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Strange capacitor question

  1. Jan 8, 2004 #1
    Imagine one plate of a capacitor connected to a terminal of a electric AC source. Imagine the other plate of the capacitor to the right is connected to the other terminal of the source.

    From the point of view of classic electronics, a capacitor with such AC signal behaves just like a wire, the signal pass without problems.

    But now imagine there is another plate to the left of the first plate of the capacitor, connected to any other point, but not to the source. (so finally, three plates, two makes a capacitor on a closed path over a source, and the other plate and one of the plates of the capacitor makes an "open path").

    I think this plate will fell also the changes on the signal and there will be charge variations on that part of the "open path".(although the plate is not connected to the source, somehow, it is coupled to the capacitor),

    Will the disconected plate follow the charge changes on the capacitor?
    Will this charge changes (if exist) do work on a load on the "open path"?
    Will the AC source have to do more work with this kind of performance?
     
  2. jcsd
  3. Jan 9, 2004 #2
    could you post a picture? I can't see what you're talking about...
     
  4. Jan 9, 2004 #3
    Here is the picture.
     

    Attached Files:

  5. Jan 9, 2004 #4
    well if open path means you mean just let it float, what you described is equivalent wit a system of capacitors. I'll try to draw an equvalent schematic quicly....
     
  6. Jan 10, 2004 #5
    I post a more simple drawing.

    My questions are:

    If we attach the load directly to the AC source (removing the capacitor) it's said that no current runs, because the circuit is open. The grounds are directly attached to the potential of the source, with no current needed.
    I don't understand that, because when you attach the load to the great potential point, it will take sometime to construct the potential on the disconected point of the load, and then, some current will take place.Can you explain me how the potential can go to the disconected point without current?

    But either that...
    When we place the capacitor, is the situation the same as if the capacitor were not there?
    For a moment (everytime the polarity of the AC source changes), I think there is current running through the branches. Is it true or is it like the previous situation?

    If there is current, the effect on the plate connected to the load will be due to induction. Then, will the source have to worry only about the capacitor, independently of the load?, or will the AC source have to do more work depending on the load conected to the capacitor?
     

    Attached Files:

  7. Jan 12, 2004 #6
    no, when the circuit is open you can't have current. It doesn't matter if you have a capacitor there or not.
     
  8. Jan 12, 2004 #7
    Ok, so no current is allowed at any time on an open circuit.

    But then, how is it possible to get the information of the voltage value instantly on the disconected points of the wires??

    I mean: Imagine a battery here, in the earth, and two disconected wires at the sun distance. Imagine you connect the battery to the ends of the wires here in the earth. The voltage will be inmediatly on the sun?? Then, we have created an instantaneous way of communication!.

    If the potential takes 8 minutes or more to take the ends of the wires on the sun, then, something is travelling through the wires. If it's not current, what is that??
     
  9. Jan 12, 2004 #8
    There are many levels of modeling electric circuits. One is that a capacitor is an open circuit to DC. This works well for many applications. But since you asked...

    The energy is really carried in an electromagnetic field surrounding the wires. If you 'turned on a battery' at the distance of the sun you would not detect anything for 8 minutes. During that 8 minutes there is a current traveling in the wire to guide the energy along the wire. There can be no current flow across the open circuit at the end so another equal and opposite direction current is established (as if by magic) to cancel the incident current at the end of the wire. This oppositely directed current travels back toward the battery.

    Ignoring the all the details on what can happen when the reflected current reaches the battery you can conclude that at the end of the wire there is:

    1. no current (simplest model)
    2. an incident and reflected current such that their sum is zero (a more complicated model but one that answers your latest question).

    If you placed a high quality voltmeter at the end of the wires you would measure the voltage of the battery there.
     
  10. Jan 13, 2004 #9
    Thanks mmwave, the fields outside the wires does the job.

    Just another question.

    Imagine the same battery here in the earth, and the long long wires to the sun.

    Imagine you connect the battery for a period less than 8 minutes (ie: 5 minutes), then disconnect it.

    What will happen on the extreme of the wires in the sun?

    Will the potential just arrive at the sun, and be there disponible for 5 minutes, or will the voltage pulse never be disponible on the sun extremes?

    If the pulse potential arrives, remember that the battery now is disconnected... could the potential do practical work on a circuit between the two wires in the sun?
     
  11. Jan 13, 2004 #10
    Hhmmm...

    Electricity in a conductor does not travel the speed of light. I believe it is on the order of about 100,000 mph under ideal conditions, making it somewhere around 4000 times slower than light.
    Regardless of the actual numbers, electron movement in a conductor is far slower than the speed of light.
    In any event, a charged conductor removed from it's source to power a load will work. But!!!! It is momentary(because it is no longer connected to the power source) and, more importantly, NO EXTRA POWER can ever be obtained, because the disconnected source provided the power "potential" before dis-connection resulting in source loss.
     
