Battery connected to capacitors in series

[matt72lsu]

Homework Statement

A 12-V battery is connected to three capacitors in series. The capacitors have the following capacitances: 4.0uF, 14uF, and 33uF.

Homework Equations

R = V/I

The Attempt at a Solution

I'm thinking you add the capacitances since they are in series (do i need to convert them from uf to F?) but I'm not sure what to do after that

 

[willem2]

you don't add them. This should be in your textbook somewhere. You can derive the equation to use with Q = UC (charge = potential difference * capacitance), together with the fact that all of the capacitors have the same charge (if they started out empty) because the same current has gone through them always.
Compute the potential difference across the 3 capacitors with Q=UC and the sum of them should be 12 V

 

[matt72lsu]

I forgot to add the actual question of what they are looking for: Find the voltage across the 33uF capacitor

 

[matt72lsu]

Ok so I know you use Q = CV. But I am still confused about how to find C.

 

[matt72lsu]

anybody?