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Capacitors Please HELP

  1. Jan 17, 2008 #1
    Capacitors...Please HELP!!!

    1. The problem statement, all variables and given/known data
    An uncharged capacitor is connected to the terminals of a 1.02 V battery, and 1.69 μC flows to the positive plate. The 1.02 V battery is then disconnected and replaced with 6.78 V battery, with the positive and negative terminals connected in the same manner as before. How much additional charge flows to the positive plate?

    2. Relevant equations

    3. The attempt at a solution
    I am not really sure how to start this problem without having the capacitance. Can someone help me get started..thanks!!
  2. jcsd
  3. Jan 17, 2008 #2
    Can I set these up as ratios?
  4. Jan 17, 2008 #3
    Capacitance is just charge/voltage. You could set up two ratios equal (capacitance isn't changing with additional voltage, but as voltage increases you must increase charge to hold the capacitance constant.
  5. Jan 17, 2008 #4
    So I could do...
    1.69x10^-9 C/1.02 V = Q/6.78 V
    Q = 1.12x10^-8 C
    Then subtract them to get my answer being...
    9.54x10^-9 C
  6. Jan 17, 2008 #5
    No thats not right...ahh what am doing wrong??
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