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vineel
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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.
http://img39.imageshack.us/img39/7147/capacitors.png
S2 is now opened after C1 has been fully charged and S1 is then closed. What is the final voltage?
[itex]C = \frac{Q}{V}[/itex]
[itex]U = \frac{1}{2} C V^{2}[/itex]
[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?
Homework Statement
http://img39.imageshack.us/img39/7147/capacitors.png
S2 is now opened after C1 has been fully charged and S1 is then closed. What is the final voltage?
Homework Equations
[itex]C = \frac{Q}{V}[/itex]
[itex]U = \frac{1}{2} C V^{2}[/itex]
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?
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