Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:


    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


    User Avatar

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This cycle will eventually converge to a stationary system.


    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.

    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

    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:


    hope you'll find it usefull.
  8. Apr 6, 2003 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.

  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:


    (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


    User Avatar

    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.

  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


    User Avatar

    Staff: Mentor

    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


    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


    User Avatar


    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

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Electrostatic paradox ?
  1. Electrostatic discharge (Replies: 11)