Capacitors are charged, disconnected from battery, and had plates swit

[penguinnnnnx5]

Homework Statement

Capacitors C₁=13μF and C₂=21μF are each charged to 10V, then disconnected from the battery without changing the charge on the capacitor plates. The two capacitors are then connected in parallel, with the positive plate of C₁ connected to the negative plate of C₂ and vice versa.

Homework Equations

Q=CV
C_eq = C₁ + C₂

The Attempt at a Solution

C_eq = C₁ + C₂
C_eq = 34μF

Q=C/V

Q₁= 13μF*10V = 130μC

Q₂= 21μF*10V = 210μC

In the new C₁, the positive plate should be 130 μC and the negative -210μC.
In the new C₂, the positive plate should be 210μC and the negative -130 μC.

I'm not sure what to do after this.

I have Googled the problem and have found a Yahoo Answers thread with the correct method, but I'm not sure why the method was correct. If someone can explain his steps and/or help me with the next step, then that would be great.

Edit: I think I may have figured it out. Is he taking the net charge of each capacitor, multiplying it with the first capacitor (to find the charge of the new C₁) and dividing by C_eq because the net charge divided by C_eq is equal to the new value for V? Which is basically taking the fraction of one capacitor over the total capacitance and finding the charge in that one capacitor? And because in a parallel, voltage is the same in both capacitors? I'm not sure if that makes sense, but hopefully it does?

 

[ehild]

What is the question at all?

ehild

 

[penguinnnnnx5]

I'm sorry, I forgot to state the questions. It's just asking for the charges in each of the capacitors after connecting the positive plate of one to the negative plate of the other and vice versa, then the potential difference in both.

 

[ehild]

OK, your method looks correct. What voltage and charges have you got?

ehild

 

[penguinnnnnx5]

~31 microCoulombs for Q1 and ~49 microCoulombs for Q2.

V1 = V2 = V = 2.4 volts.

 

[ehild]

Correct!

ehild