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Series Capacitance

  1. Jul 29, 2005 #1
    2 Capacitors, C1 = 3*10^-12 f and C2= 6*10^-12 f are connected in series and the resulting combination is connected across 1000V

    a) the equivalent capacitance
    i got this 1/Ctotal = 1/C1 + 1/C2

    b) total charge on the combination and the charge taken by each capicator

    ok here is where i am confused because im not entirely sure what equation to use

    do I use C = Q/V
    but in my notes it says that it is teh potential difference between the capicators i dont' know how to calculate how much voltage each capicator is taking...so when i do the problem i got total voltage to be 2000V which can't be right

    c) the potential difference between each capacitor
    i'm assuming this is the difference in voltage...so i guess i need part B , but im' not entirely sure what its asking

    ok really any help would be awesome :)
  2. jcsd
  3. Jul 30, 2005 #2


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    Total chatge = Ctotal *v
    in series connection the potential difference is distributed but charge on each capacitor is same Q = C1V1 = C2V2 and V1 + V2 =Vtotal
    you may use Q = Ctotal*Vtotal
  4. Jul 30, 2005 #3
    In a series connection the magnitude of charges on all plates of a capacitor are the same. Since you know the equivalent capacitance then you can calculate the charge as this charge will be the same for all the capacitors.
  5. Jul 30, 2005 #4
    arrrrggh!!! lol i hate my text book i'm seriously it doesn't say that!!! no joke and same with my notes i had to research that online

    but thank you for your help that just solidifed with what i just learned. i apprecaite it =)
  6. Jul 30, 2005 #5


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    V is the voltage across the capacitor that is the negative potential difference between the capacitor plates. You have the capacitance of the resultant capacitor which is equivalent with the two capacitors in series. You can calculate the charge on this equivalent capacitor.

    See the picture. When you connect voltage to the serially connected capacitors the external plates get charged, but no charge can flow across the plates so the net charge on the internal plates stays zero. Because of influence, the positive and negative charges will be separated however, the left inner plate gets -q charge and the right one gets q one: the charges on the serially connected capacitors are the same as the charge on the resultant capacitor.

    Last edited: Jun 29, 2010
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