# Homework Help: Capacitors question

1. Dec 1, 2008

### kidsmoker

1. The problem statement, all variables and given/known data

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 accross 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."

2. Relevant equations

Q=CV
CT = C1 + C2

3. 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 accross each capacitor, but the charge should be conserved shouldn't it? Or will they somehow discharge?

2. Dec 1, 2008

### LowlyPion

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.

3. Dec 1, 2008

### kidsmoker

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?

4. Dec 1, 2008

### turin

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).

5. Dec 1, 2008

### LowlyPion

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.)