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Electric field in coaxial cable

  1. Jan 12, 2009 #1
    Consider a coaxial cable with inner radius a and outer radius b. The potential difference between the inner and outer conducter is V. How can it be shown that the electric field at a distance r from the inner conducter is given by:

    E = \frac{V}{r ln\frac{b}{a}}

    Any help would be much appreciated!
     
  2. jcsd
  3. Jan 12, 2009 #2
    Let the inner conductor of radius a have a charge +Q and the outer conductor at radius b have a charge of -Q. Use a cylindrical Gaussian surface to calculate the electric field, E, at any point inside the cable. This will give E = Q/(εA) where A = 2*pi*r*l and l is a unit axial length. Then find V by finding the integral of E dot dr. This will give V = Q/(ε*2*pi*l)ln(b/a). But Er = Q/(ε*2*pi*l) so V = Er*ln(b/a). Solving for E gives shows the desired expression for E.
     
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