Non-symmetrical electrostatics problem I was thinking about a problem like this one

where $q$ is a point charge and the gray thing is an infinitely long and infinitely wide conducting wall, although its thickness is finite. (I mean that it is infinite up and down, and in and out, but not right to left.)
I have two questions:


*

*Is it possible to find the electric potential for any position in space? I was thinking about the method of images but the conductor is not infinitely wide, so I don't think that can be of use here.

*Also, I don't know if the field to the right of the wall would be zero or not. I think that the charge will induce a negative superficial charge on the left side of the conductor, but due to the neutrality of it, a positive superficial charge with the same distribution value will appear on the right side, so I don't know if there will be electric field to the right of the wall. I believe that, if the conductor was grounded, the field would be 0 there. But what if it isn't grounded?
 A: "Is it possible to find the electric potential for any position in space?"
Of course you can solve the problem with separation of variables in cylindrical coordinates. Separate the problem in four regions as shown in the next picture:

                           

Consider the distribution of the point charge as an infinite sheet with surface charge density:
$$\sigma_q=\frac{1}{2\pi r}\delta(r)\delta(z-z^\prime)$$
where primed variables denote the position of the charge. Also, you know that the induced charge will be found on the surface of your "conductor wall", so you can think of it as two infinite induced charge sheets with (unknown) densities $\sigma_{\text{ind}}^1$ and $\sigma_{\text{ind}}^2$. 
Now, the problem can be done via superposition principle, taking the electrostatic potential $\Phi(r,z)$ of each distribution as an expansion in cylindrical coordinates (which will be similar to the Green's function expansion in these coordinates, equation (8) given by mathworld). You will have to determine constants applying boundary conditions. In particular, you may obtain the induced surface densities (and therefore the total induced charge) using the condition of discontinuity in the normal component of the electric field at the surface of the conductor.
"What if the conductor is not grounded?"
Then, you will still have that the electric field inside the conductor will be zero, but the potential will be constant. There should be an electric field produced by the induced conductor outside of it. For a grounded conductor, the potential inside will be zero, and will not produce electric field outside.
PS: If you don't understand any of this, you need to know about Boundary-Value Problems in Electrostatics (separation of variables and Green's function), which you can find in chapters 2 and 3 of Jackson's Classical Electrodynamics 2nd Ed. book. You can also find about the subject here. 
A: In fact, an infinite non-grounded conductor is not a well-defined boundary condition, especially in your case. The electric potential at infinite point cannot be defined as zero. The only well-defined boundary condition is the finite-size of conductor, and I believe that you can find answer in textbooks.
In this case, your problem is the electric potential local distribution. Let's try a finite-size but largely enough conductor. Locally, it is close to your problem. The answer is the electric field on right side is zero.
Of course，the non-grounded conductor has a finite electric potential. Roughly speaking, the charge accumulating on the left surface is equal to $-q$, and charge density on the right side is $\sigma=\frac{q}{S}$, where $S$ is the area of your planar conductor on right surface. In the limit of large enough conductor, $\sigma=\frac{q}{S} \rightarrow 0$. Thus, the electric field at right side is $E=\frac{\sigma}{\epsilon_0}\rightarrow 0$.
