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Could you please explain why Q=C*V?

Discussion in 'Electronic Basics' started by Boki, Jul 24, 2003.

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  1. Boki

    Boki Guest

    Hi, All:

    Could you please explain why Q=C*V?

    Thanks.

    Boki
     
  2. Don Kelly

    Don Kelly Guest

     
  3. grahamk

    grahamk Guest

    The larger the Capacitor is, and the higher the Voltage is, then the larger
    the charge( Q) is.
     
  4. That's a dimensional analysis way of looking at it. For a given
    capacitor, the higher the voltage the more charge you expect,
    so:

    q proportional to V

    and for a given voltage, the bigger the capacitor is the more
    charge you expect, so:

    q proportional to c

    There don't appear to be other obvious factors at this point,
    so:

    q proportional to c*v

    or

    q = k1 * c * v

    with 'k1' being yet to be determined.

    Or, rearranged, C = k2*Q/V.

    Turns out, C is *defined* so that k2 is just 1.000 when Q is in
    Coulombs and V is in Joules per Coulomb. Capacitance is in
    Coulumb^2/Joule.

    Add to this, an Ohm is a Joule-sec/Coulomb^2 (a Joule-sec is a
    measure of angular momentum, for example, and is the units of
    Planck's constant -- it's also just kg*(m/s)*m.)

    Multiply units of R*C:

    (Joule-sec/Coulomb^2) * (Coulomb^2/Joule)

    and you get seconds! Good thing.

    Since a Henry is just a Joule-sec^2/Coulomb^2 (a kind of
    Ohm-second or Joule per amp^2), the SQRT(L*C) also works out to
    seconds. Also a good thing. And L/R is also in seconds.
    Another good thing.

    Jon
     
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