# Stored charege and potentials of two capacitors in series

1. Feb 25, 2009

### zoner7

1. The problem statement, all variables and given/known data
Two capacitors of 2.0F and 5.0F are connected in series; the combination is connected to a 1.5 volt battery. What is the charge stored on each capacitor? What is the potential difference across each capacitor?

2. Relevant equations
C = Q / (Delta)V

3. The attempt at a solution

Since the capacitors are in series, we can find the equivalent capacitance by solving:

1/Ceq = 1/2.0F + 1/5.0F
Ceq = 1.429F

I then used this value to find the charge on each plate of each capacitor

C = Q/V
CV = Q
1.429F * 1.5V = 2.143

So now I now the amount of charge on each plate.

Using this value in conjunction with a specific capacitance, I can find the potential difference across any given capacitor whose capacitance I know.

so, V = Q/C
2.143/2 = 1.072V
2.143/5 = .429V

Thank you for the help.

2. Feb 25, 2009

### Delphi51

Re: Capacitors

All correct, I think. I used a slightly different method, adding the voltages around the loop to get Q/2 + Q/5 = 1.5
which yields the same answer you got.