Calculating Charge in a Parallel Connection of Capacitors

AI Thread Summary
In a parallel connection, the equivalent capacitance of the capacitors is calculated as Ceq = 11.1uF + 1.9uF = 13uF. The total charge when connected to a 333V battery is Qtotal = 333V * 13uF = 0.004329 C. When the capacitors are disconnected and reconnected with opposite plates, the charge distribution changes, requiring the use of series capacitance principles. The original charge on the first capacitor must be recalculated based on the new configuration, as the total charge will redistribute. The problem emphasizes understanding charge conservation and the behavior of capacitors in different configurations.
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Homework Statement


Capacitors of 11.1uF and 1.9uF are charged as a parallel combination across a 333V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive to negative plate and negative to positive plate. Find the resulting charge on the first capacitor in units uC.


Homework Equations


1. Ceq = C1 + C2 parallel capacitance C1=11.1uF C2=1.9uF
2. Qtotal=V * Ceq

The Attempt at a Solution


1. Ceq = 11.1uF + 1.9uF = 13uF
2. Qtotal = 333 * 13uF = .004329 C

i tried using an equilibrium theorm:
(Qtot / Ceq) = (Qtot - q1) / 11.1uF -> .004329 / 13e-6 = (.004329 - q1) / 11.1e-6
with q1 = .0080253C but it's bigger than q= 333V * 11.1uF = .036963 so i don't think it's correct because that would be the max charge, right?

after this point do i treat it as capacitors in series or what?

part B of the problem states they are charged in series and then disconnected from the battery and connected positive plate to positive place and negative to negative plate

sorry if I'm overlooking something obvious.
thanks
david
 
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Find the original charge on each capacitor. When they are reconnected, what's the total charge on both capacitors? When reconnected, what's the same across each capacitor? Use that to figure out how the charge rearranged itself upon reconnection.
 
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