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EM Troubles - Conducting body / potential

  1. Oct 15, 2006 #1
    So I have an upcoming project that I started working on. I thought I had the foundation to actually tackle the task at hand, but now that I've started it, I realized I'm kind of weak with some fundementals. If someone would be so kind to read what I am going to type, and add to it, or tell me what I'm doing wrong, I would be very appreciative.


    Say we have a conducting plate in free space. If we set a reference point far enough away and measure the potential with respect to this point we have,

    [tex] V(\vec R) = \int_V \frac{\rho_V(\vec R')}{4 \pi \epsilon_0 R} \, dv' [/tex]

    [tex] \vec R [/tex] is the distance from the orgin to a point of interest (where we will measure the potential)
    [tex] \rho_V [/tex] is the volume charge density of the plate.
    [tex] \vec R' [/tex] is the distance from the orgin to a charge (which are located on the plate.
    [tex] R = |\vec R' - \vec R| [/tex]
    [tex] dv' [/tex] is the differential volume.

    Now since we have a conducting plate, the voltage (is this part right?) on the plate (with respect to our reference point) is constant.

    If we call this constant voltage [itex] V_0 [/itex] then we have,

    [tex] V(\vec R) = V_0 = \int_V \frac{\rho_v (\vec R')}{4 \pi \epsilon_0 R} dv' [/tex]
    If (could someone help me with this set notation here too?),
    [tex] \vec R \in S [/tex]
    Where: [itex] S [/itex] represents the surface of the plate.

    (I'm trying to say, [itex] \vec R [/itex] points to a point that is on the surface of the plate.

    Thus, as long as [itex] \vec R [/itex] points to a location on the surface, the integral will evaluate to a constant, [itex] V_0 [/itex].

    Is my understanding correct?
    Last edited: Oct 15, 2006
  2. jcsd
  3. Oct 15, 2006 #2
    conductors are supposed to be equipotential, so i think yes. that expression is correct
  4. Oct 16, 2006 #3


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    Regarding your question, you should use [tex] \sigma dA [/tex] rather than [tex] \rho_v dv[/tex]. If your plate is a conductor, then all the charges are at the surface in an electrostatic case.

    Also, you might want to consider solving the Laplace equation to find V. In many cases, this approach is much easier (and practical, as you usually have control over the values of V at the boundaries).
    Last edited: Oct 16, 2006
  5. Oct 16, 2006 #4

    Meir Achuz

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    You can't simply use that Coulomb's law integral to find V(r) in the presence of any conductor, because you can't determine rho until you have solved the problem. That is because the charge positions itself on the surface of any conductor to make the conductor an equipontential.
    You will have to use some Greens function method. If the conducting plane can be considered infinite, you could use images.
  6. Oct 16, 2006 #5
    This problem will actually be solved numerically.

    The conducting plane is finite, and it is supposed to be solved with the Method of Moments. I believe that method uses Green's function (something I have not studied). The professor gave us an example of the setup (using MoM) for a 1-dimensional case, and we are expected to come up with something similar for a 3-dimensional case.

    My understanding is limited at this point, as the project is not due for another month. However what I've read thus far, is yes [itex] \rho_S [/itex] or [itex] \sigma [/itex] or whatever you call it is NOT constant, and arranges itself (as you said) so that the voltage on the plate is constant.

    I wanted to make sure that integral expression I wrote is correct.
    So it is correct right? It's just we don't know what the charge function is at this point?

    So now if we restrict the domain to points on the conductor, the voltage will be constant. This leaves the conduction function to be solved.

    This is the part I am working on right now, and need to figure out. I wanted to make sure I was cool, up to this point.

    The little I know this far is,

    [itex] \rho_S [/itex] (I'm comfortable with this notation, I hope this is ok) is broken into a piecewise function, then we can assume [itex] \rho_{Si} [/itex] is constant on each "square" of the conducting plate. A condition must be set that says that [itex] \rho_S = 0 [/itex] if not on the plane.

    I still have more to do for sure...
    is it cool if I use this thread with more questions regarding this project?

    thanks everyone!
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