Calculating Charge with Capacitors in Series

[PeachBanana]

Homework Statement

What is the charge on the right
plate of C2 long after the switch is
closed?

1. CV
2. CV/2
3. CV/3
4. 2CV/3
5. 0

Homework Equations

Q = CeqV

The Attempt at a Solution

Q = 1/3c * 2V
so I thought
Q = 2V/3C

 

[gneill]

Homework Statement

What is the charge on the right
plate of C2 long after the switch is
closed?

1. CV
2. CV/2
3. CV/3
4. 2CV/3
5. 0

Homework Equations

Q = CeqV

The Attempt at a Solution

Q = 1/3c * 2V
so I thought
Q = 2V/3C

Is that (2V)/(3C) or (2V/3)C ?

 

[PeachBanana]

I wanted (2V/3)C but written that way it looks like

2CV/3 which is one of the answers but I'm trying to understand why "C" is in the numerator. Is it because Ceq = 1 / C?

 

[gneill]

I wanted (2V/3)C but written that way it looks like

2CV/3 which is one of the answers but I'm trying to understand why "C" is in the numerator. Is it because Ceq = 1 / C?

What's the net (equivalent) capacitance of the two given capacitors?

 

[PeachBanana]

Ceq = 1/C + 1/2C = 3/2C

 

[gneill]

Ceq = 1/C + 1/2C = 3/2C

Not quite. Two capacitors Ca and Cb in series yield:

$$C_{eq} = \frac{1}{\frac{1}{Ca} + \frac{1}{Cb}}$$

You've forgotten to take the overall reciprocal.

 

[PeachBanana]

I see what I did incorrectly now. My final answer should have been 2C/3.