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Charge on a capacitor

Discussion in 'Electronic Design' started by ChrisGibboGibson, Nov 9, 2004.

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  1. If we charge an air-spaced cap to 100 volts. Then increase the spacing between
    the plates, what happens to the voltage?

  2. It goes up. There are several ways to imagine this without math.

    The opposite charges on the plates cause them to attract each other.
    It takes work to pull them apart against this attractive force. The
    increase in distance also lowers the capacitance. The electrical
    energy stored in the cap is 1/2 * C * V^2. The only way the energy
    can go up (because mechanical work was done on the cap, conservation
    of energy and all that) in spite of the drop in capacitance is if the
    voltage increases.

    Another way to visualize this is to have two similar capacitors except
    for the difference in spacing. Apply enough voltage across each to
    force the same charge to be displaced through each. The one with the
    bigger spacing (lower capacitance) will require more voltage to force
    the same charge to move. If there were some way to get them so far
    apart (or otherwise change the plates) that there was zero capacitance
    between them it would take an infinite voltage to move that charge.
  3. PN2222A

    PN2222A Guest

    Q = C V

    C = epsilon A / D (ignoring fringing effects)

    V = D / epsilon A (where epsilon A is a constant)

    so V rises directly as D increases
  4. Thank you John and PN

    Now what, exactly, causes the potential difference between the 2 plates?

  5. The electric field caused by the separated charges and the distance
    this field is spread across.
  6. PN2222A

    PN2222A Guest

    Um, the electrons?
    I'm not sure I understand the question.

    When opposite electric charges are separated, the electric field exerts a
    force on those charges. That force is the "electromotive force" -- e.g.
    the voltage.

    ft = 300MHZ minimum.
  7. nospam

    nospam Guest

    It goes up, and so does the stored energy to balance the work done pulling
    the plates apart.
  8. Whatever it was you used to charge the cap in the first place. :)

  9. Ban

    Ban Guest

    this question would be more appropriate in s.e.b.
    The metal conductor of which the plates consist has free electrons, which
    are vibrating between the grid of the nucleusses. In an uncharged state both
    plates have the same density of free electrons. A voltage source will "pump"
    some of those electrons from one plate to the other until the resulting
    electric field is in equilibrium with the applied voltage.
    Now one plate has a higher electron density than the other. If you remove
    the voltage source, this distribution stays the same. We call this
    "potential difference" and the amount of moved electrons "charge". The
    situation creates an electrostatic field.
    It is similar when you move a mass upwards, the energy is transformed into a
    potential energy, which is released when you drop the mass. The weight of
    the mass(or the charge Q) stays the same, but the energy(or voltage V)
    augments, the higher the mass is moved.
    Compare this to the formulas of PN, and you hopefully might have a few more
  10. Robert Baer

    Robert Baer Guest

    Lemme see.. Q = C * V
    So, decrease C and V goes up.
    Seems that has been done before in a rotary fashion to make rather
    high voltages.
    There may even be a patent somewhere on them thar new-fangled
  11. Active8

    Active8 Guest

    Wrong. Force is force. The voltage potential is the work (F.d)
    needed to move the charge divided by the amount of charge,
  12. CBarn24050

    CBarn24050 Guest

    This is completely wrong Ban, it's hard to imagine a more fundamental
  13. Ban

    Ban Guest

    I'm missing your correction here!
    I do not know what exactly you refer to, but the gravitational field and the
    electrostatic field have a lot in common. Look at Newtons Law of
    F = (-G* m* M)/r^2 F=force G=const. m= mass1 M= mass2
    and Coulombs Law:
    F = (k* q* Q)/r^2 F= force k= const. q= charge1 Q= charge2
    Now charge has a sign, whereas gravity is alway attracting, but the fields
    have similarities.
  14. Marlboro

    Marlboro Guest

    Or.. becauses human calls that the voltage, the energy space for what
    human called charge :)
  15. CBarn24050

    CBarn24050 Guest

    Yes quite so Ban, but that isn't what you said at the start of your post.
  16. James Meyer

    James Meyer Guest

    Doesn't anti-matter fall UP in a gravity field produced by ordinary

  17. I don't think the experiment has been definitively done, yet.

    But it has been pretty well tested that antimatter has the same sort
    of momentum as matter. It couldn't be contained in accelerators if
    this was not the case.
  18. Tim Wescott

    Tim Wescott Guest

    "So what we would really like to have is a laboratory experiment where
    we simply drop some antimatter in a lab, and see how fast it falls. "

    But they don't mention the difficulty of keeping one's lab notebooks
    intact when it touches down...
  19. Use The Force, Luke.

  20. So, it's not really anti-"matter", but just matter with the opposite

    Actually, I'm trying to ask this seriously. But I speculate about
    stuff like, can the equations that describe a proton be mapped onto
    the equations that describe a black hole?

    And I don't have the maths, which might be a good thing, since my
    head isn't all clogged up with "what everybody already knows." ;-)

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