# I Conservation of Charge in Series Capacitors.

1. Mar 31, 2016

### Bigger than smaller

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.

2. Mar 31, 2016

### cnh1995

They are in parallel. When you close the switch, it's the voltage across them that remains same, not the charge.