Solving the Capacitor Circuit: Final Charge on 1.00 micro-F Capacitor

[azure kitsune]

Homework Statement

A 1.00 micro-F capacitor charged to 50.0 V and a 2.00 micro-F capacitor charged to 20.0 V are connected, with the positive plate of each connected to the negative plate of the other. What is the final charge on the 1.00 micro-F capacitor?

Homework Equations

Q = CV

The Attempt at a Solution

Let a subscript of 1 denote the 1.00 micro-F capacitor, and a subscript of 2 denote the other.

Intially:

Q1 = C1V1 = 50 micro-C

Q2 = C2V2 = 40 micro-C

When the two capacitors are connected together, the first one has charge Q1' and the second has charge Q2' with:

Q1' + Q2' = 50 - 40 = 10 micro-C

The total potential drop around the loop must be zero, so we have

V1' = V2'

so

Q1'/C1 = Q2'/C2

Then when I solve for Q1', I get 3.3 micro-C, which is wrong. The correct answer according to the textbook is 45 micro-C.

 

[willem2]

You're right. The capacitor formed from C1 and C2 in parallel with 3 microF capacitance gets a charge that is the difference of the charges on C1 and C2. This 10 microC charge will get divided between C1 and C2 proportional to their capacitance, so C1 gets 3.33 microC and C2 gets 6.66 microC

If you would connect the positive plates together you'd get a total charge of 90 microC with 30 microC on C1 and 60 microC on C2. I have no Idea how they got 45 microC

 

[azure kitsune]

Thank you! Time to mark some corrections into the book!