Conductance per unit length of a co-axial cable (Intro to Electromagnetics)

[peterpiper]

Homework Statement

Starting with:

G = $\frac{I}{V}= \frac{-\int_{s}\sigma\mathbf{E}\cdot d\boldsymbol{s}}{-\int_{l}\boldsymbol{E}\cdot d\boldsymbol{l}}$

Derive the conductance per unit length, G', of a coaxial cable by assuming a line charge of $\rho_{l}$ on the center conductor.

Homework Equations

$\boldsymbol{E}=\boldsymbol{\hat{r}}\frac{\rho _{l}}{2\pi\epsilon_{0}r}$

Let a be the radius of the inner conductor and let b be the radius of the outer conductor.

The Attempt at a Solution

So far my only attempt has been plugging E straight into the integrals with the region of integration being a to b on both integrals and 0 to 2$\pi$ on the double integral. I get the correct coefficient of $2\pi\sigma$ but I'm supposed to have a ln(b/a) in the denominator. I'm kinda baffled as to what I'm supposed to do. I'm assuming I'm doing something stupid in the double integral but those never were my strong suit in math. Any help would be greatly appreciated as this is the last bit of work I need to do to finish my E-Mag course. Thanks in advance for anything you may be able to offer me.

 

[klondike]

The numerator can be evaluated algebraically because of the symmetry of the E field on any coaxial cylindrical surface. The denominator is a trivial integral. It's hard to imagine how you got it wrong without seeing how your integrations were done.

Homework Statement

Starting with:

G = $\frac{I}{V}= \frac{-\int_{s}\sigma\mathbf{E}\cdot d\boldsymbol{s}}{-\int_{l}\boldsymbol{E}\cdot d\boldsymbol{l}}$

Derive the conductance per unit length, G', of a coaxial cable by assuming a line charge of $\rho_{l}$ on the center conductor.

Homework Equations

$\boldsymbol{E}=\boldsymbol{\hat{r}}\frac{\rho _{l}}{2\pi\epsilon_{0}r}$

Let a be the radius of the inner conductor and let b be the radius of the outer conductor.

The Attempt at a Solution

So far my only attempt has been plugging E straight into the integrals with the region of integration being a to b on both integrals and 0 to 2$\pi$ on the double integral. I get the correct coefficient of $2\pi\sigma$ but I'm supposed to have a ln(b/a) in the denominator. I'm kinda baffled as to what I'm supposed to do. I'm assuming I'm doing something stupid in the double integral but those never were my strong suit in math. Any help would be greatly appreciated as this is the last bit of work I need to do to finish my E-Mag course. Thanks in advance for anything you may be able to offer me.

 

[peterpiper]

For the surface integral I did this:

$-\sigma\int_{0}^{2\pi}\int_{a}^{b}\frac{\rho_{l}}{2\pi\epsilon_0r}drd\phi=-\sigma\int_{a}^{b}\frac{2\pi\rho_l}{2\pi\epsilon_0r}dr=\frac{-\sigma\rho_l}{\epsilon_0}ln(\frac{b}{a})$

For the line integral I did this:

$\int_{a}^{b}\frac{\rho_{l}}{2\pi\epsilon_0r}dr= \frac{\rho_l}{2\pi\epsilon_0}ln(\frac{b}{a})$

This leaves me with a result of $-2\pi\sigma$ correct?

 

[klondike]

The surface integral represents how many charges leaked through a cylindrical surface per unit time. Select such a surface somewhere between the inner and outer conductor by selecting a fixed r. Hence the r should be constant in the first integral. An notice that E is perpendicular and uniform to the surface everywhere.

For the surface integral I did this:
$-\sigma\int_{0}^{2\pi}\int_{a}^{b}\frac{\rho_{l}}{2\pi\epsilon_0r}drd\phi=-\sigma\int_{a}^{b}\frac{2\pi\rho_l}{2\pi\epsilon_0r}dr=\frac{-\sigma\rho_l}{\epsilon_0}ln(\frac{b}{a})$

 

[peterpiper]

I'm not sure I understand what you're getting at.

 

[klondike]

Maybe a picture helps. The surface you're trying to integrate is the blue one as shown in the picture. http://www.physics.sjsu.edu/becker/physics51/images/23_16Cylinder.JPG

since r is constant, you don't need to integrate. the E field is uniform on and perpendicular to the surface. You don't need to integrate at all.

I'm not sure I understand what you're getting at.

 

[rude man]

You don't need to integrate to get I. You do need to integrate to get V.

The reason you don't need to integrate to get I is that E(r) ~ 1/r and S ~ r so the r's cancel.

But when you do the line integral of _a∫^bE*dr you are actually integrating.

You can double-check your answer by computing the total resistance R from r = a to r = b directly, then G = 1/R = I/V.

 

[peterpiper]

Thank you so much for that. I feel ridiculous for not just drawing a picture. It's funny how PHYS212 gets overshadowed by ENGE360.

 

[rude man]

Thank you so much for that. I feel ridiculous for not just drawing a picture. It's funny how PHYS212 gets overshadowed by ENGE360.

So what's your answer?

 

[peterpiper]

$\frac{2\pi\sigma}{ln(\frac{b}{a})}$

 

[klondike]

bingo!

$\frac{2\pi\sigma}{ln(\frac{b}{a})}$

 

[rude man]

$\frac{2\pi\sigma}{ln(\frac{b}{a})}$

Bingo again! Good work!