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Displacement current between two parallel plates

  1. Aug 5, 2012 #1
    OK, I was thinking about transformers and mutual induction and then drifted onto displacement current. I then decided to build a simple test apparatus and test for the magnetic field between two parallel plates. using a variety of coil designs to have as much of the windings perpendicular to the expected field I stumbled into an interesting problem.

    Now I'll describe the arrangement so that a mental image may be formed,
    two copper plates 3" dia 12ga thickness. a flat spiral coil, a flat biflilar coil, and a torus coil wound on a .05" thick plastic disk 2.75" dia with a .5" dia hole. all coils are 22ga ptfe insulated copper wire.

    the flat bifilar coil and spiral coil would have there turns nearly parallel to the magnetic field and should have the least measured inductance.

    what I found was that the rectified voltage from the spiral coil and bifilar coil was in increase in voltage greater than the torus coil. I've exhausted all the equations I can find on displacement current and can not resolve the measured voltages.

    So my question is, if the displacement current between two parallel plates is a magnetic field then how does a coil with it's windings in parallel to the field become induced? In the examples I've looked into in regards to this, the parallel plates are always fed with the wires at the center and perpendicular to them, this introduces the possibility of an expanded magnetic field coupling between the plates and the wires. In the arrangement I built the wires feed from the edge and in plane with the plates, there is no center out "field".


    here's the worksheets as I've been working on this trying to resolve the discrepancy.
     

    Attached Files:

  2. jcsd
  3. Aug 8, 2012 #2
  4. Aug 9, 2012 #3
    Is my post in the wrong section?
     
  5. Aug 10, 2012 #4

    vanhees71

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    I think it's all fine with your posting. It's simply not such an easy question to answer since it needs a treatment of the charging process of a capcitor with the full Maxwell equations, i.e., obviously including the displacement current.

    Perhaps it is a good idea to look in textbooks, where a simple problem of this kind is solved. I guess the most simple setup is to calculate the electromagnetic field between two spherical shells ("spherical capacitor") to understand the theory behind this. Whether there are explicit measurements of the displacement current, I cannot say.
     
  6. Aug 11, 2012 #5
    Thank you, I've not found any that deal with what I'm doing. I will update with some relevant information that I'll be working with to see if it holds promise. I have an old RCA book that deals with vacuum tube design and construction, lecture notes etc.. great resource on old tech. here's a note from page 120; of Space-current flow in vacuum tube structures: B.J. Thompson.

    For most practical purposes in calculating the electric fields at cathode, anode, and the space between, except very near the grid, a potential may be assigned to the plane of the grid. In other words, it is assumed that an equipotential plane may be substituted for the grid with out altering the electric fields. This would be true only when the grid wires are small and closely spaced in comparison with the spacings between grid and cathode and anode.
    The physical basis for this analysis is that the anode can influence the field at the cathode
    only by acting through the grid plane. By definition, the grid has u times the influence of the
    anode. It is obvious that this reasoning implicitly assumes that amplification factor is
    proportional to grid-anode spacing, for we might just as well have called the cathode the anode. The quantity dgp/dgj^u is simply the reciprocal of the amplification factor of the grid with respect to the cathode.

    My only thought to look there was the arrangement made me think of tubes, took digging thru a number of books to find something that delt with space-current vs charge. In this case it's still dealing with E-volts in plane but it's a start with some useful information.
     
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