Find the individual charge (and potential difference?) on each capacitor

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Homework Statement

Four capacitors are connected as shown in the figure below. (Let C = 18.0 µF.)

Find the equivalent capacitance between points a and b

Calculate the charge on each capacitor, taking ΔVab = 17.0 V

Homework Equations

Q = CΔV

C = Q / ΔVΔV1 = ΔV2 = ΔVtot (for parallel capacitors)

Q tot = Q1 + Q2 (charge on parallel capacitors, it says)ΔVtot = ΔV1 + ΔV2 (for series capacitors)

Q = Q1 = Q2 (charge on series capacitors, it says)

The Attempt at a Solution

The only thing I could solve was the equivalent capacitance across, which is 6 x 10^-6

And I broke it up like this (and it's correct):
http://desmond.imageshack.us/Himg213/scaled.php?server=213&filename=prob1m.jpg&res=landing I have NO IDEA at all how to determine the individual charge on each of them, even with the formulas...

I can see ΔV across series is supposed to add up to 17V but no idea how.

Where should I start on that?Edit: ok used Q = CV for the (20 x 10-6 f) capacitor and got (102 x 10-6 C) because charge is supposed to be the same across series

 

[truesearch]

What you have done so far is correct...now find the total capacitance of the 2 in series... (You were able to find the equivalent of C and the 3uF in series !) then you should be able to get the total charge.

 

[Color_of_Cyan]

ok i got them all

you need to just mainly worry about using total charge Q = Q1 = Q2 for series capacitors and V tot = V1 = V2 for parallel capacitors, just to find the individual potentials / charges.

Then you need to work your way down with these and the other two formulas.

 

[truesearch]

thats correct !