has to be R=R1+R2 [connected in series].

 

[Jeff Rosenbury]

If the resistive material is uniform, it is moderately simple. Find the field[/PLAIN] distribution and resulting current flow. Integrate the conductivity to get the conductance. Invert to get the resistance.

This works for more complex resistive/time distributions as well, but the math isn't obvious or easy.

Note: There are some pedantic problems. A plane has no width, thus the resistance is infinite. Charges are not current sources, they will quickly (instantly if point charges) discharge. Of course you can re-define the problem a little better and eliminate these.

The basic idea of modeling the charge density, field lines, and current flows should work for all non-relativistic problems similar to this, though there are hopefully easier methods for situations with AC signals, etc. Finding charge distributions with magnetic fields pushing the currents around is not trivial.

 

[Baluncore]

the resistance between two close points

Points have zero circumference so they have infinite resistance.
You require two equipotential lines or areas to make a resistance measurement on a resistive sheet.

 

[Hesch]

Points have zero circumference so they have infinite resistance.

That's right, but say you have some small area, where you attach the positive ohm-meter probe, you will get equi potential curves ( the red curves ) like this:

A current will always flow perpendicular to the red curves, so within an area surrounded by two black curves ( current paths ), the current "flux" will be the same.

Now you attach the negative probe and will get this:

Again, the current flux will be the same within areas surrounded by two black curves. The resistance of such an area may be calculated:

R = σ * ₀∫^L 1/A(s) ds , where A(s) is a cross section area, s is the current path and L is the length of the path.

If the area of the plate is not infinit, the curves will of course will be distorted. I recommend a numerical calculation to solve the problem.

Anybody knows why I cannot paste two small images ?

 

[Baluncore]

That's right, but say you have some small area, where you attach the positive ohm-meter probe, you will get equi potential curves ( the red curves ) like this:

You agree it does not work for mathematical point contacts, points must always have infinite resistance.

The minimum valid electrical contact with zero area is a line. Because that line has length the field pattern can be solved from the equipotential defined by that line. That line can be open or closed, the simplest form being a circle.

Where the contact is an area, it is only the peripheral line that defines the contact. You have chosen to use two small circular areas as electrodes. The circumference of those two circles are the equipotential line contacts with the sheet.

For the OP's infinite sheet problem, the simplest solution is to select any two circular equipotentials as electrodes. You show examples as dotted red circles in your second diagram. So long as the two circles chosen do not contact each other, have a known radius ratio and centre positions, they can be mapped onto the simple bipolar field pattern you show. The circles can be separate, or one circle can enclose the other.For the OP's bounded sheet problem the simplest solution is to define two separate parts of the sheet boundary as contact electrodes. For a square or rectangle the obvious choice is two opposing edges.Alternatively, a minimum of four separate points are required to make resistance measurements of a sheet. Firstly two points are allocated to be a current source and a current sink. The second pair of points are then used to measure the potential.

 

[Roger44]

Hi
In fact the resisstance would be the same as between the metal electrodes in this image. This presentation reminds of of a similar problem I saw a few years ago where the only way was by a finite element approach. Somehow I find it hard to believe there's no pure mathematical solution to such a simple problem as mine.

Having said this, I now discover numerous replies for which I thank you, not really understanding why I was only alerted by E-Mail to the first reply, At a first glance your replies do seem to indicate that it's not so easy as that, so I'll now go through them and get back.

 

[Baluncore]

Somehow I find it hard to believe there's no pure mathematical solution to such a simple problem as mine.

That is because a mathematical point cannot both deliver a current to the surface of a sheet and measure the voltage on the sheet at the same time.

 

[Roger44]

Hesch, I've re-arranged you image to get a clearer idea in my head of what you're saying. Your formula R= ... is obviously true but may I saxact, but really is only the beginning.

 

[Roger44]

Something went wrong, I'll continue. "obviously true but geometrically doesn't help me much. I don't know if I should integrate with respect to d theta between 0 and Pi, hopefully being able to express S and A(s) as a function of theta, or with respect to dx between 0 and infinity. I'm not , or no longer able to master this problem.

 

[Hesch]

In the figure in #9, you have written a "dx" and a "dθ".

I prefer the θ because the current-flux then will be the same through all N areas. ( dI = I * dθ / 2π, seen close to the 12V-circle ).
( Hope you understand what I'm trying to express ).

 

[Jeff Rosenbury]

Perhaps integrate over dθ using the area and length determined by θ? Of course integrate form 0 to 2π (or 0 to π and double). The area (or at least the width of the area) is a function of θ I think.

I'll give some thought to finding the functions for the length and width. They don't spring to mind. Perhaps someone else knows?

