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Work Done on a Parallel Plate Capacitor

  1. Mar 22, 2008 #1
    1. The problem statement, all variables and given/known data
    An isolated 5.60 microfarad parallel-plate capacitor has 4.90 mC of charge. An external force changes the distance between the electrodes until the capacitance is 2.00 microfarads.

    2. Relevant equations
    W = Fd
    c = Q/V

    3. The attempt at a solution
    My first instinct was to solve to the potential of the system in it's initial conditions:

    C = Q/V

    5.60*10^-6 F = .0049C/V
    V = 875 V

    Now is where I become a little confused. I'm not sure how I can solve for the distance between the plates in the final condition. Also, I'm not sure what values stay constant as the plates are separated.

    Any help is much appreciated!
    Last edited: Mar 22, 2008
  2. jcsd
  3. Mar 23, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Use C = Q/V. Note that V = Ed, where E is the field between the plates. Does the field change as the plates are separated?
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