1. Limited time only! Sign up for a free 30min personal 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!

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




Similar Discussions: Charging a Capacitor
  1. Charge on capacitor (Replies: 10)

  2. Charge in a capacitor (Replies: 11)

  3. Charging of Capacitors (Replies: 0)

  4. Charge on a Capacitor (Replies: 3)

  5. Charged capacitor (Replies: 1)

Loading...