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Homework Help: Charging a Capacitor

  1. Feb 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A 4.6 mu or micro FF capacitor is charged to a potential difference of 15.0 V. The wires connecting the capacitor to the battery are then disconnected from the battery and connected across a second, initially uncharged, capacitor. The potential difference across the 4.6 mu or micro FF capacitor then drops to 7 V. What is the capacitance of the second capacitor?

    2. Relevant equations

    Q = CV

    Capacitor in series
    1/Ceq= 1/C1+1/C2

    3. The attempt at a solution

    I found the charge of the capacitor in the first scenario when the potential difference is 15 V

    Q = CV
    = 4.6E-6 F *15 V
    = 7.35E-5 C

    Since the potential drops to 7 V, you know the capacitors are in a series so that means the charge remains the same on both capacitors but potential changes

    Capacitor in series
    1/Ceq= 1/C1+1/C2

    Q = Ceq * V

    7.35E-5 C = (1/(4.6E-6 F) +1/C2)*7 V

    I solved for C2...

    C2 = -4.6E-6 C

    - which is negative so its obvious that its wrong, but could someone tell me where I went wrong that would be greatly appreciated. Thank you.
  2. jcsd
  3. Feb 11, 2009 #2
    I think there is an error in your calculation of the charge on the first capacitor. It seems that your work is correct, but when I multiply (4.6 microF)*(15V) I get the charge to be
    6.9e-5 C. Since you know that the voltage drops to 7V across the first capacitor, you can treat the circuit as if there is initially a 7V battery connected to only the second capacitor, which is unknown. In this case, you know that V=7V and Q= 6.9e-5C. From here you can manipulate the equation so that C=(Q/V). Therefore, C should equal 9.9 microF.
  4. Feb 11, 2009 #3
    Your way makes sense, and its a lot simpler than what I tried. The answer didn't work though so I'm going to try to keep double checking figures, i can't figure out what else would be wrong.
  5. Feb 11, 2009 #4
    Aren't the capacitors connected in parallel? And isn't the amount of charge calculated in the first part the total amount of charge in the 2-capacitor system in part two?
  6. Feb 11, 2009 #5
    I'm pretty sure you were doing it correctly in your first post (except for the first charge calculation, I got 6.9*10-5 C like w3390). Draw out the 2-capacitor circuit in the way it's described and I think you'll see that they are in parallel. When calculating C2, your Ceq should be Ceq = (C1 + C2) and not 1/Ceq = (1/C1) + (1/C2).

    Are you arriving at the correct solution now?
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