1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Transmission Line Question

  1. May 4, 2009 #1
    1. The problem statement, all variables and given/known data
    A transmission line consists of a cylindrical conductor of radius r at a distance d in air from a conducting plane (r >>d).

    Derive the capacitance per unit length C and the inductance per unit length L and
    check that 1/sqrt(LC) = c.

    2. Relevant equations

    3. The attempt at a solution
    I thought I could just give the cyclinder a charge -Q and the plane a charge +Q and superimpose the fields. So that along the line through the centre of the cylinder and normal to the plane, the field is:

    E = -Q/(2.pi.r.l.e0) -Q/(2.l.m.e0)

    Where l is the length of the cylinder and l, m are the dimensions of the plane.

    Then integrating to find the potential difference and dividing by Q and multiplying by l gives:

    1/C = (1/2.pi.r.e0)*ln[(d-r)/r] + (1/2e0)*(d-r)/m

    Then the B due to the cylinder would be vI/2.pi.r where I am using v as permeability of free space. And on the normal through the centre line, it would be perpendicular to the line.

    I think that on the line, the field from all the elements on the plane would superimpose to produce a field perpendicular to the line, which Ampere would then give as vI/m


    B = vI/2.pi.r + vI/m

    Then flux is the integral of that, and dividing by I and l gives L:

    L = (v/2.pi)*ln[(d-r)/r] + v(d-r)/m

    But this doesn't seem right.

    Any help? Thanks.
  2. jcsd
  3. May 4, 2009 #2


    User Avatar
    Science Advisor
    Gold Member

    You can't assume that the charge on the plane would be uniformly distributed. However, since it is an infinite plane, you can use image theory to model the effective charge distribution.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook