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What is the capacitance of the system?

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Two metal objects have net charges of +70pC and -70pC, which result in a 20V potential difference between them.

    a.) What is the capacitance of the system?
    b.)If the charges are changed to +200pC and -200pC, what does the capacitance become?

    2. Relevant equations

    Q=CV
    [itex]C=\epsilon A/d[/itex]

    3. The attempt at a solution

    a.) Q=CV
    C=Q/V = 70e-12 / 20 = 3.5pF

    b.) I know the capacitance only relies on the area and the distance between the plates. But that is not obvious from Q=CV. Why can't this equation be used to find the answer by substituting in the 200pF for the charge and 20V? It's not obvious from these equations why and the book's descriptions absolutely suck.
     
  2. jcsd
  3. Sep 20, 2009 #2

    rl.bhat

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    Re: Capacitance

    If you change the charges, potential difference will change.
    If you remove the charges completely, still there will be the capacitor.
     
  4. Sep 20, 2009 #3
    Re: Capacitance

    Doesn't really help at all.
     
  5. Sep 21, 2009 #4

    rl.bhat

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    Re: Capacitance

    In the formula C = εA/d, there is no charge term of the potential difference term. The capacity of a capacitor independent of these quantities.
     
  6. Sep 21, 2009 #5
    Re: Capacitance

    Typing mistake noted. You meant to type 200 pC, not 200 pF.

    The potential difference won't be 20 V anymore. Q=CV tells you that, when you change Q, you also change V.
     
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