Two opposite polarity charged capacitors in series

AI Thread Summary
When two opposite polarity charged capacitors are connected in series, they will eventually reach the same potential difference, despite having different charges. The initial charges on each capacitor can be calculated using the formula Q=CV, leading to Q1=10^(-4) C and Q2=3*10^(-4) C. As the circuit stabilizes, charge will flow until both capacitors have the same polarity and potential difference. The discussion emphasizes the importance of understanding charge transfer and the resulting balance in the circuit. Overall, the behavior of the capacitors illustrates fundamental principles of charge conservation and potential differences in electrical circuits.
Painguy
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Homework Statement



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Homework Equations


Q=C

The Attempt at a Solution


a) I want to say that since both have the same potentials then it is similar to a parallel circuit where the charges can differ on each capacitor. If i think of point A as an input and point B as an output then the potential difference would have to be 100V, but I am not sure.

b)
since both are at the same potential they will no longer transfer charge.

Q1= 10^(-6) * 100=10^(-4) C

Q2 = 3*10^(-6) * 100 = 3 * 10^(-4) C

Is this right?
 
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The equation should be Q=CV but I see that you did that in your calculations.. Looks alright to me though. for a), voltage doesn't chance for capacitors in parallel. Although it does seem counterintuitive, that is the case.
 
Painguy said:
a) I want to say that since both have the same potentials then it is similar to a parallel circuit where the charges can differ on each capacitor. If i think of point A as an input and point B as an output then the potential difference would have to be 100V, but I am not sure.
The polarities are opposite. When the switches close, there will be a flow of charge through each until both capacitors have the same potential difference again, but now with the same polarity.
 
haruspex said:
The polarities are opposite. When the switches close, there will be a flow of charge through each until both capacitors have the same potential difference again, but now with the same polarity.

How would that happen though? If a negative plate transfers its charge to the positive plate of the other capacitor then it just causes a chain reaction around the loop. I'm having a hard time seeing how everything will balance out.
 
Painguy said:
How would that happen though? If a negative plate transfers its charge to the positive plate of the other capacitor then it just causes a chain reaction around the loop. I'm having a hard time seeing how everything will balance out.
Let's just do the algebra and see if it becomes clearer to you.
Let the initial charges be +Q1 and -Q1 on C1, +Q2 and -Q2 on C2. Write equations for those. When the switch is closed, assume a quantity of charge Q flows from the +ve of C1 to the +ve of C2. What charge will flow on the negative side? What equations can you write now?
 

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