Find the charge?

A capacitor of C1=1.04uF is in parallel to a capacitor C2=4.83uF, C1 and C2 is then in series to a parallel connection of C3=8.13uF and C4 =3.48 uF (C1and C2 in series to C3 and C4). The potential difference V across the battery is 12 V.





a. find the charge on q1 on capacitor c1.


b. find the charge on q2 on capacitor c2.


c. find the charge on q3 on capacitor c3.


d. find the charge on q4 on capacitor c4.





i think that you would use the equation Q=CV to find the charge but its not working can anyone help me out....








thanks

Find the charge?
From the law of conservation of charge q1 + q2 = q3 + q4.


From definition of capacity q1=C1*V1, q2=C2*V2, q3=C3*V3 and q4=C4*V4. Because C1 and C2 are in parallel V1 = V2, because C3 and C4 are in parellel V3 = V4. The total potential difference V1 + V3 = V = 12V, so V3 = V - V1. Thus V2 = V1, V3 = V4 = V - V1. Then q1 + q2 = C1*V1 + C2*V1 = (C1 + C2)*V1, while q3 + q4 = (C3 + C4)(V - V1). So we have equation to find V1: (C1 + C2)*V1 = (C3 + C4)(V - V1) =%26gt;V1 = (C3 + C4)V/(C1+C2+C3+C4).


Now you can find V2 = V1 and V3 = V4 = V - V1 and finally use Q=CV to find charge of each capacitor.
Reply:First you need to find the effective capacitance of the system as a whole. Two capacitors in parallel add simply:


C(1 and 2) = C(1) + C(2)


Then two capacitors, say C$ and C*, in series add like this:


C($ and *) = 1/((1/C*) + (1/C$))


we have C(1 and 2) in series with C(3 and 4) which gives:


C(total) = 1/((1/C(1 and 2)) + (1/C(3 and 4)))


C(total) = 1/((1/(C(1)+C(2))) + (1/(C(3) + C(4))))


Now you have the total capacitance, and the charge comes from Q=CV, where you need to use C(total). The charge on the capacitor plates closest to the +12V will be negative, and the charge on the plates closest to 0V will be positive. Those capacitors will have equal and opposite charge on their other plates. It will turn out that the charge will be the same on every plate, except for the + or -.