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Finding the charge on a capacitator

  1. Apr 23, 2009 #1
    3 capacitators, C1=8μF C2=4μF C3=8μF, are connected in a colum, over a potential difference of 12V, what is the charge of C2??

    i know that connecting the capacitators in a colum, ΣC=(Ca*Cb)/(Ca+Cb)

    also i know that
    C=Q/V, which i think must be the equation i need since i am looking for Q, but i dont think the V i know is the right V, i think its the V for the whole system.

    ΣC1,2=(C1*C2)/(C1+C2)=(8*4)/(8+4)=(8/3)μF

    ΣC1,2,3=(C3*C1,2)/(C3+C1,2)=(8*(8/3))/(8+(8/3))=2μF

    so now i know Vtotal=12V and Ctotal=2μF

    C=Q/V
    Q=C*V
    Qtotal=2*12=24μC

    but how do i find Q2 from the Qtotal
     
  2. jcsd
  3. Apr 23, 2009 #2

    Cyosis

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    The V is indeed the potential difference over the equivalent capacitor.

    Do you know why capacitors in series add up like 1/C=1/C1+1/C2+....+1/Cn? Imagine at the very start the first plate of the capacitor is getting charged. What will this do to the plate below and the plate below and the plate below etc?
     
  4. Apr 23, 2009 #3
    a sort of chain reaction, 1st will charge 2nd will charge 3rd.... so they will all have the same charge?? meaning that
    Q2=(1/3)*Qtotal

    Q2=Q3=Q1=8μC

    is that right?
     
  5. Apr 23, 2009 #4

    Cyosis

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    You're almost right. The chain reaction you describe is correct and indeed all three capacitors have the same charge. However this charge is not 1/3 of Qtotal.

    Lets call the potential difference over the first capacitor V1, over the second V2 and over the third V3. Using your chain reaction argument we know that Q=Q1=Q2=Q3. We also know that in series V=V1+V2+V3 and we know C=Q/V.

    Now lets calculate them separately:
    V1=Q/C1
    V2=Q/C2
    V3=Q/C3

    V=V1+V2+V3=Q(1/C1+1/C2+1/C3). Looks familiar? Now calculate this Q and see that Q1=Q2=Q3=Q=Qtotal.
     
  6. Apr 23, 2009 #5
    so it would probably be easier to just work it out like that from the beginning instead of adding up all the C's,,
    thanks
     
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