Fundamental capacitor discharge question

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SUMMARY

The discussion centers on the behavior of capacitors when subjected to an open circuit after charging. Participants clarify that while a capacitor stores charge when a DC voltage is applied, it does not discharge immediately upon removal of the voltage due to the dielectric material's minimal conductivity. The hydraulic analogy is debated, with emphasis on the importance of understanding the principles of capacitance and charge displacement rather than relying solely on analogies. Ultimately, the consensus is that charge remains displaced within the capacitor until a conductive path is established.

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  • Understanding of basic capacitor functionality and charge storage
  • Familiarity with dielectric materials and their properties
  • Knowledge of electrical circuits and open circuit concepts
  • Basic grasp of hydraulic analogies in electrical systems
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EE4life
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Hi All,

I understand that when a capacitor is subjected to a DC voltage, it stores charge.
However, when the DC voltage is removed (open circuit) why does the capacitor not internally discharge? Shouldn't the charge just go back to where it was, leaving the capacitor with 0 stored charge?

I am trying to understand the capacitor through the membrane in a water pipe example (http://en.wikipedia.org/wiki/Capacitor#Hydraulic_analogy). Wouldn't the water be pushed back once the water pressure was relieved(open circuit)?

Thanks in advance.
 
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EE4life said:
However, when the DC voltage is removed (open circuit) why does the capacitor not internally discharge?
Condenser will discharge, but veeery slowly. Dielectric material in condenser has some small conductivity.
 
EE4life said:
Wouldn't the water be pushed back once the water pressure was relieved(open circuit)?

You may be confused by the terminology, “open switch” = “closed valve”.
Pressure is the analogue of voltage, water flow is the analogue of electric current.

The hydraulic analogy requires no water current flow when the switch is opened after charging the elastic membrane dielectric. That requires a valve that can be turned off. Pressure can then remain until valve leakage (=discharge) allows it to fall.
 
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Baluncore said:
Pressure can then remain until valve leakage (=discharge) allows it to fall.
Okay. When an open circuit is introduced, it is like the edges of the pipe are clamped.
Since the water at both sides of the membrane are at equal pressure, the forces on the membrane cancel and the membrane does not move. When the pressure is made to fall on one of the sides, the membrane pushes water/electrons causing charge to flow.

As for a capacitor, an applied voltage causes a displacement of charge in the capacitor. This internal displacement of charge attracts extra positive and extra negative charges on the opposing external leads/electrode of the capacitor. These extra charges come from the power supply. When the power supply is disconnected(open circuit) the extra charge has no where to go.The extra charges continue to displace the internal charge of the capacitor.

The extra charges cannot flow into the capacitor because charge cannot flow through a capacitor. If the ends the capacitor are connected, either through a wire or resistor, the charges at a higher potential will fall to the lower potential, ie current/spark. The extra positive charges are attracted to the negative induced charge in the capacitor, but the resistance is infinite as the conductivity of a dielectric is theoretically zero. The extra positive charges take the path of least resistance to the lower potential.

I think that answers my question.
 
EE4life said:
Okay. When an open circuit is introduced, it is like the edges of the pipe are clamped.
Since the water at both sides of the membrane are at equal pressure, the forces on the membrane cancel and the membrane does not move. When the pressure is made to fall on one of the sides, the membrane pushes water/electrons causing charge to flow.

When an open circuit is introduced, it is like both ends of the pipe are sealed.
The water on both sides of the stretched membrane are at different pressure. The volume of water on each side of the membrane is fixed, so the membrane cannot move. When the volume on both sides is allowed to change, the membrane tension can push water/electrons causing charge to flow.
The water analogy is not always useful.
 
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That danged water analogy again! It always gives people problems - at all levels.
 
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Analogies between electrical and mechanical situations are bad in general.
 
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These analogies are only meaningful after you know the principles they're illustrating. :D
 
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dlgoff said:
These analogies are only meaningful after you know the principles they're illustrating. :D
Absolutely.
In such a situation, it's best to avoid passing on the personal analog model to someone who has not 'got there yet' because it can easily add to their confusion.
Even the Mathematical model (analogy) is quite capable of letting you down at times but there's almost never a better one to use than the Maths.
 
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  • #10
zoki85 said:
Analogies between electrical and mechanical situations are bad in general.

I don't know; the second-order RLC filter as a mass-on-a-spring is a pretty good analogy.
 
  • #11
I don't know, flow of electricity through metal conductors as flow of a watter through pipes is a pretty bad analogy.
 
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  • #12
hmm... not sure what to make of that comment. I certainly agree that comparing flow of water with flow of electricity is problematic.
 
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  • #13
Well, some analogies are better than others, and some are just bad.
 
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