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Single-Wire 'Circuit' using Displacement Current as a Return Path

  1. Mar 14, 2014 #1
    So, consider an average shorted capacitor connected to an alternating current power source. This capacitor has a certain capacitance, voltage, displacement current, etc. associated with it. Assume the dielectric of this capacitor is air. Let us increase the frequency of this source. Recall that this would cause more current to leak through the dielectric in the capacitor, since the dielectric is not perfect. I believe the current leakage is in the form of displacement current. At a certain point, the frequency is so high that the capacitor is leaking current completely, i.e. the dielectric is no longer effective.

    If we take this capacitor and then increase the distance, the capacitance will go down, whereas the voltage goes up, for a given charge. Eventually we increase the distance so far that one plate is very close to the source, while the other is very close to the load.

    The voltage and frequency are both very high, and so, the circuit is complete by means of displacement current from one capacitor to another.

    So does this work, or is there something I am overlooking? If anyone needs me to explain it better, just ask and I'll try to help you understand what I am saying.
     
  2. jcsd
  3. Mar 14, 2014 #2
    Yes, it works. Not very practical though.
     
  4. Mar 14, 2014 #3

    nsaspook

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    Last edited: Mar 14, 2014
  5. Mar 14, 2014 #4

    Drakkith

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    Is this a form of single-wire transmission?
    See here: http://en.wikipedia.org/wiki/Single-wire_transmission_line

    Also, I believe this can be done using normal capacitance. With enough capacitance for the given frequency, the capacitor is unable to be completely charged during each alternation of the voltage/current, which means that charges are simply moving on and off the plates repeatedly.
     
  6. Mar 14, 2014 #5

    nsaspook

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    Single wire transmission normally involves Evanescent waves that are not self-propagating free space waves. The fields from 'currents' in the conductors or dielectric coating are reactive and store energy as they regenerated from reflections caused by the impedance boundary from conductor to dielectric while propagating down the wire using modes like we see below cutoff in a RF waveguide.
     
  7. Mar 14, 2014 #6
    What makes it impractical?

    From what I understand, the power losses would be lower than, say, your average 3-phase 60-cycle transmission line, since the losses in two wires would effectively be zero in a single-wire situation; of course, those wires aren't there.

    I think what I describe is a bit different from this, since we're still dealing with the near-field effects of the two distant capacitors... If what I describe IS indeed what you say, then wouldn't a capacitor be a special care of a dielectric resonator antenna?
     
  8. Mar 15, 2014 #7

    nsaspook

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    Maybe, I might have read too much into your 'displacement current' leakage statements. My understanding of the question was that at the separation distance the transfer was mainly from the electric field between the plates of one capacitor and not from EM radiation from widely separated plates acting like a normal radiative antenna. A normal capacitor used in a circuit is not designed as a radiative antenna but we can design efficient antennas that depend on a dielectric to be the current path instead of a metallic conductor.
     
    Last edited: Mar 15, 2014
  9. Mar 15, 2014 #8
    Well, to be specific, I am thinking of this device mostly in a situation of power transmission, so we would want to reduce the far-field effects as much as possible and keep power transfer on the near-field side of things. I think you did read into it a bit too much, I was just mentioning leakage so those reading would understand that currents need not be confined to conductors at high frequencies, thus allowing for the return path to be in the air around the conductor via displacement current.

    In simpler terms, the circuit would be completed by capacitive coupling. In fact, we could treat the return path as a 'virtual wire'. I've actually found a paper speaking specifically on what is happening on the return path, but it is in English poorly translated from Russian.

    http://tf.llu.lv/conference/proceedings2008/Papers/13_VoitkansGreivulis.pdf
     
  10. Mar 16, 2014 #9

    nsaspook

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    Maybe thinking about the capacitor as a form of transmission line in a power transmission situation instead of a 'charge/displacement current' device helps if electromagnetic radiation is unimportant. It's not my argument that "something called 'displacement' current does not exist" as a term used to calculate the magnetic field but IMO and others 'current' was a poor choice of words for it.

    Note: Appendix A.2
    http://www.physics.princeton.edu/~mcdonald/examples/displacement.pdf
     
  11. Mar 16, 2014 #10
    Actually, I agree that it is better to look at it using transmission line theory. Also, I've often heard that displacement current was poor choice of words, and it is a misnomer, but this is only due to the definitions science has taken on.

    See, Maxwell defined things like this: "All charge is the residual effect of the polarization of the dielectric", and "The variations of electric displacement evidently constitute electric currents." Hence, if a current is "moving charge", i.e. "moving residual effect of polarization of the dielectric", then displacement current is as real a current as any other; according to Maxwell, electric displacements are alternating variations of positive and negative charge, and this qualifies as 'current'. It's not really his fault, it's really our own for changing our definition from Maxwell's. The quotes are from Maxwell's Treatise, pages 167 and 65, respectively.

    This gets into an interesting discussion on definitions and how the view of electricity changed over the years, but that is beyond the scope of this topic. In short, people who aren't familiar with the original works of Maxwell find displacement current to be a misnomer, but in reality the definition of "displacement current" is completely consistent.

    All this in consideration, I want to stress once again, that I agree that it would be better to look at it in the form of transmission line. I was only using that term to simplify my meaning.
     
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