Charges on a surface Suppose we have a cylinder with uniform density and radius A. A conductor cylindrical shell with inner radius B and outer radius C is axial to the cylinder and involves it. Between the solid cylinder and the shell, the space is covered with a dielectric with permittivity $\varepsilon$. The shell has zero total charge.
I am a little confused about how to calculate the surface charge in the radius B and C. I mean, should we use $\sigma = \vec P\cdot\hat n$ for B, and so the negative of it to C? I am really confused because, if this is the case, I believe that the field between B and C would not be zero (as it should be in a conductor). But I don't know also why this is wrong.
Maybe this is only half of the answer?
 A: For problems like this, where some given charge $Q$ is enclosed somewhere, Gauss's law is always very useful. (I will assume the cylinders are very long compared to their radii.) As you state correctly, the electric field inside a conductor must always vanish.
Let's apply Gauss's law to a surface just inside the inner side of the conductor, i.e. we integrate the electric field over a cylinder of radius $r=B + \Delta$ for a small $\Delta$. The field at this position is zero, so the integral automatically vanishes. This is also equal to the total charge inside this volume, so the total charge on the inner side of the shell must be $-Q$ so that it cancels out the charge of $+Q$ on the inner cylinder. Working out the surface charge is just dividing that $Q$ by the surface area at $r=B$.
Consequently, if the inner surface at $r=B$ has charge $-Q$, the outer surface must have charge $+Q$, but it's spread out over a larger area, so the density is lower. We can also see that by Gauss's law. If we integrate the electric field over a cylindrical surface at $r=C+\Delta$, we must get a value of $Q$, since that is the enclosed charge. At $r=C-\Delta$, the field is zero (conductor), so all that electrical field must come from the surface charge at $r=C$, meaning $+Q$ is evenly spread out over the surface at $r=C$.
