Calculating Capacitance and Charge in Connected Capacitors

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The discussion focuses on calculating the equivalent capacitance and charge for a circuit of four capacitors connected in series and parallel. The relevant equations for capacitance are provided, with participants confirming the use of Ceq formulas for both configurations. For part a, the correct approach involves using the series formula for the capacitors in series and then adding the capacitance of those in parallel. In part b, the charge on each capacitor can be calculated using Q = CV, noting that in series, the charge remains constant across all capacitors. The conversation emphasizes understanding the relationships between charge and capacitance in different configurations.
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Homework Statement


Four capacitors are connected as shown here: http://i4.photobucket.com/albums/y103/thepastryman/HWa.jpg (a) Find the equivalent capacitance between points a and b (b) Calculate the charge on each capacitor, taking delta Vab= 15.0V

note: the actual numbers are not the same as those in the diagram. Stating from the top moving from left to right they are: 15.0 micro farads, 3 micro farads, 20 micro farads and 6 micro farads


Homework Equations


parallel: Ceq= C1+C2

series: Ceq=1/C1+1/C2


The Attempt at a Solution



I just used the above equations to find part a, is that right? For part b, would you just use Q=cv or is it more involved then that?
 
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tag16 said:
I just used the above equations to find part a, is that right? For part b, would you just use Q=cv or is it more involved then that?

I hope you used \frac{1}{C_{eq}}= \frac{1}{C_1}+\frac{1}{C_2}

if you have 2 parallel capacitors you can just use Q = CV for both of them.

If you have 2 series capacitors, you need the fact the charge of a both of the capacitors is equal to the charge on their equivalent capacitors. (because the same current goes through all of them)
This also assumes all capacitors started uncharged.
 
This is what I did for part a: (1/C1+1/C2)^-1= (1/15microF+1/3microF)^-1= 1.42x10^-7

Ceq= 6microF+20microF+1.42x10^-7

is that right?

For part b I know for series capacitors that the charges are all equal I just don't know how that would effect the set up to my equation to solve the problem.
 
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