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Perpetual battery

  1. May 7, 2003 #1
    From my last two topics, this is my proposal:



    Can we get a change on surface of a charged conductor with less energy than the electric energy we can obtain using the potential generated by the surface change if we put these conductor in reference with other charged conductor? (I mean, the energy to change surface depends only on the conditions of the conductor, but the electric energy extracted depends of these conditions and the conditions of the other conductor).

    Are there other methods to change the potential of a certain amount of charges in a conductor without loosing the excess charge amount?
  2. jcsd
  3. May 7, 2003 #2
    Don't forget that changing size of charged balls is work (against electric field).
  4. May 7, 2003 #3
    Yes, i've got it present, that's why i comment that the balls are only an illustrative example.

    But there is no method to change the surface of a conductor with little energy?.

    What do you think about the method to pass from a box shape to a heat dissipator shape?
  5. May 7, 2003 #4
    Any re-arrangement of charges is by definition a work (positive or negative) because you change their potential energy. If the work is zero, then change of electric potential is zero too ('cause potential is energy (=work) per unit charge).
  6. May 18, 2003 #5
    Alexander, thanks for your support.

    Just another thinking about this.

    Imagine the expanding-shrinking spheres.

    Imagine that you've got the same charge and surface spheres connected through the load.

    Expand one of the spheres cost no work, because you're increasing the surface, and the charges want to scape one from each other. The charge density and voltage of this sphere will fall down. Then, as it is connected to the other sphere, some charges will flow from the other sphere.

    Now, looking at this sphere, as you increase the surface of the other sphere, you've got charges going out of this sphere, so the charge density and voltage will fall down also. In this case, reduce the surface to keep the voltage may cost no work also.

    So finally you can do this cycle once and again: Increase the surface of one sphere without opposition, then reduce the other sphere surface meanwhile the charges redistribute to keep the voltage, also with no opposition.

    Is there a flaw?
  7. May 18, 2003 #6
    Yes, T.A.N.S.T.A.F.L.
  8. May 18, 2003 #7


    User Avatar
    Science Advisor

    Abviously you can not decrease one sphere
    in size forever and the charge is not
    infinite either.

    In addition, whatever energy you may produce
    you do have to remember that you had to charge
    the spheres originally and that "cost"
    more energy then you can produce due
    to lower than 100% effectivness (or according
    to thermodynamics I can rephraze that as
    "You can not break even.").

    Live long and prosper.
  9. May 19, 2003 #8
    J-man: Yes, T.A.N.S.T.A.F.L. ?? [?] [?] [?]

    Drag: I've posted new schemes to clarify the concepts:


    I see it as two recipients of water.

    One is "V" Volume and filled with "H" water level.
    Another is "V" Volume and "h" water filled.

    Now, if you connect them nearly on top of "H" water level, the water on "H" recipient will go to the "h" recipient. The "h" level will began to go up.

    Now imagine as the "H" filled recipient discharges water, you shrink its volume, to keep this "H" level constant. The water goes discharging to the other recipient, so you've got less water on this recipient, but also you're shrinking the volume, so the "H" water level keeps the same. You cost no "pressure" to shrink the volume to keep "H", as water is leaving this recipient.

    In the other hand, the other recipient grows from "h" level to "H" water level. Now you've got a "v" volume recipient filled with "H" water level, and another "V" volume recipient filled at "H" also.

    Now, disconnect the recipients and expands the "v" recipient to "V" volume. Then, the water level falls down from "H" to "h". You have to do no "pressure" to get the water going to a lower level.

    Now, you're on initial conditions, so you can again connect the recipients and repite the cycle, expending only the work needed to do the resizing of recipients with no water opposition. Maybe this "resizing" could cost less than the energy obtained from the water transferences, don't you think so?
    Last edited: May 19, 2003
  10. May 19, 2003 #9
  11. May 20, 2003 #10
    Pardon, I forgot an 'A'.

    TANSTAAFL: There Ain't No Such Thing As A Free Lunch.
    (The double negative doesn't negate itself in this phrase... odd.)
  12. May 22, 2003 #11
    You can also make an analogy with compressed air.

    Imagine two connected cilinders. One has compressed air, the other air at normal pressure.

    When you let the air from the compressed air cilinder go to the normal air pressure cilinder, you can reduce the size of the compressed air cilinder to keep the pressure at the same value. Why should this reduction of size cost more energy than extracted from the air movement? You are only moving one wall of the cilinder, with the energy needed to keep the pressure, not to increase or oppose more pressure.

    Can anybody clarify this concept on any of the proposed systems?
  13. May 25, 2003 #12
    please, can somebody answer that question?, this is something that really is revolving my curiosity.
  14. May 25, 2003 #13
    I am haveing a hard time understanding what your question is. If it is in referance to perpetual motion, then any energy from one source to another has resistance, this resistance will eventualy cause loss of energy in any system and will cause it to stop. I believe in general terms you have designed an oscillator. Which may appear to be perpetual cause your not seeing the loss. Yes it may run for a long time but will eventualy stop without any outside source of energy. By reducing cylinder size or anyother process will cause extra loss and only shorten its life span.
  15. May 26, 2003 #14
    Well, thanks... one answer

    I'm not forgetting that the load is a ressistance,and also that a work will be done on that ressistance, but what i'm saying is that normally, if we have Q and q amount of charges (or H and h water levels) we use them only one time, until the amount or levels become equal.
    I think with this method we are able to repite a cycle keeping the levels in some way, so maybe, finally there will be looses, but you've been doing work more times. That is not "perpetual", but it's more times work exerted with the same initial conditions. My only final question is:

    What is the work needed to maintain the voltage by reducing the shape in a device that is also naturally loosing the charge?


    What is the work needed to maintain the pressure by reducing the volume in a chamber that is also naturally loosing water or air pressure?
  16. May 26, 2003 #15
    The work needed to maintain the voltage or pressure depends on the amount of natural occurring loss of energy, A simple oscillator can be maintained indefinitely with minimal energy. You would have to measure the amount of loss verses time and find an outside source to replace it.
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