Capacitors in Series: Why No Current Flow?

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Discussion Overview

The discussion revolves around the behavior of capacitors in series, specifically addressing why current does not flow between two fully charged capacitors when connected in series. Participants explore concepts related to charge distribution, potential difference, and equipotential conditions in the context of electrical circuits.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants question why current does not flow from the right plate of C1 to the left plate of C2 when they are connected in series.
  • Others argue that for both capacitors to become charged, current must have flowed at some point.
  • One participant clarifies that connecting fully charged capacitors together should result in current flow, prompting further questions about circuit configuration.
  • Another participant suggests that if the ends are left open, it may not affect the current flow, questioning the necessity of a complete path for current to flow.
  • Some participants assert that once fully charged, no current will flow due to the lack of potential difference between the plates of the capacitors.
  • One participant explains that the combination of the right plate of C1 and the left plate of C2 becomes an equipotential, leading to zero resultant force on electrons, thus preventing current flow.
  • Another participant introduces a scenario involving a potential difference of 12 volts across the circuit and discusses how this voltage would be divided across the capacitors, raising questions about current flow under these conditions.
  • Some participants agree that if the circuit is broken and reformed without discharging the capacitors, charge could flow between them, but the total potential would remain the same.

Areas of Agreement / Disagreement

Participants express differing views on whether current flows between fully charged capacitors in series, with some asserting that it does not due to equipotential conditions, while others propose scenarios where current could flow under specific circumstances. The discussion remains unresolved with multiple competing views present.

Contextual Notes

Participants reference concepts such as potential difference, equipotentiality, and charge distribution without reaching a consensus on the implications for current flow in series-connected capacitors.

Pranav Jha
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----C1------C2------

If the left and right plates of C1 are +ve and -ve charged respectively and the left and right plates of C2 are +ve and -ve charged respectively then why doesn't the current flow from the right plate of C1 to the left plate of C2 when they are connected in series?
 
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What do you mean doesn't flow?

The only way for both to become charged is for current to flow. If they are both charged, then current has flowed.
 
I meant won't connecting the two fully charged capacitors together via a connecting wire result in a current to flow from one capacitor to another?
 
So you're talking about shorting the circuit? Or are you leaving each end open as in your picture?
 
will leaving the ends open make any difference? shouldn't a complete path between the negative plate of one capacitor and the positive plate of another capacitor be enough to result in a current flow from the negative plate of the first to the positive plate of the second?
 
If they are fully charged, no current will flow.
 
Once fully charged the combination of right plate of C1 with left plate of C2 and the connecting wire between them becomes an equipotential,that is there is no potential difference between C1 and C2.
Think of say an electron on the right plate of C1.This is repelled by the other electrons on that plate and on the right plate of C2 but attracted to the positive charges on the other two plates,the resultant force being zero.
 
suppose we have a capacitor fully charged that is connected across a p.d of 12 volts. Now, if we add another capacitor in series, the potential of 12V will be divided across the two capacitors. So, the charge is in excess in the first capacitor for the new reduced voltage across it. So, will the current now flow from the first capacitor to the new second capacitor?
 
Dadface said:
Once fully charged the combination of right plate of C1 with left plate of C2 and the connecting wire between them becomes an equipotential,that is there is no potential difference between C1 and C2.
Think of say an electron on the right plate of C1.This is repelled by the other electrons on that plate and on the right plate of C2 but attracted to the positive charges on the other two plates,the resultant force being zero.

that does make sense in the terms of electrical forces :D
 
  • #10
Pranav Jha said:
----C1------C2------

If the left and right plates of C1 are +ve and -ve charged respectively and the left and right plates of C2 are +ve and -ve charged respectively then why doesn't the current flow from the right plate of C1 to the left plate of C2 when they are connected in series?

Because you are forgetting that there is a voltage drop across each capacitor. Let's say that there is a potential difference of 12 volts across the entire circuit. If you had a voltmeter and put it on the terminals of C1 you would find a drop of 6 volts and you would find the same drop across the terminals of C2, assuming that they are of equal capacitance. The charges do not flow because there is a 6 volt potential opposing them.
 
  • #11
Pranav Jha said:
suppose we have a capacitor fully charged that is connected across a p.d of 12 volts. Now, if we add another capacitor in series, the potential of 12V will be divided across the two capacitors. So, the charge is in excess in the first capacitor for the new reduced voltage across it. So, will the current now flow from the first capacitor to the new second capacitor?

If you could break and reform the circuit without discharging the capacitor and they were of equal capacitance half of the charge would flow from one to the other. The total potential across the two would still be 12 volts. Each capacitor would have a drop of 6 volts across it.
 

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