Stored charege and potentials of two capacitors in series

[zoner7]

Homework Statement

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?

Homework Equations

C = Q / (Delta)V

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.

 

[Delphi51]

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.

 

[nlightner]

It is important to note that the potential difference across each capacitor will be different, as it is dependent on the capacitance of each individual capacitor. The higher the capacitance, the lower the potential difference across the capacitor. Additionally, the charge stored on each capacitor will also be different, as it is dependent on the potential difference and the capacitance. It is important to carefully consider the values of capacitance and potential difference in a series circuit to accurately calculate the charge stored on each capacitor.