Capacitors - Conservation of Charge vs Conservation of Energy

In summary, the conversation discusses a problem given to students in a lab quiz involving capacitors and switches. The students attempted to solve the problem using conservation of energy instead of conservation of charge, but the two methods resulted in different answers. The conversation ends with a suggestion to try a thought experiment involving a resistance in the circuit to determine which method is correct.
  • #1
vineel
2
0
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.

Homework Statement



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?

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?
 
Last edited by a moderator:
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  • #2
vineel said:
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.

Homework Statement



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?

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?

What are your own thoughts on the matter?
 
Last edited by a moderator:
  • #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.
 
  • #4
vineel said:
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.

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?
 
  • #5


Both methods are sound, but the correct method to use in this situation is the conservation of charge. This is because capacitors are devices that store charge, not energy. While the energy stored in a capacitor can be calculated using the equation U = 1/2 CV^2, this is not the most appropriate method to use in this scenario.

The conservation of charge method is more appropriate because it takes into account the fact that the total charge in the circuit remains constant. When S2 is opened, the charge on C1 will redistribute between C1 and C2, but the total charge in the circuit remains the same. Therefore, using the equation Q = CV, we can determine the final voltage. This method is also more accurate because it takes into account the capacitance of both capacitors, while the conservation of energy method only considers the capacitance of C1.

In conclusion, the conservation of charge method is the correct method to use in this situation. It is important to understand the underlying principles and properties of the components in a circuit in order to choose the appropriate method for solving a problem.
 

1. What is the difference between conservation of charge and conservation of energy in capacitors?

The conservation of charge states that the total electric charge within a closed system remains constant over time. In capacitors, this means that the amount of charge stored on the plates will not change unless a current flows in or out. On the other hand, the conservation of energy states that energy cannot be created or destroyed, only transferred or converted. In the case of capacitors, this means that the total energy stored in the electric field between the plates remains constant, even as the charge on the plates changes.

2. How do capacitors demonstrate the principle of conservation of charge?

Capacitors demonstrate the principle of conservation of charge by storing an equal amount of positive and negative charge on their plates. When a voltage is applied, electrons flow onto one plate, creating a negative charge, while an equal number of electrons are pulled from the other plate, creating a positive charge. This charge remains constant as long as the voltage is maintained, demonstrating the conservation of charge.

3. Can energy be lost in a capacitor due to resistance?

Yes, some energy can be lost in a capacitor due to resistance in the circuit. When a capacitor is charging or discharging, there is a flow of current through the circuit, which encounters resistance. This resistance causes some of the electrical energy to be converted into heat, resulting in a loss of energy. However, the total energy stored in the capacitor remains constant due to the conservation of energy principle.

4. How does the voltage affect the conservation of charge and energy in a capacitor?

The voltage applied to a capacitor affects both the conservation of charge and energy. When a higher voltage is applied, more charge can be stored on the plates, resulting in a greater electric field and more energy stored. However, the conservation principles still hold true, as the total charge and energy stored in the capacitor will remain constant, regardless of the applied voltage.

5. Why is it important to understand the principles of conservation of charge and energy in capacitors?

Understanding the principles of conservation of charge and energy in capacitors is important for the design and functioning of electronic circuits. Without these principles, we would not be able to accurately predict or control the behavior of capacitors, which are essential components in many electronic devices. Additionally, understanding these principles allows scientists and engineers to optimize the efficiency and performance of capacitors in various applications.

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