Two capacitors, one that has a capacitance of 4 μF and one that has a capacitance of 12 μF are connected in parallel. The parallel combination is then connected across the terminals of a 12 V battery. Next, they are carefully disconnected so that they are not discharged. They are then reconnected to each other - the positive plate of each capacitor connected to the negative plate of the other.
(a) Find the potential difference across each capacitor after they are reconnected.
(b) Find the energy stored in the capacitors before they are disconnected from the battery, and find the energy stored after they are reconnected.
(In parallel) Ceq = C1 + C2 + ... + Cn
C = Q/V
E (energy) = (1/2) C V2
The Attempt at a Solution
I know that in parallel, the voltage across all of the capacitors is the same (which is 12 V in this case). I also know that Ceq = C1 + C2 + ... + Cn, which in this case should turn into (4 * 10-6 ) + (12 * 10-6) = 1.6 * 10-5 F
If I plug this into the formula E (energy) = (1/2) C V2 then I get:
E = (1/2)(1.6 * 10-5)(122) = 1.152 * 10-3 J
With this, it seems that I have solved the first half of (b), which was the question about how much energy is stored in the capacitors before they disconnect from the battery. I don't know if it is right however, since this is written homework and not webassign. Furthermore, I do not know how to do (a) or the 2nd half of (b). Can someone guide me in the right direction on how to do (a) and the 2nd half of (b), and also verify whether or not my first half of (b) is correct? Thank you.