Calculate the resulting charge on each capacitor

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The discussion focuses on calculating the resulting charge on two capacitors, C1 (6.35 µF) and C2 (2.00 µF), after they are charged in parallel across a 250 V battery and then reconfigured in series. Initially, the charge on each capacitor is determined using the formula Q = C × V, leading to charges of 1587.5 µC for C1 and 500 µC for C2. When connected in series, the final charge on both capacitors becomes equal, allowing for the calculation of the resulting charge based on the equivalent capacitance and voltage across the series configuration.

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  • Understanding of capacitor charging in parallel and series configurations
  • Familiarity with the formula Q = C × V for calculating charge
  • Knowledge of equivalent capacitance in series circuits
  • Basic electrical circuit theory
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Calcguy
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Capacitors C1 = 6.35 µF and C2 = 2.00 µF are charged as a parallel combination across a 250 V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on each capacitor.
q'1 = µC
q'2 = µC

Can someone please help me with this, I don't know what to do.
 
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First you have to find the charge on each capacitor after it has been charged. Then note from the statement of the problem whether the new connection yields a series or parallel connection.

From here, note that if the caps are in series, the resulting charge (if you let the circuit stabilise) on each capacitor will be the same. If they are in parallel, then their final potential differences will be the same. From these 2 conditions you can calculate the final charge on each cap.
 

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