Potential Difference over Capacitors in Series

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Homework Help Overview

The discussion revolves around the potential difference across capacitors in series and parallel configurations, specifically addressing the behavior of capacitors when connected in a circuit. Participants are exploring the implications of voltage differences and the assumptions related to capacitor arrangements.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the assumptions about the arrangement of capacitors, particularly whether they are in parallel or series. There are discussions about the criteria for determining parallel connections and the implications of voltage differences across capacitors.

Discussion Status

The discussion is active, with participants providing clarifications and questioning the original poster's assumptions. Some guidance has been offered regarding the need for a circuit diagram and clearer definitions of parallel connections. Multiple interpretations of the problem are being explored, particularly concerning the behavior of capacitors when connected in parallel.

Contextual Notes

There is a mention of a potential typo in the original equations provided, which may affect the understanding of the problem. Additionally, some participants express concerns about the nature of the question itself, suggesting it may lead to paradoxical situations in real-world applications.

nmfowlkes
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Homework Statement
Students perform an experiment using a circuit that contains two parallel plate capacitors. Capacitor C1= 30μF is fully charged and connected to an initially uncharged capacitor C2= 60μF, as shown in the circuit. Which of the following is a correct hypothesis by the students regarding the relative magnitude of the potential difference V1 and V2 across the plates of capacitors C1 and C2, respectively, when the circuit reaches a steady-state?

For clarification, I assume capacitors that are on opposite end of a square circuit are in parallel.
Relevant Equations
What exactly is the significance of the second capacitor's initially uncharged state. I thought my solution would apply but it does not lie among the answer choices.
V1 = Q/30

In parallel: C = C1 + C2

V2 = Q/(60 + 90)

V1/V2 = 3

V1 = 3V2
 
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Hi @nmfowlkes and welcome to PF.

Please show the circuit. Also, "For clarification, I assume capacitors that are on opposite end of a square circuit are in parallel" is not clear enough to convey what your assumption is. Two capacitors are in parallel if they have the same potential difference across their terminals. That's what you should look for as a criterion.

Edit: Please also include the answer choices.
 
Looks OK to me. Except for a typo in line #3: V2 = Q/(60 + 90) → V2 = Q/(60 + 30). I assume you mean V1 is before and V2 after, since there is only one voltage across parallel capacitors. So, perhaps there is a misunderstanding of the question?
 
Yes Dave, the question was asking about the difference across the two capacitors. V1 = V2 as they are in parallel. I did not understand it.
 
nmfowlkes said:
Yes Dave, the question was asking about the difference across the two capacitors. V1 = V2 as they are in parallel. I did not understand it.
Yea, something's wrong. There is no voltage difference after they're connected in parallel, according to me.

Anyway, slightly off-topic, but.. I dislike this question (even though it's always asked) a circuit with two capacitors at different voltages suddenly connected in parallel (zero resistance switch) is fundamentally ill-defined; a paradox with infinite currents, etc. The real answer is an oscillation that has no steady state. But that's a bit beyond this level, I think. You can find a bunch of discussion of this in older posts.
 
Last edited:
To @nmfowlkes :
Can you please post the circuit that goes with this question for the benefit of those of us whose imagination is not vivid enough to reconstruct it from the problem statement?
 
DaveE said:
The real answer is an oscillation that has no steady state. But that's a bit beyond this level, I think.
Depends upon your level of reality. An ideal capacitor connected with ideal wires will oscillate because there is inductance (even for "ideal" wires I think) and it will oscillate ideally. A "real" capacitor has both inductance and resistance and so will oscillate with a time decay to steady state asymptote. @DaveE knows this but I find it amusing.
 
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