Induced magnetic field produces electric field and vice versa forever! So here are the two of Maxwell's laws that I am interested in:
So we have the simple circuit (from google):

So, before the system goes into steady-state we know that charge slowly accumulates on the plates of the conductor. So the charge on the plates gets bigger and bigger while the charges that carry the current get smaller and smaller, so the current gets weaker.
Applying Ampere's law on the wire we find the induced magnetic field due to the current $I(t)$ that penetrates the surface $\Sigma$ (see the integral of $\mathbf{J}\mathrm{d}\mathbf{S}$) and not due to an electric field.
Now, this induced magnetic field is changing with respect to time (because current is changing). But from the Maxwell-Faraday equation we conclude that this changing magnetic field will produce an electric field which again changes with respect to time. And then we have another induced magnetic field due to that changing electric field. And the cycle goes on.
So, am I right? And if I am, when does this stop? And how does it changes the way I calculate each induced field? Does it have to do with electromagnetic waves?
 A: You are roughly correct. However, you must be careful because the surface you would choose for finding the magnetic field from $\mathbf{J} \cdot \mathrm{d}\mathbf{S}$ is NOT the same surface you would use to find the electric field. 
The concept you want to solve this problem is self-inductance. Defining the magnetic flux $\Phi=\int\mathbf{B} \cdot \mathrm{d}\mathbf{S}$ and the electromotive force $\varepsilon = \oint\mathbf{E} \cdot \mathrm{d}\mathbf{S}$, we can rewrite the Maxwell-Faraday equation as 
$$\varepsilon = -\frac{\mathrm{d}\Phi}{\mathrm{d}t}$$
and note that the current produced is 
$$I=\frac{V+\varepsilon-\frac{Q}{C}}{R}$$
i.e., the sum of voltages around the whole circuit.
In general the total magnetic flux $\Phi$ through a circuit will depend in some complicated way on the geometry of the circuit, and it is hard to solve except in a few simple cases like solenoids. However, we can see from the second Maxwell equation that it will always be proportional to the current. (The second term is zero since there is no electric field perpendicular to the circuit in this problem.) Let's call the complicated geometric dependence the self-inductance $L$, and rewrite the second Maxwell equation as 
$$ \Phi = LI$$
You can now write that 
$$ RI = V-L\frac{\mathrm{d}I}{\mathrm{d}t}-\frac{Q}{C}$$
Observing that $I=\frac{\mathrm{d}Q}{\mathrm{d}t}$ lets you rewrite the expression in terms of a single differential equation with one variable $Q$. Once you have the solutions for $Q(t)$, you can find the behavior of, say, $\varepsilon$ through the equations we already defined and appropriate initial conditions. If you have not seen equations of this type before, it may help to search for "damped harmonic oscillator."
