Polarization of Coaxial Cable with Compound Dielectric

[sportfreak801]

Homework Statement

A coaxial cable of circular cross section has a compound dielectric. The inner conductor has an outside radius a, which 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. The outer conducting shell has an inner radius c. If a potential difference \varphi_0 is imposed between the conductors, calculate the polarization at each point in the two dielectric media.

Homework Equations

D = \epsilon_0 E + P

\epsilon_1 = K_1\epsilon_0 <br /> \epsilon_2 = K_2\epsilon_0

\oint D\cdot nda = Q_e

The Attempt at a Solution

\oint D\cdot nda = Q_e

D\oint da = Q_e

D(2\pi rl) = Q_e

D = \frac{\lambda}{2r\pi}

The potential from a to b over the first dielectric K_1

\Delta\varphi_1 = D\frac{a_1}{\epsilon_1}

a_1 = \pi(b^2 - a^2)

\Delta\varphi_1 = \frac{\lambda}{r\pi}\frac{\pi(b^2 - a^2)}{\epsilon_1}

Lets call \Delta\varphi_1 = \varphi_1

E_1 = - \nabla\varphi_1

E_1 = -(\frac{\partial\varphi_1}{\partial r} + \frac{1}{r}\frac{\partial\varphi_1}{\partial\theta} + \frac{\partial\varphi_1}{\partial z}

E_1 = -\frac{\partial\varphi_1}{\partial r}

E_1 = -\frac{\partial}{\partial r} \frac{\lambda}{r}(\frac{b^2 - a^2}{\epsilon_1})

E_1 = -\lambda(\frac{b^2 - a^2}{\epsilon_1}) \frac{\partial}{\partial r} \frac{1}{r}

E_1 = -\lambda(\frac{b^2 - a^2}{\epsilon_1}) (\frac{-1}{r^2})

E_1 = \frac{\lambda}{r^2}(\frac{b^2 - a^2}{\epsilon_1})

P_1 = D - \epsilon_0 E_1

P_1 = \frac{\lambda}{r\pi} - \epsilon_0(\frac{\lambda}{r^2}(\frac{b^2 - a^2}{\epsilon_1}))

P_1 = \frac{\lambda}{r\pi} - \frac{\epsilon_0}{\epsilon_1}(\frac{\lambda}{r^2}(b^2 - a^2))

P_1 = \frac{\lambda}{r\pi} - \frac{\lambda}{K_1 r^2} (b^2 - a^2)

P_1 = \frac{\lambda}{r} (\frac{1}{\pi} - \frac{b^2 - a^2}{K_1 r})

However, this does not seem correct since \lambda is not given in this problem. Any help would be greatly appreciated. Thanks in advance.

 

[sportfreak801]

\oint D\cdot nda = Q_e

D\oint da = Q_e

D(2\pi rl) = Q_e

D = \frac{\lambda}{2r\pi}

E_1 = \frac{D}{\epsilon_1} = \frac{\lambda}{2r\pi \epsilon_1}

The potential from a to b over the first dielectric

\Delta\varphi_1 = E_1 (b - a)

\Delta\varphi_1 = \frac{\lambda (b-a)}{2r\pi \epsilon_1}

The potential from b to c over the second dielectric

\Delta\varphi_2 = \frac{\lambda (c-b)}{2r\pi\epsilon_1}

\varphi_0 = \Delta\varphi_1 + \Delta\varphi_2

\varphi_0 = \frac{\lambda (b-a)}{2r\pi \epsilon_1} + \frac{\lambda (c-b)}{2r\pi \epsilon_2}

\varphi_0 = \frac{\lambda}{2r\pi}(\frac{b-a}{\epsilon_1} + \frac{c-b}{\epsilon_2})

\lambda = \frac{\varphi_0 2r\pi}{(\frac{b-a}{\epsilon_1} + \frac{c-b}{\epsilon_2})}

E_1 = \frac{1}{2r\pi\epsilon_1} ( \frac{\varphi_0 2r\pi}{(\frac{b-a}{\epsilon_1} + \frac{c-b}{\epsilon_2})})

E_1 = \frac{1}{\epsilon_1}(\frac{\varphi_0}{(\frac{b-a}{\epsilon_1} + \frac{c-b}{\epsilon_2})})

P_1 = \chi_1 E_1

P_1 = (\frac{K_1 - 1}{K_1})(\frac{\varphi_0}{(\frac{b-a}{\epsilon_1} + \frac{c-b}{\epsilon_2})})

Does this make sense? Any help would be greatly appreciated. Thanks in advance.