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

AI Thread Summary
The discussion focuses on deriving the conductance per unit length, G', of a coaxial cable using the electric field and line charge on the inner conductor. The participant initially struggles with integrating the electric field and expresses confusion over obtaining the correct logarithmic term in the denominator. Clarifications are provided regarding the integration process, emphasizing that the electric field is uniform and perpendicular on the surface of the cylindrical conductor. Ultimately, the correct formula for conductance is confirmed as G' = 2πσ / ln(b/a). The conversation highlights the importance of visualizing the problem and correctly applying integration techniques in electromagnetics.
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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.
 
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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.
peterpiper said:

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.
 
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?
 
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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.

peterpiper said:
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})
 
I'm not sure I understand what you're getting at.
 
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.


peterpiper said:
I'm not sure I understand what you're getting at.
 
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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 abE*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.
 
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Thank you so much for that. I feel ridiculous for not just drawing a picture. It's funny how PHYS212 gets overshadowed by ENGE360.
 
peterpiper said:
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?
 
  • #10
\frac{2\pi\sigma}{ln(\frac{b}{a})}
 
  • #11
bingo!
peterpiper said:
\frac{2\pi\sigma}{ln(\frac{b}{a})}
 
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  • #12
peterpiper said:
\frac{2\pi\sigma}{ln(\frac{b}{a})}

Bingo again! Good work!
 
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