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Stored charege and potentials of two capacitors in series

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Two capacitors of 2.0F and 5.0F are connected in series; the combination is connected to a 1.5 volt battery. What is the charge stored on each capacitor? What is the potential difference across each capacitor?

    2. Relevant equations
    C = Q / (Delta)V

    3. The attempt at a solution

    Since the capacitors are in series, we can find the equivalent capacitance by solving:

    1/Ceq = 1/2.0F + 1/5.0F
    Ceq = 1.429F

    I then used this value to find the charge on each plate of each capacitor

    C = Q/V
    CV = Q
    1.429F * 1.5V = 2.143

    So now I now the amount of charge on each plate.

    Using this value in conjunction with a specific capacitance, I can find the potential difference across any given capacitor whose capacitance I know.

    so, V = Q/C
    2.143/2 = 1.072V
    2.143/5 = .429V

    Thank you for the help.
  2. jcsd
  3. Feb 25, 2009 #2


    User Avatar
    Homework Helper

    Re: Capacitors

    All correct, I think. I used a slightly different method, adding the voltages around the loop to get Q/2 + Q/5 = 1.5
    which yields the same answer you got.
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