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Redistributed Charge (Capacitors)

  1. Apr 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Two capacitors, C1 = 2910 pF and C2 = 2230 pF, are connected in series to a 12.0 V battery. The capacitors are later disconnected from the battery and connected directly to each other, positive plate to positive plate, and negative plate to negative plate. What then will be the charge on each capacitor? (Answer needs to be in units of Coulombs)

    2. Relevant equations
    Q = C*V
    V = Q/C
    (Total capacitance): CTotal = 1/C1 + 1/C2
    (Charge of capacitors in series is equal) Q of C1 = Q of C2



    3. The atempt at a solution
    I found the total Capacitance by doing the following:
    1/C1 + 1/C2 = 1/( 1/2910 + 1/2230) = 11262.509 PF
    I found the Charge of the Capacitors by doing the following:
    Q = C*V = (1262.509pF)*(12V) = 1.5150108*10^-8 Coulombs (after conversion)
    From here on is where I'm a bit confused. So what I did is:
    V = Q/C =(1.5150108*10^-8 Coulombs)/(1.2625*10^-9F) = 12.00008554 Volts (this doesn't make sense to me because it is basically the voltage that was given in the problem)
    From there:
    q1 (Charge after disconnecting) = C*V = 1.2625 (12) = 1.5150108*10^-8 Coulombs (which is what the charge was before disconnecting)

    I am assuming that the charge on the Capacitors changes after disconnecting? It doesn't make sense that my calculations say that the charge is still the same. Can someone please help me out and point me in the right direction?
     
    Last edited: Apr 11, 2015
  2. jcsd
  3. Apr 11, 2015 #2

    Simon Bridge

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    After you remove the battery and separate the capacitors, but before reconnecting then in parallel, what is the charge and voltage on each cap?

    The first equation is false as you wrote it.
    You write that the combined capacitance is 11262pF but use 1262pF in the next calculation... is that a typo?
    The next equation is false as written... another typo? ...what value did you use for capacitance and why?
     
  4. Apr 11, 2015 #3
    Q=C*V is an equation straight out of my textbook, and so are the rest of the equations actually
    and yes that was a typo, 1/C1 + 1/C2 = 1/2910 + 1/2230 = 1262.509 pF is what it should have said.
     
  5. Apr 11, 2015 #4
    Maybe I need to clarify what the equation stands for:
    Q=C*V
    Q=charge of capacitor, C is Capacitance, V is Voltage

    The second equation, V=Q/C is to find voltage, and it's just Q=C*V rearranged because I'm answering for V
     
  6. Apr 11, 2015 #5

    Simon Bridge

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    When you use an equation out of your text book, you have to adapt it to your needs. Leaving it in the standard form can be confusing to the reader.

    1/C1 + 1/C2 = 1/2910 + 1/2230 = 1262.509 pF still evaluates as false. The units don't match up.
    probably you mean:
    1/(1/C1 + 1/C2) = 1/(1/2910pF + 1/2230pF) = 1262.509 pF

    On the third line you wrote:
    Q/C =1.5150108*10^-8 = 12.00008554
    ... this is also false.
    I'm guessing that there is a number missing, probably the capacitance. Is that correct?

    You seem to be using the same variables to mean different things and some different variables to mean the same things... its hard to tell.

    Excuse if I sound terse, Im on mobile.
     
    Last edited: Apr 11, 2015
  7. Apr 11, 2015 #6
    Sorry about that. For the first part, yes you're right, I just calculated the reciprocal of my answer which I didn't show work for and I see how that was confusing.

    Q/C =(1.5150108*10^-8 Coulombs)/1.2625*10^-9 Farads = 12.00008554 Volts
     
  8. Apr 11, 2015 #7

    Simon Bridge

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    Im a little slow and bits of our posts crossed. Ive added more see above...

    I can figure out common stuff like leaving out when you take the reciprocal but I have to guess and I don't like guessing.
    Whats more important is that you get into the habit of care when writing down physics as maths expressions... it helps with troubleshooting as well as communication.
    Make sure your expressions evaluate true. Make sure you use different variable names for different things.

    So what did you use for capacitance?
     
  9. Apr 11, 2015 #8
    Yeah I understand where you are coming from. Thanks for the advice, I'll try to keep my units consistent in my problems and when trying to solve problems.

    My capacitance that I used was 1.2625*10^-9 Farads.
     
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