# Electrical Point on a Plane

Discussion in 'Mathematics and Physics' started by MrAl, Mar 24, 2015.

1. ### MrAlWell-Known MemberMost Helpful Member

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Hello there,

This question comes up from time to time, and i was just taking another look at it recently.
The main question is that we have a square conducting sheet of uniform thickness and we apply one or more voltages to the sheet and then measure a point (or two) inside the square. We hope to be able to figure out where the point is located on the square.

Starting with two strips of conductive material, longer than wide, we have a resistance along the length of each one. The resistances are equal, and we measure by placing 1v at one end of each strip and 0v at the other end of each strip.
Any voltage measured along the strip shows that the gradient is linear, so if we measure half way across the strip we see 1/2 the voltage.

But now place the two strips side by side so that their edges touch electrically and perfectly. Do we still get the same results? Maybe so for this over simplified example, but take many strips and place them side by side and the Laplace equation says that we dont get the same results because there is action along the horizontal as well as along any line where the two potentials are placed. In other words, the field spreads out.

What do you think?

2. ### JimBSuper ModeratorMost Helpful Member

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OK, I understand that.

But I do not understand this, I just cannot visualise what you are doing.

Can you draw a picture?

JimB

3. ### cowboybobWell-Known MemberMost Helpful Member

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Since charges tend to move to the edges of a conductor, one would assume that (in your example) there would be disparities in the potential(s) observed as one probed from the middle of the connected strips towards one or the other of the outside edges, i.e., the charges would all attempt to move to the edges, thus reducing the charge(s) in the center, increasing as the edges are approached. The LaPlace Equations predict what those changes in charge should be (ideally) and more or less where.

Those measurements, however, should still be linear if the probes traced a perfectly straight line placed exactly perpendicular to the line between the o and 1V connection points.

<<EDIT>> As I recall, the effect is what allows waveguides to perform as they do (in AC, specifically microwave, applications).

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5. ### moty22Member

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There was a discussion about making coil from iron wire, this question relates and has the same answer.
The theory is that conductor creates a magnetic field around it that relates to current. That magnetic field in turn generates voltage on the conductor. At the middle of the conductor the field is more dense and the resistance is higher, this causes the current to flow away from the middle of the conductor.
This theory sounds totally crazy at first but after developing induction heaters for several years I accepted it.

6. ### MrAlWell-Known MemberMost Helpful Member

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Hello again,

Well i am not sure we are talking about the same thing, because the Laplace solution shows that the field spreads out and so the field toward the edges gets weaker, but maybe that's the same thing.
As a side note, one solution might be to generate a wave function that is the inverse (or something like that) of the solution, and that might force all the voltage points to form a perfect gradient from top to bottom. This would require pretty high frequency harmonics however.
One solution i found that actually does work however, and is static, is to implement boundary conditions on the sides as well as top and bottom. Unfortunately this isnt that easy either, as the boundary conditions have to actually be the same as the gradient we are after. This could be quite hard to do, although it is possible. It is made more difficult however when we reverse the roles of the sides and top and bottom, so we can actually determine what point the probe is at (swapping roles allows pinpointing the x,y position if we do in fact have a linear gradient surface for both roles: one north to south, and one east to west).

JimB:
Yes, i included a diagram which i'll explain below.

cowboybob:
Well the solution says that the center has a higher voltage, but you have to realize this is not a 3d solution but a 2d solution where we take the thickness of the sheet to be negligible. This is also using a DC source not AC of any kind. Does that make sense?

moty22:
The permeability of the material in this case is assume to be low, like air, so around 1 (relative).

Explanation of attachment:
The illustrations are numbered where you can see the number of the illustration right under the object.

#1 is a single strip of some conducting non ferrous metal like copper or maybe graphite.
There is a voltage applied at the top, and zero volts at the bottom. The strip is thin, so we expect to see 0.5v near the center. This is a typical non lumped resistor where the resistance varies with the length, and we have to take the length into consideration (unlike a lumped resistor) to find any information about it.
The material is uniform and thin.

#2 is simply two of the strips placed end to end so that their ends contact each other all along the edge, so they form a single strip. The voltages are still 1v at the top and 0v at the bottom, and since they are thin there is not much width and so the voltage near the center is 0.5 again, even though there are two strips. This is how regular resistors would work too, so nothing new here yet.

#3 just shows two strips in parallel. The edges connect perfectly forming a single block of material, but still thin like the copper on a copper clad PC board. With 1v at the top and 0v at the bottom the center is still roughly 0.5v.

#4 just shows that we can put several of these in parallel, and they form a single sheet with no joints although the joints are shown here. A better drawing would show this as one complete sheet like a PC board.

Now we come to #5. This has two strips in series, then several of those sets in parallel, forming a larger sheet where we can see that the width of the entire sheet now is not insignificant anymore. A better drawing would show this sheet as one continuous solid sheet, and it would look like a PC board from the top seeing the copper clad, and the thickness of the sheet would still be thin like copper clad.
As shown, 1v applied at the top again and 0v at the bottom, but it is assumed that the 1v is at every sub section not just in the very middle of the top of the sheet, and 0v is at the bottom of every sub section too.
When we measure the voltages at points across the center, we dont see 0.5 all the way across (following that central line from left to right) but we see what looks like an unusual pattern of voltages that start low, increase to a max near the very dead center, and then decrease to a low value again that matches the other side voltage. If we analyze the sheet using differential elements, we end up with the Laplace equation:
Uxx+Uyy=0
and the solution follows the Laplace equation solutions in two dimensions. Written out in more formal notation:
∂^2(u)/∂x^2+∂^2(u)/∂y^2=0

At first it makes sense to think that across the center we would see 0.5v everywhere, but apparently that's not what happens because there is some side to side action as well because there is side to side resistance as well as top to bottom resistance. As mentioned above, we can force this solution into a perfectly linear gradient from top to bottom, but one solution found so far requires gradient boundary conditions, which are not that easy to generate and even harder to swap when it comes time to reverse roles. So other solutions would be interesting to see.
There's really no rule here either, for what might work, but whatever it is it has to be practical too, say within the confines of say hobby electronics where almost anybody can do it.

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7. ### cowboybobWell-Known MemberMost Helpful Member

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Thanks, MrAl. I had not considered a 2D (theoretical vs. actual) scenario.

8. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

No problem, but i may have jumped the gun a little in stating that the "inverse solution or something like that" could work as a single edge boundary condition. What i should have realized right off is that the analytical solution is a function of both x and y, and since on any single border we can only control one variable, we'd have to be very lucky to be able to find a function that would counter the non linearity over the entire sheet perfectly. Maybe to an approximation perhaps. I found several functions that 'almost' work, but still fail short to correct the entire sheet. Also, the edge could be curved, which would help generate a correcting pattern, but we'd still have to be able to control three sides with the fourth held at zero volts.

So since the analytical solution for the sheet is a function of both x and y, we'll probably have to control both the top and sides of the square at the same time, as mentioned previously. If we can figure out how to generate a gradient on both sides we've got it made. Unfortunately i think this means we have to use some higher resistance material for the sheet, as copper would not allow the generation of gradients i dont think because it's already too conductive. I've never tried that however. Perhaps a curvy trapezoidal shaped sheet (like a witch), but with considering the angled parts as not being part of the useable part of the sheet.

Maybe this is just too impractical for the hobbyist, and so may require advanced manufacturing techniques using special materials.