1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Delphi51

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Stored charege and potentials of two capacitors in series
Loading...