# Homework Help: Polarization of Coaxial Cable with Compound Dielectric

1. Mar 3, 2010

### sportfreak801

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

$$D = \epsilon_0 E + P$$

$$\epsilon_1 = K_1\epsilon_0 \epsilon_2 = K_2\epsilon_0$$

$$\oint D\cdot nda = Q_e$$

3. 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.

2. Mar 4, 2010

### 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.