Assuming the capactior is intially charged with voltage $V$, then detached from the voltage source
I know mathematically by using:
$$C\frac{dV_c}{dt}+\frac{V}{R}=0$$
I can get an expression of the current and see the increase of it.
But what happens microscopically? With the classical model (assume only 1 type of particle), the current is defined as (taken from Purcell): $$I = \rho {\left\langle {\vec u} \right\rangle _{drift}}\vec a = Ne{\left\langle {\vec u} \right\rangle _{drift}}\vec a = {e^2}\left( {\frac{{N\tau }}{m}} \right)\vec f\vec a = \sigma \vec E_p\vec a \\ $$\begin{cases} \text{$\rho$ is the charge density per unit volume} \\ \\ \text{$\left\langle {\vec u} \right\rangle _{drift}$ is the average velocity of electron}\\ \\ \text{${\vec a}$ is the cross section area}\\ \\ \text{$N$ is the number of particles per unit volume}\\ \\ \text{$\tau$ is the characterastic time of collision} \\ \\ \text{$m$ is the mass of electron} \end{cases}
There is also the electrostatic field inside of a parallel capacitor: $${\vec E_c}=\frac{\sigma}{2\epsilon_0}\hat{n.}$$
At first glance, since $I \propto {\left\langle {\vec u} \right\rangle}_{drift} \propto \vec{E}_p $, as the current increasing, there should be an increasing eletrostatic field, which is not the case inside of the capacitor.
I then realized that $\vec E_c$ inside of the capacitor is not the same as $\vec E_p$ which gives momentum to charged particles in the wire:
$$\vec E_c \ne \vec E_p$$
But what is $\vec E_p$ then? It doesn't seem like an electrostatic field since the voltage $V(t)$ provided by the capacitor is time dependent.
What is the mechanism that keeps accelerating the particles, so drifting velocity increases to its maximum, as the voltage drops to zero? And how does this mechanism correspond to $\vec E_p$?
