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Electric Potential, Work, and Capacitance

  1. Feb 10, 2008 #1
    1. The problem statement, all variables and given/known data
    A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 2.60 µC.

    How much work is required to quadruple the separation?


    2. Relevant equations

    [tex]C = \frac{Q}{V}[/tex]

    [tex]V = Ed[/tex]


    3. The attempt at a solution

    In previous parts of the question, I was able to find, correctly, the potential difference before ([tex]V_i = 2.83\times 10^3\ \textsc{V}[/tex]) and after ([tex]V_f = 1.13 \times 10^4\ \textsc{V}[/tex]).

    What's confusing me is what equation to use for work.

    I don't think I'm supposed to use [tex]W = \frac{1}{2}CV^2[/tex] because that finds the work needed to charge the capacitor.

    The only other equation I can think of is [tex]W = -\frac{QV}{d}[/tex], but I'm not sure how to implement it. I've tried changing it to [tex] W = -QE[/tex], but beyond that, I'm stuck.
     
  2. jcsd
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