Capacitor transient response, what happens in the dielectric?

[Lavabug]

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?

 

[ashishsinghal]

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.

 

[Lavabug]

Not sure I understand you, could you please elaborate a bit more?

 

[Evil Bunny]

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...