How Does Charge Redistribute When Capacitors Are Connected in Series?

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SUMMARY

When capacitors C1 (4.0 µF) and C2 (2.0 µF) are connected in series across a 95 V battery, they initially store charges of 380 µC and 190 µC, respectively. Upon disconnection from the battery and reconfiguration with positive plates connected together, the equivalent capacitance is calculated as 4/3 µF. The total charge remains conserved, leading to a redistribution of charge between the capacitors, which requires further calculation to determine the final charge on each capacitor.

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  • Understanding of capacitor charge and voltage relationships (Q = CV)
  • Knowledge of series and parallel capacitor configurations
  • Familiarity with equivalent capacitance calculations (1/Ceq = 1/C1 + 1/C2)
  • Basic principles of charge conservation in electrical circuits
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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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