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Voltage Across Two Capacitors In Series

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data
    photo.jpg


    2. Relevant equations
    Q=CV


    3. The attempt at a solution
    For part A, I've tried working out the algebra, with the assumptions that V1 + V2 = V and Q1 + Q2 = Q

    Basically, I solved Q by finding the equivalent capacitance, then said Q1=C1*V1 and Q2=C2*V2
    Then by my second assumption, said Q - Q2=C1*V1.

    Next I eliminated Q2 by combining those equations. At this point the only two unknowns were V1 and V2 which I solved for using the first assumption. As you can imagine this resulted in an algebra nightmare. I'm afraid that I am completely off track with this problem. I there an easier way to solve this?

    Thanks
     
  2. jcsd
  3. Sep 19, 2010 #2

    Delphi51

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    Homework Helper

    You are missing the key: the charges on the two capacitors are equal. That is because any electrons taken away from one go onto the other.

    (a) is a strange question, assuming you can't use the C1 = 10C2 info given in (b). I guess the (a) answers will be messy and include Q, V, C1 and C2 in their expressions.
    The (b) answers will be very simple.
     
  4. Sep 19, 2010 #3
    Thank you very much for your reply. That makes the problem MUCH simpler. I had actually been stumped before about how current was even able to flow across a capacitor at all, then I read about how charge collecting on one plate would cause charge to move off the opposing plate, causing a current. I just didn't realize that this would imply that the charge across the two capacitors would be the same.

    I did have one other question, and that was if my assumption that V1 + V2 = V is correct. I can't think how this could be proven, and I know that it's not true if the capacitors are in parallel, but it seems like it should be, just by intuition. But it seems that my intuition and physics are rarely the same.

    Thanks so much for your help.
     
  5. Sep 19, 2010 #4

    Delphi51

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    Homework Helper

    No doubt about that.
     
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