Capacitor charged then Connected to another Capacitor.

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Homework Help Overview

The problem involves two capacitors, one initially charged with an energy of 4.0 J, and the other uncharged, which is connected in parallel after the charging battery is removed. The discussion centers on the redistribution of charge and the implications for energy storage in the capacitors.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the effects of connecting a charged capacitor to an uncharged one, questioning how the charge distribution affects the electric field and total energy stored. There is uncertainty about the implications of energy loss during this process.

Discussion Status

Participants are actively discussing the redistribution of charge and its impact on energy storage. Some have suggested that the energy stored in each capacitor decreases, while others are questioning where the lost energy goes and considering the role of resistance and inductance in the connection.

Contextual Notes

There is an ongoing exploration of the physical principles involved, including the behavior of electric fields in capacitors and the effects of connecting them in parallel. Participants are considering real-world factors that may affect energy transfer, such as resistance in the connecting wires.

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Homework Statement


One capacitor is charged until its stored energy is 4.0 J, the charging battery then being removed. A second uncharged capacitor is then connected to it in parallel. (a) If the charge distributes equally, what is now the total energy stored in the electric fields?(b) Where did the excess energy go?


Homework Equations


u=1/2*ε*E2 Where E is the electric field and u is the energy density(U/Volume)


The Attempt at a Solution


Okay so I'm somewhat lost. Since the charge is spread equally, and they are in parallel, so I assume they have the same capacitance. So for that I assumed that the total energy stored in each is half of that? However my reasoning is not correct since energy disappears? How can I approach the problem correctly?
 
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Jonnyto said:

Homework Statement


One capacitor is charged until its stored energy is 4.0 J, the charging battery then being removed. A second uncharged capacitor is then connected to it in parallel. (a) If the charge distributes equally, what is now the total energy stored in the electric fields?(b) Where did the excess energy go?


Homework Equations


u=1/2*ε*E2 Where E is the electric field and u is the energy density(U/Volume)


The Attempt at a Solution


Okay so I'm somewhat lost. Since the charge is spread equally, and they are in parallel, so I assume they have the same capacitance. So for that I assumed that the total energy stored in each is half of that? However my reasoning is not correct since energy disappears? How can I approach the problem correctly?

The charge will redistribute between the two capacitors. Each of two capacitors will have half the charge of the original capacitor. How will that affect the electric field in each capacitor?
 
Would that cut the electric field in half?
 
Jonnyto said:
Would that cut the electric field in half?

Yes, it would. What will that do to the total stored energy?
 
1/4 the original energy on each.
 
Jonnyto said:
1/4 the original energy on each.

Right again. So now you've got half the energy you started with. Where did it go?
 
Hmm that's what I'm unsure of now. Was this energy lost pushing the charge to the other capacitor?
 
Jonnyto said:
Hmm that's what I'm unsure of now. Was this energy lost pushing the charge to the other capacitor?

Well, it has to flow through a wire. What kind of property of wire might make that not free?
 
Jonnyto said:
Hmm that's what I'm unsure of now. Was this energy lost pushing the charge to the other capacitor?
Yes, but how exactly?
In the real world, the connection will have some inductance and some resistance, however small. What will that lead to?
 

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