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Capacitors - Conservation of Charge vs Conservation of Energy

  1. Feb 28, 2012 #1
    I am a TA for a physics teacher. I wrote a problem that the students did in the lab quiz. The students tried to use conservation of energy instead of conservation of charge, which I used. Both methods seem sound to me, but they produce different answers. I need help figuring out which method is correct, and why the other one is incorrect.

    1. The problem statement, all variables and given/known data

    http://img39.imageshack.us/img39/7147/capacitors.png [Broken]

    S2 is now opened after C1 has been fully charged and S1 is then closed. What is the final voltage?

    2. Relevant equations

    [itex]C = \frac{Q}{V}[/itex]

    [itex]U = \frac{1}{2} C V^{2}[/itex]

    3. The attempt at a solution

    [itex]V_{0} = \text{initial voltage} = 3V[/itex]
    [itex]C_{0} = \text{initial capacitance} = C_{1} = 470 \mu F[/itex]
    [itex]Q_{0} = \text{initial charge}[/itex]
    [itex]Q_{1} = \text{final charge on } C_{1}[/itex]
    [itex]Q_{2} = \text{final charge on } C_{2}[/itex]
    [itex]V^{'} = \text{final voltage}[/itex]

    Conservation of Charge:

    [itex]Q_{0} = C_{0} V_{0}[/itex]

    [itex]Q_{1} = C_{1} V^{'}[/itex]
    [itex]Q_{2} = C_{2} V^{'}[/itex]

    [itex]Q_{0} = Q_{1} + Q_{2} = C_{1} V^{'} + C_{2} V^{'} = (C_{1} + C_{2}) V^{'}[/itex]

    [itex]V^{'} = \frac{Q_{0}}{C_{1} + C_{2}} = \frac{C_{0} V_{0}}{C_{1} + C_{2}} = \frac{(470 \times 10^{-6} F) (3V)}{470 \times 10^{-6} F + 100 \times 10^{-6} F} \approx \textbf{2.474 V}[/itex]

    Conservation of Energy:

    [itex]U_{1} = \frac{1}{2} C_{1} (V^{'})^2[/itex]
    [itex]U_{2} = \frac{1}{2} C_{2} (V^{'})^2[/itex]

    [itex]U_{0} = U_{1} + U_{2} = \frac{1}{2} C_{1} (V^{'})^2 + \frac{1}{2} C_{2} (V^{'})^2 = \frac{1}{2}(V^{'})^{2}(C_{1} + C_{2})[/itex]

    [itex]\frac{1}{2} C_{1} V_{0}^{2} = \frac{1}{2}(V^{'})^{2}(C_{1} + C_{2})[/itex]
    [itex]C_{1} V_{0}^{2} = (V^{'})^{2}(C_{1} + C_{2})[/itex]
    [itex]V^{'} = V_{0} \sqrt{\frac{C_{1}}{C_{1} + C_{2}}} = (3V) \sqrt{\frac{470 \times 10^{-6} F}{470 \times 10^{-6} F + 100 \times 10^{-6} F}} \approx \textbf{2.724 V}[/itex]

    The answers are inconsistent. Which one (if at all) is correct?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 28, 2012 #2

    gneill

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    Staff: Mentor

    What are your own thoughts on the matter?
     
    Last edited by a moderator: May 5, 2017
  4. Feb 28, 2012 #3
    gneill,

    I personally feel that my method (conservation of charge) is the correct one. I feel as if conservation of energy does not apply for some reason.

    I am not sure at all, so I am posting this question.
     
  5. Feb 28, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    We can't just give you an answer here (Forum rules), but we can help to a solution...

    I suggest that you try a thought experiment. Suppose your circuit was the same as before but includes a resistance R between the two capacitors. Clearly some energy will be lost as heat in the resistor when S1 is closed and the current flows.

    Write the equation for the current I(t) in the circuit and then find the energy lost by integrating I2R from time 0 to infinity. What's the resulting expression? What does it depend upon?
     
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