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Unformatted text preview: Ph12a Solution Set 6 November 10, 2009 4.7 Use Gauss’s law to get the electric field for r 1 < r < r 2 . Make a “Gaussian pillbox” of cylindrical shape, parallel to the axis of the transmission line: integraldisplay vector E · dvectora =  vector E  2 πra = Q ǫ o , where a is the length of the pillbox and Q is the charge enclosed. So, vector E = Q 2 πǫ o ra ˆ r. Then the potential is: V = integraldisplay r 1 r 2 vector E · d vector l = Q 2 πǫ o a ln( r 2 /r 1 ) . Then the capacitance per unit length is C a = Q V a = 2 πