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Electrostatic paradox ?

  1. Apr 6, 2003 #1
    Hello everybody!

    Many changes since the last time i was here!. Now, I've got a question that i can't stop think about. I hope you could help me.

    We all know the excess of charges on connected conductors always tends to distribute equally... ALWAYS?

    Take a look to this animation:

    http://www.geocities.com/k_pullo/system3.htm

    I will explain how it works:

    1 - Initially, the two connected spheres are pre-charged with Q electrons, and the skin conductor is 2Q pre-charged.

    2 - Now, we disconnect the two spheres, and connect the inner sphere to the skin. Then, all the electrons goes to the surface of the skin conductor, so the skin charges from 2Q to 3Q, and the inner sphere discharges from Q to 0.

    3 - Now, we disconnect the inner sphere and the skin, and connect the skin and the outter sphere. Then, the skin will discharge from 3Q to 2Q and the outter sphere will charge from Q to 2Q. We can use this discharging proccess to obtain voltage and current to load a circuit.

    4 - Then, we disconect the outter sphere and the skin conductor, and connect the outter and the inner sphere. We've got the outter sphere discharging from 2Q to Q and the inner sphere charging from 0 to Q. Also, we can use this proccess to obtain voltage and current as before to load a circuit.

    Now, the skin conductor is charged at 2Q, and each one of the spheres charged at Q, that's initial conditions, so we can make another cycle.

    We can obtain voltage and current to power a circuit always with the same charges, so we can use this "battery" forever!.

    Normally, the charges distribute equally when you connect two differently charged conductors... But there's one EXCEPTION: when one conductor is inside the other conductor. In this case, all the charge pass to the surface of the outter conductor, so the charges re-order without work to expend.

    We can use this two steps of order and disorder the charges to get current and voltage indefinitely. (voltage and current to power a circuit when connect two differently charged conductors, and re-charging of "battery" when connect the inner sphere and the skin conductor).

    Also take a look to www.geocities.com/k_pullo/system.htm

    What do you thik about this?.
     
    Last edited: Apr 6, 2003
  2. jcsd
  3. Apr 6, 2003 #2

    emu

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    Perpetual battery eh? Sounds like modern quackery to me.
     
  4. Apr 6, 2003 #3
    Hehe, please, just use physical arguments to refute the idea.
     
  5. Apr 6, 2003 #4

    Hurkyl

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    This cycle will eventually converge to a stationary system.

    E.G.

    For simplicity of the arithmetic, assume that your three pieces (the two spheres and the circle with a bar on it) all have the same size.


    I will label your three pieces A, B, and C, letting a, b, and c represent the net electric charge on each piece.

    A is the bottom sphere.
    B is the top (inner) sphere.
    C is the odd shaped thing.


    I will also assume that your connectors have neglegible size and take no work to connect, and I will neglect secondary electromagnetic effects.


    So, from your starting point: a = b = q, c = 2q

    We connect B and C. The charges on B and C will equalize to yield:

    a = q, b = 1.5q, c = 1.5q

    We connect A and C.

    a = 1.25q, b = 1.5q, c = 1.25q

    We connect A and B:

    a = 1.375q, b = 1.375q, c = 1.25q

    et cetera

    a = 1.375q , b = 1.3125q, c = 1.3125q
    a = 1.34375q , b = 1.3125q, c = 1.34375q
    a = 1.328125q, b = 1.328125q, c = 1.34375q
    ...

    eventually the charges will stabalize to:

    a = b = c = (1.3333...)q

    and rearranging the connectors will yield no current.

    Hurkyl
     
    Last edited: Apr 6, 2003
  6. Apr 6, 2003 #5
    Urkil, when you say:

    "So, from your starting point: a = b = q, c = 2q

    We connect B and C. The charges on B and C will equalize to yield:

    a = q, b = 1.5q, c = 1.5q"

    is not exactly what happens. If so, we couldn't charge a Van de Graaff generator.

    You're forgetting the effect of a conductor inside another conductor, the E field must be 0 inside, so when the inner conductor is not connected, induced possitive charges appear on the inner surface of the skin conductor, and negative induced charges appear on outter surface. When the inner sphere touch the skin, the induced charges dissapear, and the skin gets real negative charges of the inner sphere, and the inner sphere totally discharge (This is simply the FARADAY ice cube experiment).

    so the sequence is:

    a = b = q, c = 2q

    We connect B and C. ALL the INNER charges on B goes to SURFACE on C (because at this moment, they are in fact the same conductor):

    a = q, b = 0q, c = 3q
     
    Last edited: Apr 6, 2003
  7. Apr 6, 2003 #6
    Cala!!

    i solute your idea but i'll have to disapoint you if you were expecting perpetual circulation of that charge while exporting energy from the system.i mean if there is no export of energy then it is only adding zeros or this system has no practical usage.to make an export you'll have to connect a consumer of that charge in the circuit. even if you don't attach consumer the inner resistance of the conducters will vaperize the charge.

    here is my slightly different idea on perpetuum mobile:

    http://members.fortunecity.com/dock0609/deda.htm [Broken]

    hope you'll find it usefull.
     
    Last edited by a moderator: May 1, 2017
  8. Apr 6, 2003 #7

    Hurkyl

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    If your odd shaped thingy is a sufficient enclosure of the inner sphere to suck all of the charge out of it, it will also be a sufficient cage to prevent the charge from entering it in the first place.

    Hurkyl
     
  9. Apr 7, 2003 #8
    Urkil, i think you're misunderstunding what is happening. The amount of charge on the skin has nothing to do with the charging proccess on the inner sphere, it's only a matter of geometry.

    The skin charge doesn't affect the inner sphere charging proccess, but the inner sphere charging proccess affects the charges on the skin:

    While the inner sphere is charging from the outter sphere, a positively induced charge appears on the inner surface of the skin conductor, and also a negative induced charge on the outter surface.

    It's better explained here:

    http://www.sc.ehu.es/sbweb/fisica/elecmagnet/campo_electrico/cubeta/cubeta.htm

    (it's in spanish, but the schemes should help).

    If you draw a gaussian surface rounding the inner sphere, you'll see the skin conductor is not present. Only the charges on the inner sphere counts.

    If you draw a gaussian surface that engloves the inner surface of the skin conductor, the E field must be 0, so a positive induced charge appears on that surface of the skin.

    If you take a gaussian surface engloving all the skin conductor, the excess of charge is present as an induced negative charge at the outter surface of the skin.

    About Dock's refutation: i think the charges are not "wasted" on electric circuits, i think they only "disorder" an existing order of charges present on their terminals.
    Normally, we need the same energy to order the charges than the energy we can extract from that order (normal batteries). But there is a way to create order of charges with little energy to apply, by geometry. Once you order the charges, you've got the energy present again.
    You're not wasting the electrons, you're wasting their ordenation, and disordering or equalizing their amount on terminals of the circuit, but once you've re-ordered the same electrons (with my method or another one: with or without energy to apply), you've got the energy ready to be used, because it appears from the order of charges, not by the charges by themselves.
    The energy of electric circuits is extracted from the order of charges, not from the charges.
     
  10. Apr 7, 2003 #9
    Urkil, i post a more accurate sequence of what is happening:

    a=outter sphere
    b=inner sphere
    c=skin conductor
    co=skin conductor outter surface
    ci=skin conductor inner surface

    INITIAL CONDITION: a=q-, b=q-, c=2q-

    A and B connected:

    a=q-, b=q-, co=3q-, ci=q+

    B and C connected: ("battery" charging proccess)

    a=q-, b=0, co=3q-, ci=0

    C and A connected: (usable voltage and current)

    a=2q-, b=0, co=2q-, ci=0

    A and B connected: (usable voltage and current)

    a=q-, b=q-, co=3q-, ci=q+ (INITIAL CONDITIONS)
     
  11. Apr 7, 2003 #10

    pmb

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    re "We all know the excess of charges on connected conductors always tends to distribute equally... ALWAYS?"

    No. In fact that will hardly every happen. It only happens when the conductor is spherical and there are no other charges around which are not on the sphere itself.

    Pmb
     
  12. Apr 7, 2003 #11
    What i wanted to say was that the charges move through the connected conductors until the potential equalyze.

