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Parallel Plate Capacitors Being Separated

  1. Sep 25, 2008 #1
    1. The problem statement, all variables and given/known data

    A parallel-plate air capacitor is made by using two plates 10 cm square, spaced 4.7 mm apart. It is connected to a 15 V battery. If the plates remain connected to the battery and the plates are pulled apart to a separation of 9.9 mm.

    What is the new capacitance? What is the new charge on each plate?

    2. Relevant equations

    C = Q/V = (permittivity of freespace)(A/D)

    3. The attempt at a solution

    I attempted to solve for the new capacitance by simply substituting in the new separation value and it isn't working. Why?
  2. jcsd
  3. Sep 26, 2008 #2


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    Homework Helper

    Hi Fizzicist,

    What numbers did you use in the formula? and what answer did you get?
  4. Sep 26, 2008 #3
    What about your solution doesn't work? Are you using

    [tex]A = \pi (0.1m)^2[/tex]

    [tex] \epsilon_{0} = 8.854 \times 10^{-12} F/m[/tex]

    [tex] C = \frac{\epsilon _{0} A}{0.0099m}[/tex]


    [tex] Q = C \times 15V [/tex]?

    These are the equations you listed in your relevant equations section, and they should work!
  5. Sep 26, 2008 #4


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    Hi Clairefucious,

    I don't think this is correct here.

    I was not certain about what the poster meant by "two plates 10 cm square". I interpreted that to mean that the area is

    A= 10 \ {\rm cm}^2

    (and converting cm2 to m2 is a common place to make a mistake) but if the problem wasn't quoted directly it could mean a square with sides of length 10cm. Maybe the poster will clarify the problem.
  6. Sep 26, 2008 #5
    Yes, I used 10 cm^2. This might be the problem.
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