Two opposite polarity charged capacitors in series

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SUMMARY

The discussion focuses on the behavior of two opposite polarity charged capacitors in series, specifically analyzing the charge transfer when the circuit is closed. The participants confirm that the voltage remains constant at 100V across both capacitors, despite differing charges, due to their parallel-like configuration. The calculations provided indicate that the charge on the first capacitor (C1) is 10^(-4) C and on the second capacitor (C2) is 3 * 10^(-4) C, using the formula Q=CV. The conversation emphasizes the importance of understanding charge flow and potential differences in capacitor circuits.

PREREQUISITES
  • Understanding of capacitor fundamentals, including charge (Q), capacitance (C), and voltage (V).
  • Familiarity with the equation Q=CV for calculating charge in capacitors.
  • Knowledge of series and parallel circuit configurations and their implications on voltage and charge.
  • Basic algebra skills for manipulating equations related to electrical circuits.
NEXT STEPS
  • Study the behavior of capacitors in series versus parallel configurations.
  • Learn about charge conservation principles in electrical circuits.
  • Explore the effects of closing switches in capacitor circuits and the resulting charge redistribution.
  • Investigate real-world applications of capacitors in electronic devices and energy storage systems.
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Students studying electrical engineering, electronics enthusiasts, and anyone looking to deepen their understanding of capacitor behavior in 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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