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.

Thanks in advance.


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.


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