# Electrostatics - two capacitors in parallel

1. Aug 29, 2007

### zimo

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

We begin with two capacitors c(1)=6 microFarad and c(2)=3 microFarad in parallel which are connected to V=24Volts. The charged capacitors are disconnected and is being connected to each other on opposite signs (+ to -). How much Energy is got wasted in the process?

2. Relevant equations

Energy=(1/2)CV^2=(1/2)QV

3. The attempt at a solution

I calculated the first array for energy and got 2.592X10^-3 Joules, and then the second array, which I am not sure how to get it right (and I will be more then happy if you can offer a general interpretation of the second state I mentioned) and I got 2.304X10^-3 Joules so the diff. is about 0.288X10^-3 Joules.
a) How should I calculate such a case (e.g. how the Charges/Capacities/Volts are being distributed after reconnection)?
b) What have I done wrong?

Thank you

2. Aug 29, 2007

### Staff: Mentor

When you reverse the capacitors, use Q=CV to tell you what the final voltage is....

3. Aug 29, 2007

### zimo

The new Q is just the same as the old one? (2.16X10^4 C)

4. Aug 29, 2007

### Staff: Mentor

No, some of the charge gets cancelled out when you reverse the orientation of the caps and reconnect them. If you have 2 coulombs on one cap and 3 columbs on the other (just making up round numbers here as an example), and then you reverse the caps, what it the net charge left on the parallel combination of those caps?

5. Aug 29, 2007

### zimo

1 Coulomb.
As a matter of fact, I realized it 5 minutes after posting... but then got stuck again,
I think I really missing the logics behind the whole thing :-(

6. Aug 29, 2007

### Staff: Mentor

I'll give one more hint, and then we need to see some more work from you. Initially you have the two caps in parallel, and you can calculate the charge on each based on the capacitances and voltage. Then when you disconnect them and invert one and reconnect them, some of the charge will cancel, and you will be left with the rest, now on the parallel combination of the caps. This new Q value and same total C value will give you a new voltage across the caps. Calculate this voltage, and then you can proceed to the energy change question.