# Finding electric field, electric displacement and polarization inside the dielectrics of a coaxial cable. Is this approach right?

A coaxial cable (infinite) is filled in with two dielectric materials. The core conducting cylinder has radius $a$ and is surrounded by a dielectric material $1$ with dialectric constant $\kappa_1$ and has radius $b$, dielectric $1$ is surrounded by a dialectric $2$ with dialectric constant $\kappa_2$ and has radius $c$, finally there is a conducting layer covering it. A difference of potential $V_0$ is established between the conductors. Find $\vec{E},\vec{D},\vec{P}$ inside dialectrics $1$ and $2$

I don't feel very confident with the way I solved it, so I would like you to tell me if my approach is fine.
First of all I think I can assume there is a net charge in the core conducting cylinder and it is distributed on the surface with uniform density $\sigma$. I would like to apply Gauss' law for dielectrics using a gaussian cylinder with radius $r$, where $a<r<b$ and using and argument of symmetry I would get $\vec{D(r)}=\frac{a}{r}\sigma$. Now if I do the same for $b<r<c$ I get the same result because (I think I can assume) there is no free charge inside the dialectrics, which seems consistent with the boundary condition between the two dialectrics: $$(\vec{D_2}-\vec{D_1}) \hat{n}= \vec{D_2}-\vec{D_1} =\sigma_f=0$$ where $\sigma_f$ is the surface density of free charge on the boundary. But if I take the boundary condition between the core cylinder and the dialectric, the electric displacement in the conductor must be zero (because the electric field inside is zero) and now $\sigma_f=\sigma$ so I get: $$\vec{D}\hat{n}=\vec{D}=\sigma_f$$ which does not depend on $r$, so I assume this last condition is wrong (but I don't realize how, so I would like some explanation).
Assuming the first condition is right I would use the equations
$\vec{D}=\epsilon_0\kappa\vec{E}$
$\vec{D}=\epsilon_0+\vec{P}$
to find the electric field and the polarization in both dialectrics as a function of $\sigma$. Finally I would use the equation $$V_0=\int_0^c\vec{E}\vec{dl}$$ to find $\sigma$ and have $\vec{E},\vec{D},\vec{P}$ as functions of $V_0$