  12. Jan 14, 2004 #11
    The voltage pulse would be measurable for 5 minutes, then gone. During that time energy is available to do work. Your computer is doing this all the time, although the amount of work is miniscule it works on short pulses. If you really want to understand pulses you'll want to star learning about electromagnetic waves. I've been at it for 20 years now and there is still much to learn.
     
  13. Jan 14, 2004 #12
    We need to be more specific here. If the conductor is surrounded by air then the field travels at the speed of light. If the surrounding medium has a relative dielectric constant of er, then they travel at c/sqrt(er). It sounds like you are talking about the drift speed of electrons inside the conductor. It's confusing to say electricity travels at 100,000 mph because most people are interested in the energy (traveling a c/sqrt(er)) and not the drifting electrons.
     
  14. Jan 14, 2004 #13
    WOW! to me that you are talking about sound very very interesting.

    Mmwave is right, i was talking on transmision of the potential field of the source outside the wires, not about the electrons inside the conductor. In fact, with the device from the earth to the sun working like i said, no current (of electrons) is allowed, only the potential fields changes will travel following the wires.

    Again think about the earth and sun experiment.

    Imagine you have the battery in the earth disconnected, and the load in the sun connected. Then, you connect the battery.

    How can the battery know there is a load connected? It will take 8 minutes to the potential to take there.

    But that is just exciting! Imagine you put the battery 5 minutes, giving potential, then disconnect the battery on the earth, and just wait the 5 minutes potential to pass through the load. What will happen?

    How could the load know there is no source connected?
    The load see the pass of the potential field. Will the load do work?

    If so, will the resistance value of the load affect the potential arriving? I mean, you could put the load you want, the potential is coming on the same amount!

    to me, the system behaves like a strange radio or TV station emisor-receptor system, but on potential. It seems like the wires act like antennas first to the source, and later to the load, using the delay of propagation of potential, and without current from the source.

    Could such a system do work???
     
  15. Jan 14, 2004 #14

    russ_watters

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  16. Jan 14, 2004 #15
    OK, now I understand your question better.
    What you are describing is a phenomenon that I called "the free-space field effect" in a series of letters to a Fermilab acquaintance.
    Basically, FSFE postulates that an unusual but expected potential exists under the conditions whereby a field source is allowed to propagate but, after a declared amount of time, terminated at the source.
    When terminated the field no longer propagates, but retains the potential as if the source were still operating.
    The "catch" #1: The effect is very short-lived, as the terminated field will collapse at the rate of C in the absence of the source supply. So, extracting work from the free-space field CAN be done during this time, but is brief and problematic.
    The "catch" #2: Energy is required to establish the field in the first place, so any extraction, free-space or not, will only be equal to or LESS than the energy used to create the effect.

    Therefore, the FSFE and its variants can not be used in the hopes of "free energy"
    However, as discussed with Dr. Weems at Fermilab, the FSFE has one other unusual potential: an action without a reaction. This is made possible due to the complete separation of a non-particle force carrier from the force source due to delayed field-source termination. Useful interaction is extremely difficult, as the speed of FS field collapse is C, and the extracted interaction must occur during the collapse.
    Fermilab is not directly engaged in this pursuit as I understand, but I believe another National Lab, BNL I think, is persuing it under some other name(not FSFE).
    Again, this is not nor will ever be free energy(not that it was suggested, rather a novel propulsion method requiring large amounts of energy if it pans-out.
    In any event I salute your thoughts, as I was myself thinking on the same line some time ago. Apparently, those greater than you or I are thinking on the same terms and we may well see a revolution in propulsion physics based on this effect.
     
  17. Jan 15, 2004 #16
    Thanks Russ and Pallidin!

    You helped me a lot to understand better my doubts.

    I sometime ago talked also about the method to get unidirectional thrust with magnetic devices, using magnetic field propagation delay.
    I post only some links related:

    www.geocities.com/k_pullo/fields.html
    http://dsp7.ee.uct.ac.za/~mdevill/emthrust.html
    http://my.voyager.net/~jrrandall/MagThrust.htm

    and another related concept here:
    www.geocities.com/CapeCanaveral/Lab/9222/ele.htm

    To me, that way of thinking has nothing wrong in theory (not physical flaw), but maybe some problems of technical matter. But all people tell me JUST BUILD IT, not thinking on the examples as a matter to physic study, they only say me it can't be constructed.

    It's just as if when i ask with the example of the battery and the wires to the sun, they told me JUST BUILD IT to avoid the physical discussion on the example!.

    So thanks again, finally i'm not the only person that sees that maybe some of the thinks or ideas i find or i think about free energy are a really possible way to take, or at least, to study.
     
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