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Capacitance and distance between plates

  1. May 16, 2009 #1
    1. The problem statement, all variables and given/known data
    The question is at
    http://dl.getdropbox.com/u/175564/capacitance.JPG [Broken]


    2. Relevant equations
    Q = CV
    C = [tex] \epsilon_{0} A / d [/tex]

    3. The attempt at a solution

    The right solution is Q^2 / C that is (h).

    I know that C is proportional to the inverse of the distance between plates.
    Thus, I suggests that the capacitance should one third of the original value.

    The other problem is to have a relation between capacitance and work.
    I know W=Fd, where we get a relation to capacitance by d.
    This suggests me that the work should one third of the work done with the initial distance between plates in the capacitor.

    How would you solve the problem.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 16, 2009 #2

    tiny-tim

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    Homework Helper

    Hi soopo! :smile:
    Yes, that's right! :smile:
    You're making it very complicated :redface:

    work done is energy …

    the https://www.physicsforums.com/library.php?do=view_item&itemid=75" getting from one state to another is the difference in energy between them (in a conservative system)

    So, although, as you say, work done = ∫force.distance (= ∫charge.voltage), it's much easier to use the standard energy equation for a https://www.physicsforums.com/library.php?do=view_item&itemid=112", which is … ? :smile:
     
    Last edited by a moderator: Apr 24, 2017
  4. May 16, 2009 #3
    The equation seems to be

    W = .5 CV^2

    It seems to be obtained by integrating CV with respect to V.
    The problem is now to know the upper and lower bounds for the voltage in the integral.
    I suggest that the upper one is infinity and the lower one is zero.

    If the formula is right, then
    V^2 = 2Q / C

    However, I cannot see how to get the result.
    This suggests me that the equation for the work done in the capacitor is not correct.
     
    Last edited by a moderator: Apr 24, 2017
  5. May 16, 2009 #4

    tiny-tim

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    Hi soopo! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    (whyever are you using W for energy? :confused:)

    No, it's obtained either by integrating ∫(1/2)QEdx or ∫QVdQ … see the PF Library on https://www.physicsforums.com/library.php?do=view_item&itemid=112"

    Anyway, since Q = CV, the energy can also be written .5 Q2/C, which should give you the result almost immediately. :wink:
     
    Last edited by a moderator: Apr 24, 2017
  6. May 16, 2009 #5
    Thank you for your answers!
     
    Last edited by a moderator: Apr 24, 2017
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