I'm not sure that's the easiest method, but hopefully it gives an analytically solvable solution.

 

[Roger44]

Are we sure that the current-flux will be the same through all N areas for the same dθ seen close to the 12V-circle ?

 

[Jeff Rosenbury]

Are we sure that the current-flux will be the same through all N areas for the same dθ seen close to the 12V-circle ?

It is not. It varies with the changing width and length, going up as the width increases and down as the length increases. That assumes an even conductivity. This should be covered by the equation -- which I still don't have.

If it helps anyone else, the field strength at any point is the linear addition of the two fields. I plan on working that into an equation for the length and width of the geometric lines. It could take some time.

I used to be smarter.

 

[Hesch]

It is not. It varies with the changing width and length, going up as the width increases and down as the length increases.

I don't agree in that. The current density will change, but the current flux will not: No current will cross a current path ( black curve ), and due to KCL the current flux will be constant within an area.

If the conductivity is not homogeneous, the equi potential curves will be distorted, and thus the current paths and the shape of the areas.

 

[Jeff Rosenbury]

I don't agree in that. The current density will change, but the current flux will not: No current will cross a current path ( black curve ), and due to KCL the current flux will be constant within an area.

If the conductivity is not homogeneous, the equi potential curves will be distorted, and thus the current paths and the shape of the areas.

You are correct. The term "flux" can mean several things, and I meant current density. Thank you for pointing that out.

 

[Roger44]

Please Hesch, what does "and due to KCL" mean?

Secondly, I still don't see why the current fluxes I1 and I2, which we can express in total ampères crossing the straight line, should be the same for a same angle at the centre..

 

[Hesch]

what does "and due to KCL" mean?

KCL means Kirchhoffs Current Law ( the 1. law ):

The sum of currents ( calculated with signs ) flowing toward a node = 0.

In case of your attached figure, a "node" could be a boundary "I1" in the leftmost area. Now KCL states that the current flowing toward this boundary = the current flowing away from the boundary, which means that the current flux will be the same as for both shown "I1" boundaries.

If the area/diameter of the "12V" circle is very small compared to the distance between "12V" and "-" poles, the "12V"-pole cannot "sense" the "-" pole and the current density will be the same close to the "12V"-area in all directions. Thus the current flux = I_dens*dθ will be the same through alle areas.

At some distance from the "12V"-pole the length of the boundaries will increase and the current density will be decreased, but the current flux = (boundary length) * (current density ) will be constant.

 

[Roger44]

"If the area/diameter of the "12V" circle is very small compared to the distance between "12V" and "-" poles, the "12V"-pole cannot "sense" the "-" pole and the current density will be the same close to the "12V"-area in all directions. "

I hope you're right on this, as it considerably simplifies the problem. I thought about this quite some time, I really remain only half convinced!

 

[Jeff Rosenbury]

"If the area/diameter of the "12V" circle is very small compared to the distance between "12V" and "-" poles, the "12V"-pole cannot "sense" the "-" pole and the current density will be the same close to the "12V"-area in all directions. "

I hope you're right on this, as it considerably simplifies the problem. I thought about this quite some time, I really remain only half convinced!

You are right to be skeptical. This is definitely an approximation based on the distance to the near part of the electrode and the far part being the same distance from the other electrode and thus seeing the same voltage. But since you started with point charges, it is likely valid. It would likely not be valid for most real world problems.

My vector analysis class was many moons ago, but many of the concepts seem to apply here. The current paths will be perpendicular to the equal potential curves.

I get the feeling there should be a simple coordinate/unit transformation, but I can't think of it. For small electrodes.
Let p = qd;
p - dipole vector
q - ½ charge (the charge on one electrode)
d - displacement vector from negative to positive.

Let the origin bisect the displacement vector with the vector pointing in the z direction.

For any point r (with r as a unit vector), the electrostatic potential Φ(r) = (p⋅r) ÷ 4πε_οr². (Does this hold for the nearfield?)

The electric field, E is the gradient of Φ. Finally, J = σE where σ is the conductivity and J the current density.

I hope this helped.

 

[Roger44]

Hi
I read in a theoretical study on the Hall Effect that for a point electrode on an infinite plane, the current does emanate uniformally in all directions. So be it.

In this case, we can say that the ratio (A+dA)/dS between equipotential lines will the same for all d theta shade areas, ie (A1+dA1)/dS1 = (A2+dA2)/dS2. I'm not sure this helps much.

 

[Roger44]

Could we therefore say that the ration of curve lengths AB/BO (O= origin) will be equal all round an equipotential line?

