How Does Charge Redistribute When Capacitors Are Connected in Series?

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When capacitors C1 (4.0 µF) and C2 (2.0 µF) are charged in series across a 95 V battery, they hold charges of 380 µC and 190 µC, respectively. After disconnecting them from the battery and each other, they are reconnected positive to positive and negative to negative. The equivalent capacitance is calculated as 4/3 µF, but confusion arises in determining the final charge on each capacitor after reconnection. The total charge should remain constant at 380 µC, but the user struggles to find the correct distribution of charge between the capacitors in this configuration. Assistance is requested to clarify the charge redistribution process in this series setup.
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1. Capacitors C1 = 4.0 µF and C2 = 2.0 µF are charged as a series combination across a 95 V battery. The two capacitors are disconnected from the battery and from each other. They are then connected positive plate to positive plate and negative plate to negative plate. Calculate the resulting charge on each capacitor. ___ µC (C1)
___ µC (C2)

2. Q = CV , 1/Ceq = 1/C1 + 1/C2
3. First, I calculated the charge on each capacitor while they were in series connected to the battery-- Which is 380 and 190, respectively. I then calculated the equivalent capacitance to be 4/3, and think that Q1 + Q2 (the charge after they are hooked together) should equal 380, but I can't ever seem to get a correct answer. I know how to solve this if they are in parallel, but this series configuration is screwing me up.Any help is appreciated.Thanks so much,
Kyle
 
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