# Coaxial cable with compound dielectric

I'm trying to solve a problem from Reitz and Milford's Foundations of Electromagnetic Theory (3rd ed, problem 4-8), and don't know how to start:

A coaxial cable of circular cross section has a compound dielectric. The inner conductor has an outside radius $$a$$; this is surrounded by a dielectric sheath of dielectric constant $$K_1$$ and of outer radius $$b$$. Next comes another dielectric sheath of dielectric constant $$K_2$$ and outer radius $$c$$. If a potential difference $$V_0$$ is imposed between the conductors, calculate the fields $$\vec{E}(\vec{r}), \, \vec{D}(\vec{r}), \, \vec{P}(\vec{r})$$ in both dielectrics.

I'm assuming I have to use the solution to Laplace's equation in cylindrical coordinates, but I'm not sure about how to use the border conditions.

Well, solved it by myself after a couple of hours, so I'm posting the solution just in case someone happens to have the same problem:

Let $$Q$$ be the charge on a length $$l$$ of the inner conductor. Applying Gauss's Law to that piece of cable: $$\oint \vec{D} \cdot d\vec{a} = D 2 \pi r l = Q \implies D = \frac{Q}{2 \pi r l}.$$ By definition, $$\vec{D} = \varepsilon_0 \kappa \vec{E}$$, so $$E_1 = \frac{Q}{2 \pi \varepsilon_0 \kappa_1 l r }, \, (a < r < b),$$ $$E_2 = \frac{Q}{2 \pi \varepsilon_0 \kappa_2 l r}, \, (b < r < c).$$

There is a potential difference $$V_0$$ between the conductors, so $$V_0 = \int_{a}^{b} \left( \frac{Q}{2 \pi \varepsilon_0 \kappa_1 l} \right) \frac{dr}{r} + \int_{b}^{c} \left( \frac{Q}{2 \pi \varepsilon_0 \kappa_2 l} \right) \frac{dr}{r} = \frac{Q}{2 \pi \varepsilon_0 l} \left[ \frac{1}{\kappa_1} \ln \left( \frac{b}{a} \right) + \frac{1}{\kappa_2} \ln \left( \frac{c}{b} \right) \right].$$ $$\implies Q = \frac{2 \pi \varepsilon_0 l V_0}{\frac{1}{\kappa_1} \ln \left( \frac{b}{a} \right) + \frac{1}{\kappa_2} \ln \left( \frac{c}{b} \right)} = \frac{2 \pi \varepsilon_0 l V_0 \kappa_1 \kappa_2}{\kappa_2 \ln \left( \frac{b}{a} \right) + \kappa_1 \ln \left( \frac{c}{b} \right)}.$$ Using $$\sigma = \dfrac{Q}{2 \pi r l}$$, we get: $$\vec{D} = \left( \frac{\varepsilon_0 V_0 \kappa_1 \kappa_2}{\kappa_2 \ln \left( \frac{b}{a} \right) + \kappa_1 \ln \left( \frac{c}{b} \right)} \right) \frac{1}{r} \hat{e_r},$$ and $$\vec{D}_1 = \vec{D}_2 = \vec{D}.$$

As $$\vec{D} = \varepsilon_0 \kappa \vec{E}$$, we get: $$\vec{E}_1= \dfrac{\vec{D}_1}{\varepsilon_0 \kappa_1}, \, \vec{E}_2 = \dfrac{\vec{D}_2}{\varepsilon_0 \kappa_2}:$$

$$\vec{E}_1 = \left( \frac{\kappa_2 V_0}{\kappa_2 \ln \left( \frac{b}{a} \right) + \kappa_1 \ln \left( \frac{c}{b} \right)} \right) \frac{1}{r} \hat{e_r},$$ $$\vec{E}_2 = \left( \frac{\kappa_1 V_0}{\kappa_2 \ln \left( \frac{b}{a} \right) + \kappa_1 \ln \left( \frac{c}{b} \right)} \right) \frac{1}{r} \hat{e_r}.$$

Finally, $$\vec{P} = \vec{D} - \varepsilon_0 \vec{E}:$$

$$\vec{P}_1 = \frac{(\kappa_1 -1) \varepsilon_0 \kappa_2 V_0}{\kappa_2 \ln \left( \frac{b}{a} \right) + \kappa_1 \ln \left( \frac{c}{b} \right)} \frac{1}{r} \hat{e_r},$$ $$\vec{P}_2 = \frac{(\kappa_2 -1) \varepsilon_0 \kappa_1 V_0}{\kappa_2 \ln \left( \frac{b}{a} \right) + \kappa_1 \ln \left( \frac{c}{b} \right)} \frac{1}{r} \hat{e_r}.$$

I hope I haven't made any mistakes. If anyone finds something please let me know.