Capacitors Question Homework: Find Final Charge on 4.10 uF Capacitor

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Homework Help Overview

The problem involves two capacitors, 4.10 uF and 5.90 uF, connected in parallel to a 660-V supply. The discussion centers on the scenario where the capacitors are disconnected from the voltage source and then reconnected with terminals of unlike signs together, specifically focusing on finding the final charge on the 4.10 uF capacitor.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of disconnecting the capacitors from the voltage source and the conservation of charge. Questions arise about how the charges will interact when the capacitors are reconnected with opposite terminals together, and whether the final charge will be split in the same ratio as the original charges.

Discussion Status

Some participants have offered insights into the behavior of the charges when the capacitors are reconnected, suggesting that the total charge will be reduced and that the new voltage will influence the charge on the capacitors. There is an ongoing exploration of how the voltages across the capacitors relate to each other and how this affects the final charge distribution.

Contextual Notes

Participants are considering the effects of charge conservation and the relationship between voltage and capacitance in the context of the problem. There is an acknowledgment of the need to understand how the new configuration alters the charge on the capacitors.

kidsmoker
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Homework Statement



Two capacitors, with capacitances of 4.10 uF and 5.90 uF are connected in parallel across a 660-V supply line.

The first questions all relate to finding the charges on each capacitor and the voltage across each one. I can do all of these with no problems.

But then it asks "The charged capacitors are disconnected from the line and from each other, and then reconnected to each other with terminals of unlike sign together.
Find the final charge on the 4.10 uF capacitor."


Homework Equations



Q=CV
CT = C1 + C2

The Attempt at a Solution



I don't really understand what will happen in this situation. The battery is now disconnected so the voltage can change across each capacitor, but the charge should be conserved shouldn't it? Or will they somehow discharge?

Please help!
 
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The problem is asking you to consider that removing the capacitors from the voltage source will leave them with charges of the number of Coulombs that you figured in part a).

Then they are asking what happens when you reverse one with respect to the other. As you suggest the total charge will be reduced as the lesser subtracts from the larger.

They now have the same equivalent capacitance as when connected to the voltage source, but the voltage across the pair is now reduced. Assuming the reduced voltage now what is the charge then on the 4.1uF cap.
 
LowlyPion said:
Then they are asking what happens when you reverse one with respect to the other. As you suggest the total charge will be reduced as the lesser subtracts from the larger.


So the positive charge and the negative charge will cancel out somewhat, leaving you with the difference. And is this then split in the same ratio as the original charges?
 
Hmm. I hadn't ever thought of it that way, but I think yes. There is a different way to approach this problem, though, that would explain why this is. Consider the voltage, and how the voltages across the two capacitors are related (hint: in parallel).
 
kidsmoker said:
So the positive charge and the negative charge will cancel out somewhat, leaving you with the difference. And is this then split in the same ratio as the original charges?

Well, it has a new voltage across it now, so yes the equivalent capacitance would determine the new voltage and it would hold a charge that would be determined by the new voltage. (Which as it turns out would be in the same ratio as the original circuit.)
 

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