Conservation of Charge in Series Capacitors.

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SUMMARY

The discussion centers on the conservation of charge in series capacitors, specifically when two differently charged capacitors are connected. The user presents a scenario with capacitors C1 and C2, having initial voltages V1 and V2, and charges Q1 and Q2. The user calculates the total charge as Qtot=Q1+Q2 and suggests that each capacitor should have Qtot/2. However, using voltage and capacitance relations, they derive a different charge for the system, leading to confusion regarding the conservation of charge principle. The consensus is that in a series configuration, the charge on each capacitor is equal, while the voltage across them is additive.

PREREQUISITES
  • Understanding of capacitor fundamentals, including capacitance, voltage, and charge.
  • Familiarity with series and parallel circuit configurations.
  • Knowledge of the formula for equivalent capacitance in series: Ceq=(1/C1+1/C2)^-1.
  • Basic principles of electric charge conservation.
NEXT STEPS
  • Study the principles of charge distribution in series and parallel capacitors.
  • Learn about the implications of Kirchhoff's voltage law in capacitor circuits.
  • Explore practical applications of capacitors in electronic circuits.
  • Investigate the effects of initial charge conditions on series capacitor behavior.
USEFUL FOR

This discussion is beneficial for physics students, electrical engineers, and anyone interested in understanding capacitor behavior in series circuits.

Bigger than smaller
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Hi. I have a question about conservation of charge when two differently charged capacitors are connected in series. I know this is like a homework problem of introductory level of physics, but since this is not my homework, I decided to post it here.

So, here is the story. There are two capacitors connected in series with offed switch (no battery, and ideal wire). One has (C1,V1,Q1) and the other has (C2,V2,Q2) as its capacitance,voltage and charge respectively. Now switch is on, and charge on one of the capacitor flows into the other capacitor.

What I am curious is charge on each capacitor after this process. For capacitors in series, they are supposed to be occupied the same amount of charge. Therefore, total charge is Qtot=Q1+Q2 and charge on each capacitor should be Qtot/2.

However, if I approach this problem using voltage, capacitance relations, then I got quite different answer. In other words, total voltage of series capacitor is Vtot=V1+V2. and equivalent capacitance is Ceq=(1/C1+1/C2)^-1. Thus, charge on this set of capacitors is Q=Vtot*Ctot=(V1+V2)C1C2/(C1+C2). This should be also the same as charge occupied in each capacitor.

Through both calculations, I realized that the answer is completely different. I think the second calculation is making sense logically, (because the first method seems quite intuitive) but it seems violating conservation of charge. Could you give me any advice for this? Any help will be appreciated.
 
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Bigger than smaller said:
There are two capacitors connected in series with offed switch (no battery, and ideal wire).
They are in parallel. When you close the switch, it's the voltage across them that remains same, not the charge.
 

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