    Only the case of a conductor surrounded by another conected conductor does not follow this premise.

    Usually, we need energy to order the charges to get potential and then use the energy stored. But what i'm talking about is something like a catalyst. Geometry acts like a catalyst to order the charges!

    Imagine a system with two stable situations. Certain conditions put the system under one stable situation, and other conditions make the system go to the other stable situation. That's what i think it's happening here. We've got two different stable steps, and we change from one to another simply by connecting some elements or another ones in time. What is stable condition at a time is not at the next one...
     
  13. Apr 7, 2003 #12

    russ_watters

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    cala, I'm afraid this is just another permutation of the self-charging battery hoax/myth or in this case a self-charging capacitor. It doesn't work quite simply because ALL of the energy must go into charging to maintain the charge. You can increase the efficiency all you want, but regardless all you have is the electrical equivalent of a pendulum. And a pendulum will NEVER increase its oscillations.
     
  14. Apr 7, 2003 #13
    What do you mean when you say "ALL of the energy must go into charging to maintain the charge"?

    When you plug a circuit to a battery, you waste a little of it's potential difference, and get current, but the charge is the same, only re-distributed to a lower energy level.


    Also, i don't know where the energy to move all the charges from the inner sphere to the skin conductor came from, but that's what finally happen. I supose this is the more stable situation of charges on these conditions.

    And finally, i'm not trying to increase nothing. If you force a pendulum with a opposite force to the oscillations, i supose it will oscillate slowly...
     
  15. Apr 7, 2003 #14
    to Cala!!

    energy E = force F times distance X

    but it is also

    energy E = electric potential V1 times charge Q2

    when you use energy from closed system it is dE<>0 but it is also

    dE = V1dQ2 when V1=const

    altering energy is simultaneous with altering charge.

    have you checked my idea?
     
  16. Apr 8, 2003 #15
    Dock. I think there is dQ on my system. Maybe the total amount of charge remains constant, but there are local variations at each time, and thus dQ. Imagine you do:

    one place: 3Q to Q
    other place: Q to 3Q

    and

    one place: Q to 3Q
    other place: 3Q to Q

    The amount of charge remains constant (no increase or decrease of total charge) but there are local variations and thus dQ, isn't it?

    It's like the sea, maybe the amount of water remains constant(forget the rain, evaporation and bla bla bla...), but you've got water flux as moving waves.

    Also, dE = V*dQ when V constant, but dE=dV*Q when Q constant, isn't it?

    Respect your system, i think it's a great idea to obtain energy changing the dielectric properties of a capacitor (is it not called parametric power conversion?). The callenge is to do it with less energy than needed to do the change. I don't know if the static configuration with 2 different dielectrics will work.

    Have you looked at: http://jnaudin.free.fr/html/paraconv.htm ?
     
  17. Apr 8, 2003 #16
    tell this to Hurkyl,please.
    i've been arguing with him about the same thing in different field.

    yes it is true but V=-kq/R where q<>Q tush dV=-kdq/R (when dk=0 and dR=0)so you have again dischargement.
     
  18. Apr 8, 2003 #17

    dg

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    SORRY ALMOST NEVER!

    How about connecting two charged conductive spheres of different radius with a thin wire?

    Doesn't look like equality of potential give equality of charge unless we have conductors with the same capacitance!!!!
     
  19. Apr 8, 2003 #18
    Dg: i thought it was clarified on a topic i posted. You're right, what goes equal when you connect two conductors is potential.

    I literally said:

    "What i wanted to say was that the charges move through the connected conductors until the potential equalyze. Only the case of a conductor surrounded by another conected conductor does not follow this premise."


    Dock: Yes, i've got discharge, but to my system, discharging is right.

    The two connected conductors at a time feels discharging procceses, as i said, but the total amount of charge on all the conductors (connected or not) remains constant.

    Actually, i don't know where do you want to go telling me my system have discharging procceses, yes, it has partial discharging proccesses.

    The potential on my system comes from the collection of charges of the same kind, more on one conductor than on the other connected one, so we have variable V and variable amount of charge at a time.

    But counting all the parts and all the proccess along time, the amount of charges is the same.

