Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Charge conservation in capacitors

  1. Jan 30, 2012 #1
    In the circuit, all elements are ideal.
    For time t<0, S1 remained closed and S2 open.
    At t=0, S1 is opened and S2 is closed.
    If the voltage Vc2 across the capacitor C2 at t=0 is zero, the voltage across the capacitor combination at t=0+ ??

    I tried this problem.
    Obviously, Vc1(0) = 3V ; Given that Vc2(0) = 0V.
    Then C1 = 1μF and C2 = 2μF are connected at t=0.
    So, Q1(0) = 3 C. Q2(0) = 0 C.
    At t=0+ how charges are distributed and hence the voltage ??
    I was thought (according to charge conservation) that, charges are distributed to two cap. such that voltage across them is equal. So, Q1 = 1 C , Q2 = 2 C. Hence Vc1 = Vc2 = 1 V.
    All these things happen at steady state. How about at t=0+ ?????
     

    Attached Files:

  2. jcsd
  3. Jan 30, 2012 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    charge one cap to 3V, then use it to charge another cap - what is the voltage?
    that the problem?

    see:
    https://www.physicsforums.com/showthread.php?t=234221

    you have no resistances (ideal components remember) so the steady state will be all-but immediate. What is your question?
     
  4. Jan 30, 2012 #3
    You are correct to say that the total charge on the 2 combined capacitors is the same as the charge on C1 at t=0.
    when S2 is closed you now have 2 capacitors connected in parallel so you should calculate the new capacitance, then you should be able to calculate V and the charge on each of the capacitors.
     
  5. Jan 30, 2012 #4
    Dear Simon,
    C1 is charged to 3V before t = 0.
    C1 and C2 are connected at time t = 0.
    As you say, C2 is to be charged by C1 (having a voltage of 3V).
    Now, pls tell me
    what will happen to the voltage across C1 and C2 at t=0+ (immediately after t=0)
     
  6. Jan 30, 2012 #5
    Dear technician,
    At time t = 0, the voltage across C1 is 3V ; while the voltage across C2 is 0V.
    Different voltages. How do u say that they are in parallel ?!?!
     
  7. Jan 30, 2012 #6

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Since you have ideal components, and no resisters, then at the exact time the switch is closed, the voltage drops in an instant to 1V. It is a step function. v(t)=3-2u(t) where u(t) is the unit step function.

    If you have some resistances in the circuit, i.e. the caps are not ideal, then v(t) drops exponentially to 1V ... see the link in my post for an example of how to do the calculation.
     
  8. Jan 30, 2012 #7

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    If you charge the first C then use it to charge the second, you have an insoluble problem unless you include some finite resistance and you 'admit' to some series Inductance in the connecting wires.
    Without resistance in circuit, you have energy conservation as well as charge conservation. The energy must exist as magnetic energy in the L, giving you a continuous oscillation of energy from C to C and back.
    If there is resistance, the 'excess' energy is dissipated in a simple exponential charge transfer.
     
  9. Jan 30, 2012 #8

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Oh I forgot about the oscillation ... lone ideal reactances in a circuit - one of those classical fudges like "rigid" bodies.
     
  10. Jan 30, 2012 #9

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    It can be a serious problem in some applications. Ringing involves over-voltage and EM interference.

    Not a "classical fudge" but a result of inadequate 'Classical" modelling by some teachers.
     
  11. Jan 30, 2012 #10

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yeh - we don't expect models to give good pictures in some situations.
    I have to be alert for that in questions.
     
  12. Jan 30, 2012 #11

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    The 'two capacitor' paradox is a good one for students at Christmas / fun times, just when they are getting too damned smart!
     
  13. Jan 30, 2012 #12
    This is an easy problem to solve in practice by doing the experiment!! You may not be able to get your hands 1F and 2F capacitors but it is not difficult to get capacitors that will give answers.
    Are these capacitors in parallel ?....yes.
    The symbol for a capacitor represents the parallel plates of a capacitor. When S2 is closed the 'top plates ' of C1 and C2 are connected together to give a plate of larger area. The bottom plates are already connected as shown in the diagram.
    The combined capacitance of this arrangement is 3F
    The charge on C1 at the start is Q = CV = 1 x 3 = 3Coulombs
    The same charge is there when the switch S2 is closed so now a capacitor of 3F has a charge of 3Coulombs. The voltage across the combination = 1Volt.
    The charge on C1 is now 1Coulomb and the charge on C2 is 2Coulombs
     
  14. Jan 30, 2012 #13
    Another good 'paradox' for students is to ask what happens to energy stored on capacitors when they are connected.
     
  15. Jan 30, 2012 #14

    jim hardy

    User Avatar
    Science Advisor
    Gold Member
    2016 Award

    ""How about at t=0+ ?????
    ""
    i think OP is asking what happens when you divide by zero.
    In world of math thinking, you have a discontinuity and it's indeterminate.
    Its limit approaches infinity . But "it" is not real anymore.

    3 volts across zero ohms would be infinite amps , were it not for Sophie's real world inductance and resistance of the wires. When you force them out you make "it" not real anymore.

    It's analogous to the thermodynamics experiment of a gas confined then the wall immediately removed. You can only define the end states.

    observe with zero ohms, charge is concerved but energy is not.


    energy in a cap is 1/2 (C*V^2)
    when open S1, C1= 1uf at 3volts , 1/2C*V^2 = 1/2 X 1 X 9 = 4.5 joules
    after close S2, C1 + C2 = 3 uf at 1 volt, 1/2 X 3 X 1^2 = 1.5 joules
    Those are the end states. 3 joules are gone.

    what happened at instant of closing S2? musta been some energy radiated into space .

    EDIT ahh,, Tech was here already..
     
  16. Jan 30, 2012 #15
    I hope I have not mis-read the original post !!!! I thought that an explanation of the final voltage with some reference to charge conservation was required.
    I hope there is no doubt or misunderstanding about what would happen if you got 2 capacitors and a battery and did this exercise for real!!!!
     
  17. Jan 30, 2012 #16

    jim hardy

    User Avatar
    Science Advisor
    Gold Member
    2016 Award

    i thought he was asking about instant of switch closure
    but i could be wrong.
    maybe i spoke to wrong question.

    indeed charge is conserved as you pointed out
    and energy goes someplace.
     
  18. Jan 30, 2012 #17
    Jim:
    I think you could be right.... his last sentence....'all these happen at steady state' suggests that.
    It will be interesting to see what he thinks of this discussion.
    Cheers
    If this is the case then it is not a simple circuit with 2 capacitors and a battery !!
     
    Last edited: Jan 30, 2012
  19. Jan 30, 2012 #18

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Whatplace? ;-)
     
  20. Jan 30, 2012 #19
    Heat in any connecting wires ? Electro-magnetic radiation from connecting wires?
    (this is way off the original OP and I hope it does not cause confusion for the OP)
     
    Last edited: Jan 30, 2012
  21. Jan 30, 2012 #20

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    I think it is bang on topic.
    If you calculate the resulting charges and then the resulting energy, you find you've lost energy. It's a small step from that to the ensuing discussion and it's a salutary lesson about the shortcomings of 'ideal' models.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Charge conservation in capacitors
  1. Charging the capacitor (Replies: 4)

  2. Capacitor Charge Time (Replies: 5)

Loading...