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Resistance between two point voltages on infinite plane

  1. Sep 16, 2015 #1
    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?
     
  2. jcsd
  3. Sep 17, 2015 #2
    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].
     
  4. Sep 17, 2015 #3
    If the resistive material is uniform, it is moderately simple. Find the field[/PLAIN] [Broken] 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.
     
    Last edited by a moderator: May 7, 2017
  5. Sep 17, 2015 #4

    Baluncore

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    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.
     
  6. Sep 18, 2015 #5

    Hesch

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    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:
    equiv3.gif

    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:
    equiv5.gif
    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 * 0L 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 ?
     
    Last edited: Sep 18, 2015
  7. Sep 18, 2015 #6

    Baluncore

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    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.
     
  8. Sep 19, 2015 #7
    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.

    750968pourforum.jpg
     
  9. Sep 19, 2015 #8

    Baluncore

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    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.
     
  10. Sep 19, 2015 #9
    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.

    237084pourforum2.jpg
     
  11. Sep 19, 2015 #10
    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.
     
  12. Sep 19, 2015 #11

    Hesch

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    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 ). :smile:
     
  13. Sep 19, 2015 #12
    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.
     
  14. Sep 20, 2015 #13
    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 ?
     
  15. Sep 20, 2015 #14
    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.
     
  16. Sep 20, 2015 #15

    Hesch

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    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.
     
  17. Sep 20, 2015 #16
    You are correct. The term "flux" can mean several things, and I meant current density. Thank you for pointing that out.
     
  18. Sep 20, 2015 #17
    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..


    189167pourforum3.jpg
     
    Last edited: Sep 20, 2015
  19. Sep 21, 2015 #18

    Hesch

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    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 = Idens*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.
     
  20. Sep 21, 2015 #19
    "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!!!
     
  21. Sep 21, 2015 #20
    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) = (pr) ÷ 4πεοr2. (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.
     
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