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Parallel Plate Capacitor

  1. Sep 15, 2010 #1
    Given: There are two parallel plates. The left plate of a parallel plate capacitor carries a positive charge Q, and the right plate carries a negative charge -Q. The magnitude of the electric field between the plates is 100 kV/m. The plates each have an area of 2 × 10-3 m2, and the spacing between the plates is 6 × 10-3 m. There is no dielectric between the plates. Calculate the energy stored in the capacitor.

    I found C is 2.95 x10^-12 by taking ((8.85x10^-12)(2 x10^-3))/(6x10^-3).

    I know:
    the electric field is uniform and that V=E*d
    The potential difference is just the magnitude of the electric field times the distance you move in the electric field

    When I try to solve for V or PE, I keep getting the Power of ten error.

    Can someone please help me?

    Thank you:)
     
  2. jcsd
  3. Sep 15, 2010 #2

    Delphi51

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    The value you have for C looks good to me.
    What formula are you using for the energy of the capacitor?
    Note that in V = E*d, the E is not energy. Also, the potential V is not the energy; it is the energy per charge.
     
  4. Sep 15, 2010 #3
    I'm not exactly sure what formula to use. I am kind of stuck.
     
  5. Sep 15, 2010 #4

    Delphi51

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  6. Sep 15, 2010 #5
    Thank you so much for your help:)

    I figured it out. I had to find the Voltage which is 600. Then plug it into the equation: PE=1/2CV^2
     
  7. Sep 15, 2010 #6

    Delphi51

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    Most welcome!
     
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