# Homework Help: Calculating Charge with Capacitors in Series

1. Oct 4, 2012

### PeachBanana

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

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

2. Relevant equations

Q = CeqV

3. The attempt at a solution

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

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2. Oct 4, 2012

### Staff: Mentor

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

3. Oct 6, 2012

### 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?

4. Oct 7, 2012

### Staff: Mentor

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

5. Oct 7, 2012

### PeachBanana

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

6. Oct 8, 2012

### Staff: Mentor

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.

7. Oct 8, 2012

### PeachBanana

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