Capacitor transient response, what happens in the dielectric?

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Discussion Overview

The discussion revolves around the transient response of a capacitor in an RC circuit when connected to a DC source, specifically focusing on the behavior of the dielectric during the initial moments after the circuit is closed. Participants explore the relationship between current, charge buildup, voltage across the capacitor, and the polarization of the dielectric.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant describes the initial behavior of the capacitor as acting like a short circuit, with zero voltage across the terminals and maximum current flowing, while questioning the role of the dielectric during this phase.
  • Another participant suggests that the charge on the capacitor tends to zero initially, leading to a corresponding potential of zero, indicating a buildup of charge over time.
  • A later reply clarifies that charge is not flowing through the capacitor but rather building up on the plates, with the dielectric transitioning from an equilibrium state to a polarized state due to this charge buildup.
  • There is a comparison made to a short circuit, where a high current results in almost no potential difference, further illustrating the initial conditions of the capacitor.
  • Participants express uncertainty about the terminology used, particularly regarding the flow of charge versus charge buildup on the capacitor plates.

Areas of Agreement / Disagreement

Participants generally agree on the initial behavior of the capacitor as acting like a short circuit, but there is some disagreement or confusion regarding the terminology and the precise nature of charge movement and dielectric polarization.

Contextual Notes

Some assumptions about the behavior of the dielectric and the definitions of charge flow versus charge buildup remain unresolved, as do the implications of these behaviors on the overall circuit dynamics.

Lavabug
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I'm trying to marry the concepts from my EM course with what I've been doing in my lab course. Correct me if I'm wrong, in a plain RC circuit, when closed with a DC source, the capacitor initially acts as a short (voltage is zero across the terminals, while I is max), until after a few microseconds the voltage increases exponentially to the one I'm administering (consequently I drops down to 0 ideally).

What's happening to the dielectric in the capacitor in these first few moments? The current through the cap is initially nonzero, meaning charge is flowing through it and the dielectric is quickly becoming polarized, but why is the voltage zero at the terminals initially? There's a circulation of E between the capacitor plates, so why isn't there any difference in potential on the terminals?
 
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Initially ie just after switch is closed, the current develops and the charge on the capacitance is very low (it builds up with time). Since charge tends to 0 initially potential also tends to zero.
 
Not sure I understand you, could you please elaborate a bit more?
 
Lavabug said:
What's happening to the dielectric in the capacitor in these first few moments? The current through the cap is initially nonzero, meaning charge is flowing through it and the dielectric is quickly becoming polarized, but why is the voltage zero at the terminals initially? There's a circulation of E between the capacitor plates, so why isn't there any difference in potential on the terminals?

I'm not sure it's completely accurate to say that charge is flowing through it. It's more like charge is building on the plates and the dielectric is shifting from an equilibrium state to a polarized state in response to the charge buildup.

You kind of hit on it yourself... It acts like a short initially. So when you short something out, for example you place a wire between the two terminals on a battery, you have a great deal of current and (almost) no potential difference between the terminals.

When the voltage builds in the cap to the applied voltage, the source can't "push" any more charge to the plates, so it then acts like an open.

It sounds like you know most of this, but I hope maybe that helped a little...
 

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