Electric field in a resistive, shorted coaxial cable

Click For Summary
SUMMARY

The discussion focuses on solving for the electric and magnetic fields in a shorted coaxial cable with a resistive inner conductor and a perfect outer conductor. The potential at the input surface is given as a function of ln(r), leading to the application of Poisson's equation, which simplifies to a Laplace equation due to the absence of charge in the dielectric region. The solution involves using cylindrical coordinates and separation of variables, with boundary conditions defined at the inner and outer conductors. The method of images is suggested as a practical approach to simplify the problem.

PREREQUISITES
  • Understanding of Poisson's equation and Laplace's equation
  • Familiarity with cylindrical coordinates in electrostatics
  • Knowledge of boundary conditions in electromagnetic problems
  • Experience with the method of images in electrostatics
NEXT STEPS
  • Study the method of images for solving electrostatic problems
  • Learn about boundary value problems in cylindrical coordinates
  • Explore the derivation of electric fields from potential functions
  • Investigate the implications of fringing fields in coaxial cables
USEFUL FOR

Electrical engineering students, researchers in electromagnetism, and professionals working with coaxial cable design and analysis.

EEGrad
Messages
1
Reaction score
0

Homework Statement


Consider a shorted coaxial line with a resistive inner conductor with radius a, and a perfect outer conductor located at radius b. A DC voltage is applied at the input end. Find the electric and magnetic fields in the dielectric region (a<r<b), assuming the battery sets up a potential that varies as ln(r) on the input surface.
[/B]
wXpuytH.png

Homework Equations


Not exactly sure if there is a better way, but I started by looking at Poisson's equation
\nabla^2 V = \frac{\rho}{\epsilon}

The Attempt at a Solution


Since there is no charge in the dielectric, Poisson's equation becomes

\nabla^2 V(r,z) = 0

where V has rotational symmetry. In cylindrical coordinates,

\frac{1}{r} \frac{\partial }{\partial r} \left( r \frac{\partial V(r,z)}{\partial r} \right) + \frac{\partial^2 V(r,z)}{\partial z^2} =0

The boundary conditions look like:

V(r=a) = V_0 - \frac{V_0 z}{L}\\<br /> V(r=b) = 0\\<br /> V(z=L) = 0\\<br />I'm not sure how to solve this differential equation. Can you use separation of variables to get:

V(r,z) = R(r)Z(z) \\<br /> \frac{1}{R}\frac{1}{r} \frac{\partial}{\partial r}\left( r \frac{\partial R}{\partial r} \right) = - \frac{1}{Z} \frac{\partial^2 Z}{\partial z^2}\\<br />

Even if I do that though, I'm not quite sure how to solve the R portion. Any nudge in the right direction would be appreciated. Thanks!
 
Physics news on Phys.org
Hi EEGrad. I notice it’s been quite a while since you posted your question - have you found a solution, yet? If not, here are some suggestions. You’ll have to think about it and decide for yourself whether this is a nudge in the right direction or not.

Firstly, you’re correct in that a full solution of Poisson’s equation would enable you to find the E field everywhere, including the fringing fields at the input face, but in real-world problems solving Poisson’s equation is often fiendishly difficult. So I like to look for an easier solution. When I see an electrostatics problem involving lots of highly symmetrical perfect conductors I think to myself “That looks like a problem which might be amenable to solution by the method of images.”

Reading the question a little more carefully I see: “assuming the battery sets up a potential that varies as ln(r) on the input surface.” That is the potential you would find near an infinite charged cylinder: C1 - C2 * ln(r), where C1 and C2 are constants. In other words the question is telling you to assume that the potential (and therefore the radial component of E) at the input face looks like you would find for an infinite cylindrical conductor. An invitation to ignore fringing fields at the input, perhaps?

What I would do is start with an infinite ohmic cylinder and calculate the field due to that ignoring everything else. Then wrap the outer conductor around it and find the field due to the outer conductor by the method of images.

Hint: it might help if you define your z-axis so that V(z) passes through zero at the origin - that may be handy when you come to “close off” the shorted end of the cable by the method of images.

Can you see how to calculate B?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
12
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
3K
Replies
10
Views
2K