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!

Disconnected and Reconnected Capacitors

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data
    A 100pF and a 400pF capacitor are both charged to 2kV. They are then disconnected from the voltage source and are connected together, positive plate to negative plate.

    (a). Find the resulting potential difference across each capacitor
    (b). Find the energy lost when the connections are made


    2. Relevant equations

    Q=C.V

    3. The attempt at a solution

    So when they are reconnected they are in parallel. However the charge is redistributed because positive plate is hooked to negative plate.

    I have calculated the total charge using Q=C.V= (100pF+400pF)*2kV=1[tex]\mu[/tex]C

    The I use Q1/V1=Q2/V2 and Q1+Q2=1[tex]\mu[/tex]C to calculate the new charge and get Q1= 0.2[tex]\mu[/tex]C and Q2=0.8[tex]\mu[/tex]C.

    However, the new voltage calculated using V=Q/C is 2*10^9 V which is larger than initial voltage. It does not make sense at all. Where did I screw up ?
     
  2. jcsd
  3. Nov 22, 2009 #2
    Do I have to calculate the initial charge of each capacitor and average them to 0.5microC and then calculate the voltage from there ?
     
  4. Nov 22, 2009 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Calculate the initial charges. As the capacitors are connected positive plate to negative, the total charge will be the difference of the initial charges. The new capacitor has capacitance C1+C2, the charge is Q2-Q1, from that you get the voltage.

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