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2 capacitors

  1. Feb 23, 2005 #1
    Hi,

    I understand generally how to do problems with capacitors, but I am faced here with a problem about 2 capacitors and I am not sure what the appropriate calculations are for combining their capacitance, etc. Here is the problem:

    A 7.7 uF capacitor is charged by a 125-V battery and then is disconnected from the battery. When this capacitor is the connected to a second (initially uncharged) capacitor, C2, the final voltage on each capacitor is 15 V. What is the value of C2?

    So here is what I did:

    Q = C V
    Q = 7.7E-6 F x 125 V
    Q = 9.625E-4 C per plate (first capacitor)

    PE = 1/2 Q^2/C
    PE = 1/2 x (9.625E-4)^2/7.7E-6 F
    PE = .06

    From this, I used PE = 1/2 CV^2
    .06 = 1/2 (7.7E-6 F + x)(15^2)
    To get: 5.3E-4 F which I am pretty sure is way too large a number
     
  2. jcsd
  3. Feb 23, 2005 #2

    dextercioby

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    Are u sure it's not [itex] C_{2}\sim 5.3\cdot 10^{-5}F [/itex] ...?

    I got that by applying the law of conservation of electric charge (the battery is disconnected when the second capacitor is brought in)...

    Daniel.
     
  4. Feb 24, 2005 #3
    Hi,

    How did you get the answer you are giving? I am not sure I understand. If the battery is disconnected, C1 retains the charge it has from the battery, or 9.625E-4 C (per plate, right?). From what you said, I would do the following:

    The charge is going to be the same now because there is no other battery. So then that would be:

    Q = CV

    Would you multiply times 2 because there are two plates?

    .001925 = C x (15 + 15)

    So then C = 6.42E-5

    Is that right or totally wrong?
     
  5. Feb 24, 2005 #4

    dextercioby

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    Nope.The charge would move from one capacitor to another,as to insure equal potentials...

    Daniel.
     
  6. Feb 24, 2005 #5
    Hi,
    OK, I'm not sure I understand. So they would have equal potentials. So the total charge would be 1.625E-4. The voltage on each battery would be 15 (total = 30). So then I just need to find C2.

    Ctotal = C1 + C2
    C1 =7.7 uF
    So C2 = Ctotal - C1

    Is that it?
     
  7. Feb 24, 2005 #6

    dextercioby

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    Q=Q1+Q2 Charge conservation.What is C total...?

    Daniel.
     
  8. Feb 24, 2005 #7
    OK, concentration of electric charge. So you have the electric charge from the first part, and then you have that total charge for the second part, right? Because there is no other source, because there is no battery.

    I was thinking that each capacitor has its own capacitance, for C1 it is 7.7 uF and for C2, well that is what we are trying to find.

    What is Q = Q1 + Q2? So each capacitor gets half of the charge, so each capacitor gets 9.625E-4/2. So do I do:

    Q = CV
    (9.625E-4/2) = C2 x 15V

    ??
     
  9. Feb 24, 2005 #8

    dextercioby

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    No.Thay would get halp of the charge,iff the capacitors would have the same C,but my guess it they don't.

    Daniel.

    P.S.Provide a number for C_{2}
     
  10. Feb 24, 2005 #9
    How can I provide a number for C2 if that is what we are looking for?

    So if they don't each get half the charge (I thought they did?) then don't we have two variables?

    Because then we have Q = Q1 + Q2 AND C = C1 + C2 (I am assuming that there are two capacitances, yes?). So we know V = 30.

    We know that C2 is initially 0. And we know that C1 is initially 7.7 pF. And we know that the final voltage. And we know the final TOTAL charge but not the charge for each individual capacitor.

    I also don't understand what the voltage thing means. If the voltage is the same on each, does that mean they have the same capacitance?

    Sorry I'm starting to guess a little at this point. I am trying to think about it but I don't really understand.
     
  11. Feb 24, 2005 #10

    dextercioby

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    Q1=C1U
    Q2=C2U

    Add,factor U and extract C2.

    Daniel.
     
  12. Feb 24, 2005 #11
    Thanks Daniel. Sorry I'm a pain.
     
  13. Feb 24, 2005 #12

    dextercioby

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    There's not a problem,whatsoever;as a side thing,i gave you the hint in post #2...:tongue2:

    Daniel.
     
  14. Feb 24, 2005 #13
    Now that I get what you were saying,

    Q1 + Q2 = V (C1 + C2)
    9.625E-4 + 0 = 15 (7.7E-9 F + x)
    x= 6.4E-5?
     
  15. Feb 24, 2005 #14

    dextercioby

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    No.Check the numbers again.And the power is [itex] 10^{-6} [/itex]...:wink:

    Daniel.
     
  16. Feb 24, 2005 #15
    Right, I was thinking pF ... this is just getting worse and worse. Anyhoo, let's try this again:

    I know 9.625E-4 is right. And 15 is right, right? Because each of them gets multiplied by 15.

    So 9.625E-4 + 0 = 15 (7.7E-6 + C2)

    5.6E-5??????

    Please let that be right argh :cry:
     
  17. Feb 24, 2005 #16

    dextercioby

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    It looks okay.

    Daniel.

    P.S.What's with that "+0"...?
     
  18. Feb 24, 2005 #17
    Well the 0 was supposed to be for Q2, since Q2 is 0 (that is, the charge brought to the system by the second capacitor, that is how I am thinking about it even though that's not very physics-y, sounds more like a tupperware party).
     
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