Capacitors Question Homework: Find Final Charge on 4.10 uF Capacitor

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Two capacitors, 4.10 uF and 5.90 uF, are connected in parallel to a 660-V supply, and the discussion focuses on finding the final charge on the 4.10 uF capacitor after they are disconnected and reconnected with unlike terminals. The charge conservation principle is emphasized, indicating that the total charge will be reduced when the capacitors are reversed, as the lesser charge subtracts from the larger. The new voltage across the capacitors will be determined by their equivalent capacitance, which remains the same as when connected to the voltage source. The final charge on the 4.10 uF capacitor will reflect this new voltage, maintaining the original charge ratio. Understanding the relationship between voltage and charge in parallel circuits is crucial for solving this problem effectively.
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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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