# Potential Difference over Capacitors in Series

• nmfowlkes
In summary, Dave's circuit has two capacitors at different voltages connected in parallel, which is not a good idea because there is no voltage difference after they're connected in parallel and the circuit is fundamentally ill-defined.
nmfowlkes
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

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.

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.

DaveE

## 1. What is potential difference over capacitors in series?

Potential difference over capacitors in series refers to the difference in voltage between the positive and negative terminals of a series of capacitors connected together. This difference in voltage is caused by the accumulation of electric charge on the plates of each capacitor, which creates an electric field between them.

## 2. How is potential difference over capacitors in series calculated?

The potential difference over capacitors in series can be calculated by adding the individual potential differences of each capacitor. This can be done by using the formula V = Q/C, where V is the potential difference, Q is the charge on the capacitor, and C is the capacitance of the capacitor.

## 3. What happens to potential difference over capacitors in series when more capacitors are added?

When more capacitors are added in series, the total potential difference increases. This is because the capacitors act as individual voltage sources, and the potential difference is the sum of the individual voltages. However, the overall capacitance of the circuit also increases, which can affect the total potential difference.

## 4. How does the arrangement of capacitors affect potential difference over capacitors in series?

The arrangement of capacitors does not affect the potential difference over capacitors in series. The potential difference is determined by the individual capacitance and charge of each capacitor, not their physical arrangement in the circuit. However, the arrangement can affect the overall capacitance and thus impact the potential difference.

## 5. What is the significance of potential difference over capacitors in series?

Potential difference over capacitors in series is important in understanding the behavior of capacitors in a circuit. It helps determine the amount of charge stored on each capacitor and the overall voltage of the circuit. It is also essential in calculating the energy stored in capacitors and their discharge rates.

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