Voltage after connecting two capacitors

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Homework Help Overview

The problem involves two capacitors, a 2.00 µF capacitor charged to 12 volts and a 4.00 µF capacitor charged to 30 volts, which are disconnected from their respective batteries and then connected together. The discussion centers on determining the final charge on the 2 µF capacitor after this connection.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss whether to apply conservation of charge or conservation of energy to solve the problem. Some calculations yield different final voltages and charges, leading to questions about the correct approach. There is also a mention of key terms in the problem statement that may hold significance.

Discussion Status

Participants are exploring different methods to approach the problem, with some providing calculations based on conservation of charge. Others express confusion about the conceptual application of conservation of energy and its relevance in this context. The discussion is ongoing, with no clear consensus reached on the preferred method.

Contextual Notes

There is mention of a textbook example that uses conservation of energy, prompting participants to consider the differences between their problem and the example provided. This raises questions about the assumptions and definitions being applied in the current discussion.

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Homework Statement


A 2.00 [tex]\mu[/tex]F capacitor is initially connected to a 12 Volt battery. In a separate circuit, a 4.00 [tex]\mu[/tex]F capacitor is connected to a 30 Volt battery. The capacitors are then disconnected from their respective batteries and connected to each other (positive sides are connected together, and negative sides are connected together). What is the final charge on the 2[tex]\mu[/tex]F capacitor


Homework Equations


Qtot = Q1 + Q2
.5(C1 V12) + .5(C2 V22) = .5(C1+C2)V2


The Attempt at a Solution


I'm not sure if you use conservation of charge or conservation of energy here. If I use conservation of charge, I get a common V of 24V and charge of 48[tex]\mu[/tex]C on the 2[tex]\mu[/tex]F capacitor.
If I use conservation of energy I get a common V of 25.5 V and a charge of 51[tex]\mu[/tex]C on the 2[tex]\mu[/tex]F capacitor.
The question has some words italicized which means they are probably key words, but I don't really get it. As always, any help is greatly appreciated!
 
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You have a conservation of charge.

Figure the total charge.

Figure the equivalent C.

Figure the new Voltage, from the equivalent C and total charge.

Figure the charge across the C at the new voltage calculated.
 
LowlyPion said:
You have a conservation of charge.

Figure the total charge.

Figure the equivalent C.

Figure the new Voltage, from the equivalent C and total charge.

Figure the charge across the C at the new voltage calculated.
Qtot = (C1*V1) + (C2*V2) = 1.44 *10-4C
Ceq = 6[tex]\mu[/tex]F
Qtot/Ceq = V = 24V

Q = C1*V = 4.8 * 10-5

I can do the math, but I don't understand it conceptually. When would you use conservation of energy?
 
cashmoney805 said:
I can do the math, but I don't understand it conceptually. When would you use conservation of energy?

There are only so many charges once everything is charged. You can't create them. They got to go somewhere. So apply the conservation of mass - as the conservation of charges.

Can't think of an example for using conservation of energy for this kind of problem. Voltage sources are usually doing work to keep things charged, and current flowing and resistors are dissipating power.
 
Hm. In my textbook there is a similar problem. One capacitor is charged by one source, and a second one is charged by a second source. They are disconnected from their batteries and then connected to each other. Find potential difference across the capacitors and charge on each capacitor.

The book solves that problem by using conservation of energy. Anyone see the difference?
 

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