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Connected capacitor

  1. Sep 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A 2.50 µF capacitor is charged to 857 V and a 6.80 µF capacitor is charged to 652 V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each ?


    2. Relevant equations
    Q = CV


    3. The attempt at a solution
    I think I got it right

    Q before connected = Q after connected
    C1V1+C2V2 = (C1+C2) V

    (2.5 x 10-6) (857) + (6.8 x 10-6) (652) = (2.5 x 10-6+6.8 x 10-6) V

    V = 707.11 Volt


    My question is : how about if the condition "the positive plates are connected to each other and the negative plates are connected to each other" is reversed so that positive plate of one capacitor is connected to the negative plate of the other. What will happen? Is it still the same?

    Thanks
     
  2. jcsd
  3. Sep 9, 2009 #2

    kuruman

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    If the capacitors were charged to the same charge (Q1 = Q2 = Q) and you connected them positive-to-negative, the charge on each would be zero [=Q+(-Q)]. So what do you think will happen if the positive and negative charges are not the same?
     
  4. Sep 9, 2009 #3

    rl.bhat

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    No.
    In that case
    C1V1 - C2V2 = (C1 + C2) V.
     
  5. Sep 9, 2009 #4
    I have the same solution! If you got a course in circuit analysis you can imagine that the capacitors are put in parallel. So the new capacitor, which results out of the parallel connection of C_1 and C_2 has the capacitance C_(entire) = C_1 + C_2 and the whole charge saved in C_1 and C_2 is distributed until equilibrium. V = Q_(entire) / C_(entire)!


    If the positive plate of one capacitor is connected to the negative plate of the other, all positve charges, which find a negative partner are going to be electric neutral, so the Voltage is than V = (Q_2 - Q_1) / C_(entire)!

    But aware of doing it at home, cause there is no resistance which limits the current flow! If the potential difference is vast enough the capacitors will burst!
     
    Last edited: Sep 9, 2009
  6. Sep 9, 2009 #5

    rl.bhat

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    V = C_(entire) / (Q_2 - Q_1)!
    It should be V = (Q_2 - Q_1)/C_(entire)
     
  7. Sep 9, 2009 #6
    embarrassing !!

    thanks for the hint rl.bhat
     
  8. Sep 9, 2009 #7
    Hi kuruman, Mr. rl.bhat, and saunderson
    Do you mean that my work is wrong?

    So, when positive plate of one capacitor is connected to the negative plate of the other :
    V = (Q_2 - Q_1)/C_(entire)
    = ((6.8 x 10-6) (652) - (2.5 x 10-6) (857)) / (2.5 x 10-6+6.8 x 10-6)
    = 246.35 volt

    Do I get it right?
     
  9. Sep 9, 2009 #8

    rl.bhat

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    Yes. It is right.
     
  10. Sep 9, 2009 #9
    Hi Mr. rl.bhat

    Can you please explain what you mean by C1V1 - C2V2 = (C1 + C2) V ?

    Thanks :)
     
  11. Sep 9, 2009 #10
    Yes, your result is correct! If you want to experiment later on, download qucs from http://qucs.sourceforge.net/" [Broken] and test your results ;) (takes some time to break in)
     
    Last edited by a moderator: May 4, 2017
  12. Sep 9, 2009 #11

    rl.bhat

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    C1V1 is the charge Q1 in capacitor C1 and C2V2 is the charge Q2 in capacitor C2.
     
  13. Sep 9, 2009 #12
    Hi saunderson and Mr. rl.bhat

    Oh sorry, I misinterpret your post Mr. rl.bhat. I thought you said that when the positive plates are connected to each other and the negative plates are connected to each other, the equation should be C1V1 - C2V2 = (C1 + C2) V.

    Thanks a lot to all of you ! (kuruman, saunderson and Mr. rl.bhat ) :smile:
     
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