    Distribution, movement of charges and potential changes, but not amount of charge.

    viewing ALL the system ALONG time, nothing change from the point of view of charge (no charges enter or leave the system), but we can obtain dQ between two places only moving the existing charges. You don't need more charges to enter or to leave to obtain a real dQ.

    Please, be more specific, i don't see where is your doubt about this method.

    Imagine the sea, and a big wave (You know the water of the wave will go back to the sea later)... but now, it's moving, so you've got "dwater" between two regions, don't you?
     
  20. Apr 8, 2003 #19
    Cala!!

    no i didn't meant that kind of discharging but discharging when the summary energy in the system changes (not some local energy but the whole energy).This you have to do (change the summsry energy) in order to make your system practical/usefull/(doing job).
    then d(summary energy)=(some potential) times d(summary charge)=(job done by the system against resistance outside)
     
  21. Apr 8, 2003 #20

    Hurkyl

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    This is what I'm considering:

    A closed conductor (like your odd shaped thing) is a Faraday cage.

    Faraday cages prevent EM waves from penetrating into the interior.

    Current flows because EM forces cause the electrons to redistribute themselves.


    If your odd shaped thing behaves as a Faraday cage, then the EM waves caused by current flow outside of the cage cannot penetrate into the cage, this there will be no change in force in the interior; no current can flow through the cage, so the charge on the inner sphere would tend to remain constant when connected to the outer sphere.

    Hurkyl
     
  22. Apr 9, 2003 #21
    I will try to resume all the disapoints you are making to the system, and explain myself better:

    1 - The charges don't distribute equally, the voltage is what is the same on connected conductors:

    That is true, the negative charges distribute from low voltage to high voltage conductor, until the voltage is the same on both conductors. But this doesn't affect the working principle of the system. I'll put a sequence that confirms this fact:

    a=outter sphere
    b=inner sphere
    c=skin conductor
    co=skin conductor outter surface
    ci=skin conductor inner surface

    Supose the equilibrium when A and B are connected is 2q- on A and q- on B. (B is smaller than A)

    Supose the equilibrium when A and C are connected is 3q- on A and 4q- on C. (A is smaller than C)


    INITIAL CONDITION: a=2q-, b=q-, c=4q-

    A and B connected:

    a=2q-, b=q-, co=5q-, ci=q+

    B and C connected: ("battery" charging proccess)

    a=2q-, b=0, co=5q-, ci=0

    C and A connected: (usable voltage and current)

    a=3q-, b=0, co=4q-, ci=0

    A and B connected: (usable voltage and current)

    a=2q-, b=q-, co=5q-, ci=q+ (INITIAL CONDITIONS)

    So the asymmetry of the amount of charges on equilibrium on each step may not affect the working principle (the system keeps working).

    2 - There are energy difference (thus job), but no total energy level losses.

    Dock pointed out that maybe there is no total energy change to do job. I think it's different to talk about energy difference, and energy levels. On discharging processes, ENERGY DIFFERENCE exist, so a job can be extracted. On charging proccess, the energy level from the exccess charges on B adds to energy on C to put the ENERGY LEVELS as on initial conditions. I'll try to explain how it's done:

    On a capacitor, you've got ordered charges, the negative plate has E1, and the positive plate same E1 but inverted. When you connect it to a circuit, the charges re-distribute, then, the ordenation is lost, and then the E=0 on both plates, so we have :

    from E=E1+E1 we go to E=0, so we made E1+E1 job on the circuit.

    On discharging phase of my system, C conductor has E1 energy from negative excess charges, and A is E2 negative also. When we connect by circuit, the charges goes from C to A until the V is the same, when this proccess stops:

    from E1-E2 we go to E3-E4 (near zero), so we made E1-E2-E3+E4 job (near E1-E2 job) on the circuit.

    BUT THE ZERO OF THIS SITUATION IS NOT A REAL ZERO IF YOU CHANGE THE CONDITIONS:

    Now, we connect A and B, then A has E4 (from the step before), and B has E=0.

    From E4-0 we go to E5-E6 (near zero), so we made E4-E5+E6 job (near E4 job) on the circuit. Then A keeps on E5 energy level, that is the same as E2.

