Four connected capacitors

1. Oct 11, 2009

tag16

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

2. Relevant equations
parallel: Ceq= C1+C2

series: Ceq=1/C1+1/C2

3. 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?

2. Oct 11, 2009

willem2

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.

3. Oct 11, 2009

tag16

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.