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.