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Parallel Plate Capacitor (work and potential difference)

  1. Sep 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Professor Milly Coulomb finds it takes 3.0 J of work to drag, at constant speed, a -9.0 mC charge between the plates of a parallel plate capacitor.

    (a) What is the work done by the electrical force in taking the charge between the plates?

    (b) What is the potential difference between the plates of the capacitor?

    2. Relevant equations

    PE = (QV) / 2
    V = 2PE / Q

    3. The attempt at a solution

    b) V = 2PE / Q
    V = (2 * 3) / -.009
    V = -666.667 V

    For part b, I tried both positive and negative values but they were both wrong.
    For part a, I tried the values of 0 and 3 but they were wrong also.

    Thanks for anyone who helps! :wink:
  2. jcsd
  3. Sep 26, 2008 #2


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    a) You have to be more explicit as to what you tried here. Positive and negative values of what?

    b) Why did you multiply it by 2?
  4. Sep 27, 2008 #3
    a) +/- 666.67 as answers for the problem when changing the charge. They were both incorect.

    b) I based it off the formula PE = .5QV
  5. Sep 27, 2008 #4


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    a) Work is done to move charge across a potential difference. This work is done against the field. So what is the work done by the field?

    b) The formula 1/2QV gives you the amount of energy stored in the capacitor, with Q being the amount of charge on the capacitor plates, not as you have interpreted it, the amount of charge the person has moved from one plate to the other.

    So for this, you have to think back to what constitutes the work done in moving the charge across the plates. In particular, think of what is the work done in moving a charge across a potential difference V?
  6. Sep 27, 2008 #5
    I figured out part a, but part b is confusing given the problem. I found a formula in my textbook about potential difference between the two capacitor plates as: V = Ed. However this cannot be found as E and d are not given in the problem. I also know that V = Q/C, but capacitance is unknown as well.
  7. Sep 27, 2008 #6


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    So, in this case, you can't rely on that formula to do it. As I said earlier, think in terms of the formula you have for work done to move a charge q over a potential difference V. This definition is more fundamental than the concept of capacitance. You don't need to understand capacitance to come up with that.
  8. Sep 28, 2008 #7
    Okay, I finally got it now. Thanks a lot!
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