Stored charege and potentials of two capacitors in series

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SUMMARY

In the discussion, two capacitors with values of 2.0F and 5.0F are connected in series to a 1.5V battery. The equivalent capacitance (Ceq) is calculated to be 1.429F. The charge stored on each capacitor is determined to be 2.143C, with the potential difference across the 2.0F capacitor being 1.072V and across the 5.0F capacitor being 0.429V. The calculations confirm the correct application of the formula C = Q / (Delta)V and the principles of series circuits.

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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.
 
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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.
 

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