 

[Jeff Rosenbury]

Could we therefore say that the ration of curve lengths AB/BO (O= origin) will be equal all round an equipotential line?

Not as far as I know. That would be true for concentric circles which these are not. It could be true, but I can't prove it.

 

[Baluncore]

Could we therefore say that the ration of curve lengths AB/BO (O= origin) will be equal all round an equipotential line?

The red equipotential lines are all circles. The black current lines must also all be circles. The circles are eccentric.

If AB/BO was constant for a particular circle then BA/AO must also be constant.
That would require A=B which is clearly not the case for eccentric circles, so the supposition must be false.

 

[Roger44]

By the use of the words circle and eccentric together, are you implying that the equipotential lines are ellipses? It might help if they were, but offhand I can't see why they should follow the strict curvature imposed by a and b of ellipses.

My supposition was that, for example in this image, B1 O/B1A1 = B2 O/B2A2

 

[Roger44]

See Excel image. The 6 red lines are ellipses having the same two focal points. The two black ellipses pass through these two focal points. It looks very much like the red and black lines cross at 90°. So the focal points could be the point contacts, the red ellipses equipotential lines and the two black ellipses the current paths.

 

[Jeff Rosenbury]

See Excel image. The 6 red lines are ellipses having the same two focal points. The two black ellipses pass through these two focal points. It looks very much like the red and black lines cross at 90°. So the focal points could be the point contacts, the red ellipses equipotential lines and the two black ellipses the current paths.

An interesting graph. But of what?

 

[Baluncore]

They are all perfect circles. But none of them have the same centre position.
Because equipotentials are circles, they can be used to model the fields between cylindrical conductors of different diameters.

 

[Roger44]

Jeff, graph of this image from the preceeding page :

Baluncore,
"they are all perfact circles" See image below. They are obviously not for a closed surface, what is the proof that they should they be so for an infinite surface ?

"they can be used to model the fields between cylindrical conductors of different diameters" Easy to say, do please carry on, assuming they are ellipses which should greatly simplify the problem.

 

[Baluncore]

Thread Title = "Resistance between two point voltages on infinite plane".

They are obviously not for a closed surface, what is the proof that they should they be so for an infinite surface ?

Years of experience. Attached are three pages from Electromagnetics. 2nd Edn. Kraus and Carver. 1973.
Start at bottom of page 81 with sect 3-19.

 

[Roger44]

That's more than good enough for me, equipotential lines on an infinite surface ARE elliptic. Thanks for this major contribution.

So now I can define such a typical line with value a and any old value b, less than a of course. I now wish to define a second line, twice as far out, so 2a, having the same focal point, but I cannot give this second line any old b value. It must have a precise b value conditionned by the a and b eccentricity of the first line. But how?

The current lines, assuming they are also ellipses:
1. pass through the 2 focal points of the mirror pair of equipotential lines
2. cross them at 90°.

In order to meet these 2 conditions, I imagine their b values are also conditionned by the eccentricity of the very first equipotential line defined. But how,

Baluncore's book extract should enable to define the a and b eccentricities of equipotential lines from source voltage, resistance and dialectric. Less is said about the current lines (field lines).

 

[Roger44]

I believe I was the first person to use the word ellipse, but in fact the equipotential lines in Hetsch's original image and in the document Baluncore posted are quite simply round circles. They are applicable to the electric field around two opposite current carrying parallel wires in a medium of known dielectric constant. The centres of these circles do not correspond to the wires.

Aren't they only applicable to non-conductive media?

 

[Baluncore]

That's more than good enough for me, equipotential lines on an infinite surface ARE elliptic.

NO.
Half way down page 83; Eqn (9); demonstrates that the equipotentials are circles. Eqns (10) & (11) let you draw those circles.
At the bottom of that section it concludes; [ocr] Field lines are also shown in Fig. 3-16. These are everywhere orthogonal to the potential circles and also are circles with their centers on the y axis. [/ocr]
Attached is better copy of that last page.

 

[Baluncore]

Aren't they only applicable to non-conductive media?

That mathematical pattern is on a plane section through two unbounded opposite polarity filaments in an infinite 3D space.
It is also applicable in your case of a plane in 2D space that includes two isolated point charges.

The filamentary or circular equipotential conductors are assumed to be much better conductors than the surrounding medium. The material can be free space or be a resistive medium. It must not have zero resistance because the current would then be infinite.

 

[Roger44]

In total agreement fo the No, that was the gist of my previous message, that I started talking about ellipses and nobody corrected me. I discovered my error going closely thru' the calculations in your PJ.

"The filamentary..." Could you be more explicit on these last two sentences. Thanks