Parllel Plate Capacitors Connected Opposite

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Two capacitors, C1 (9.8 µF) and C2 (4.0 µF), are initially connected in parallel to a 30 V battery. After disconnecting them from the battery and each other, they are reconnected with opposite plates together, prompting a need to calculate the energy decrease in the system. The initial charge on each capacitor is calculated using Q = CV, and the user seeks guidance on whether to recalculate the voltage after reconnection. It is noted that when capacitors are connected in series, they share the same charge, necessitating a new charge calculation for both capacitors. The discussion emphasizes the importance of determining the new stored energy to find the energy decrease.
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Homework Statement



Two capacitors are connected parallel to each other and connected to the battery with voltage 30 V. Let C1 = 9.8 micro F,C2 = 4.0 micro F be their capacitances. Suppose the charged capacitors are disconnected from the source and from each other, and then reconnected to each other with plates of opposite sign together.By how much does the energy of the system decrease?

Homework Equations



U = (1/2)QVFor PPC connected in parallel,

Q1 = C1V and Q2 = C2V

Ceq = C1 + C2 + C3...

The Attempt at a Solution



I found the charge stored on the plates initially, and was trying to find the charge stored on them after they were reconnect, but failed. I really don't know what to do. Do I have to find the new voltage too?
 
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If someone could just tell me if I have to recalculate a new potential difference, that would be great. I really have no clue how to proceed from here.
 
Can anyone help me with this? It's very frustrating, and I'm not quite sure how to approach it.
 
Note that when you connect the capacitors in series, they have the same charge on each other. So now you have to find the new charge on both of them and add up the stored energy in both.
 
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