    Now, B has E6 from the energy of negative charges. C has an energy of E3 at the moment. When you connect B and C, all the charge pass to C, and we got then E3+E6=E1 energy from the negative charges on C, and 0 on B, That is initial conditions of energy levels.

    3 - The charges will not enter into C due to its configuration

    That's not like that. C is not a closed cage, it has a hole to introduce the conductor that plugs on B. Then, as excess charge enters on B, induced positive charge appears on inner surface of C (CI), and induced negative charge appears on outter surface of C (CO).

    B "feels" no opposition to enter the charges, they are passing from A to B (Va is smaller than Vb).

    On the interior of C, V is ALWAYS 0 (CI compensates the charging of B with the induced positive charges.)

    When you connect B and C, The excess electrons of B annul the induced positive charges of CI, but then, the induced electrons of CO transform into real excess electrons on CO, That's the same to say that the energy level of B added to the energy level of excess electrons on C.
     
    Last edited: Apr 9, 2003
  23. Apr 9, 2003 #22
    I've posted a new diagram of how potential changes on the system, and so, how energy or job can be done:

    www.geocities.com/k_pullo/system3.htm

    I think there it's quite clear how can we use the excess-induced charge interaction to get a method of re-ordering charges without energy to do so. (If you look carefully, you'll see the excess electrons ALWAYS climb to a higher potential).

    Any comments are welcome
     
    Last edited: Apr 9, 2003
  24. Apr 9, 2003 #23

    Hurkyl

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    But C is nearly a closed cage; a fact upon which you relied in order to argue that connecting B and C will cause all of the charge to evacuate B.

    Thus, C will nearly prevent all currents from flowing from A to B, and a tiny positive charge (which I would be entirely unsurprised if it turned out to be equal to the positive charge remaining on B when C sucks away the charge) on B would be sufficient to prevent further current flow.

    Hurkyl
     
  25. Apr 10, 2003 #24
    Hurkyl, reading the document posted on "Geometry and Surface Density of Electric Charge" you'll find some answer to your doubts about my system.

    The document is The Feynman Lectures on Physics, Volume II. The relevant material occurs in section 6, pages 8 to 14. This material is posted without permission; it is copyrighted by Addison-Wesley. It is for educational fair use only.

    http://users.vnet.net/warrenc/V2_Ch06_1962-10-18_Electric_Field_in_Var_Circumstances.pdf [Broken]

    As you can see, the charge tends to distribute to surface of conductors and edges.

    And also, if you try to introduce charge into the cavity of a rounded conductor, it will have no opposition, because induced positive charges will appear to make the E=0 on this rounded conductor. In fact, the more charge you introduce, the more possibilities you can obtain a breaking current from the inner sphere to the rounded conductor. (The more negative charges in the cavity, the more ATTRACTION to the induced positive charges).
     
    Last edited by a moderator: May 1, 2017
  26. Apr 10, 2003 #25

    Hurkyl

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    The link doesn't work, it has the ellipsis in it.


    The reason that extra electrons accumulate on boundaries of surfaces is because internal electrons push others towards the boundary and there's nothing past the boundary to push them back.


    However, there is something to push back when the electrons are on the connecter between the inner and outer spheres. In the skin (which acts as a faraday cage) there would be currents induced which oppose the flow of charge from outer to inner spheres.

    Also, since you brought up edge effects, I've thought about them more and realize that they have a more important role than I previously thought; charge will tend to accumulate on the skin around the hole, which would provide a sizable clump of negative charges perfectly capable of providing that external push-back to impede the ordinary spreading of electric charges along a boundary. Let me restate this:

    If we ignore the charges on the skin, the net negative charge on the outer sphere will push electrons towards the inner sphere. Since there's nothing to oppose this, current will flow from the inner sphere to the outer sphere (i.e. electrons go from outer to inner).

    However, if we have a nice chunk of electric charge surrounding the connection between the outer and inner spheres, they will provide electrostatic repulsion that tends to push charges back towards the outer sphere, thus preventing the idealized situation where the charge divides evenly between the spheres.


    And in any of the cases, I wonder how much would collect on the connection.

    Hurkyl
     
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