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Homework Help: Capacitance of capacitor

  1. Jan 27, 2007 #1
    1. The problem statement, all variables and given/known data
    The two plates of a parallel-plate capacitor initially carry equal amount of positive charge. If some charges are transferred from one plate to another, the charges on the plates are respectively +900uC and +100uC. The potential difference across the plates becomes 4V. What is the capacitance of the capacitor?


    2. Relevant equations

    C=Q/V

    3. The attempt at a solution
    I just pick randomly pick +900uC instead of +100uC:
    C=900/4=225uC

    4. Then I check the answer:

    Totally 400uC moves away from one to the metal plate and move to another plate, so:
    C=400/4=100uC

    According to what I have learn, Q is the charge of one of the metal plate. In here, why should I subsitute 400uC, the amount of charge that is transferred?
     
  2. jcsd
  3. Jan 27, 2007 #2

    Hootenanny

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    Consider the relative charge...

    Have you met Gauss' law?
     
    Last edited: Jan 27, 2007
  4. Jan 28, 2007 #3
    I haven't heard about it before. I will try to learn it by myself. Thank you:rolleyes:
     
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