Polarization of Coaxial Cable with Compound Dielectric

  • #1

Homework Statement


A coaxial cable of circular cross section has a compound dielectric. The inner conductor has an outside radius [itex] a [/itex], which is surrounded by a dielectric sheath of dielectric constant [itex] K_1 [/itex] and of outer radius [itex] b [/itex]. Next comes another dielectric sheath of dielectric constant [itex] K_2 [/itex] and outer radius [itex] c [/itex]. The outer conducting shell has an inner radius [itex] c [/itex]. If a potential difference [itex] \varphi_0 [/itex] is imposed between the conductors, calculate the polarization at each point in the two dielectric media.


Homework Equations



[tex] D = \epsilon_0 E + P [/tex]

[tex] \epsilon_1 = K_1\epsilon_0
\epsilon_2 = K_2\epsilon_0 [/tex]

[tex] \oint D\cdot nda = Q_e [/tex]



The Attempt at a Solution


[tex] \oint D\cdot nda = Q_e [/tex]

[tex] D\oint da = Q_e [/tex]

[tex] D(2\pi rl) = Q_e [/tex]

[tex] D = \frac{\lambda}{2r\pi}[/tex]

The potential from a to b over the first dielectric [itex] K_1[/itex]

[tex] \Delta\varphi_1 = D\frac{a_1}{\epsilon_1} [/tex]

[tex] a_1 = \pi(b^2 - a^2) [/tex]

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

Lets call [itex] \Delta\varphi_1 = \varphi_1 [/itex]

[tex] E_1 = - \nabla\varphi_1 [/tex]

[tex] E_1 = -(\frac{\partial\varphi_1}{\partial r} + \frac{1}{r}\frac{\partial\varphi_1}{\partial\theta} + \frac{\partial\varphi_1}{\partial z} [/tex]

[tex] E_1 = -\frac{\partial\varphi_1}{\partial r} [/tex]

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

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

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

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

[tex] P_1 = D - \epsilon_0 E_1 [/tex]

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

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

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

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

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

Answers and Replies

  • #2
[tex] \oint D\cdot nda = Q_e [/tex]

[tex] D\oint da = Q_e [/tex]

[tex] D(2\pi rl) = Q_e [/tex]

[tex] D = \frac{\lambda}{2r\pi}[/tex]

[tex] E_1 = \frac{D}{\epsilon_1} = \frac{\lambda}{2r\pi \epsilon_1} [/tex]

The potential from a to b over the first dielectric

[tex] \Delta\varphi_1 = E_1 (b - a) [/tex]

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

The potential from b to c over the second dielectric

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

[tex] \varphi_0 = \Delta\varphi_1 + \Delta\varphi_2 [/tex]

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

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

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

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

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

[tex] P_1 = \chi_1 E_1 [/tex]

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

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

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