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**potential difference in parallel -- capacitors**

Two capacitors C1 = 3 µF and C2 = 9 µF are connected in parallel across a 11 V battery. They are carefully disconnected so that they are not discharged and are reconnected to each other with positive plate to negative plate and negative plate to positive plate (with no battery).

(a) Find the potential difference across each capacitor after they are connected.

(b) Find the initial and final energy stored in the capacitors.

related formulas:

C=Q/V

so far...

so this means that they are in parallel, store up charge, and then are disconnected and reconnected in series without battery yes?

since C(total)=12uF and Voltage = 11v (for both capacitors) I figured that the Q(total) = 1.32*10^-4 C

12uF = Q(tot)/11v

when I apply this total charge to the capacitors in a series (inverting everything) I get 1/3uF=V(1)/1.32*10^-4, V(1)= 44v

before I went on to solve for the other capacitor I submitted this but it was wrong. Where did I go wrong?