Microscopically (Classically), How does Current keep Increasing with Decreasing Electrostatic Field inside of Capacitor in a $R(L)C$ circuit? 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 $\vec E_p$?
 A: Your assumption that the current "keeps increasing" while the capacitor electric field is decreasing does not seem to correspond to what happens.
Let's start with a charged capacitor at a voltage Vo, in a circuit where the discharging resistor is separated by a switch.
Before the switch is flipped, the capacitor has a charge +Qo on one plate and a charge -Qo on the other one. Well, more or less because the wire and the switch contact will become part of the conducting body and some charge will be present there, too, but let's not look into this now.
When you flip the switch, the opposite charges on the two plates will start to flow (it's only electrons, in reality), and the plate with an excess of electrons will start to lose them while the opposite plate will start to gain them. All of this through the switch and the resistor which limit the maximum current to Vo/R.
So the current starts at this value (note 1) with an abrupt step, and then as the plates get depleted it decreases as the voltage across the capacitor decreases.
Note 1: if your question is about the initial rising step in the current, you need to leave circuit theory and solve Maxwell's equations and take into account the role of surface charge - not only in the capacitor, but in the wires as well.
Things get fairly complicated pretty quick.
The small field E = j/sigma inside the conducting wires is the result of the superposition of the electric fields due to all surface and interface charges, and it is quite different from the field between the plates of the capacitor.
Moreover, in the initial transient at the flipping of the switch, current in the wires starts locally at the switch location (where charge had previously accumulated) and then propagates along the wires (but can also appear elsewhere due to the fields propagating in space) at a very high speed until the current is the same along the entire wire (what is takes for granted in circuit theory). The initial transient, can be complicated and I would love to see a simulation of the electric and magnetic fields of an RC circuit in the first picoseconds after the switch has been closed. (Recently an online discussion about how energy is transferred has resulted in some good simulations of this transient for a battery-resistor circuit, which is close enough to get an idea.)
I am not sure you were asking about this, tho.
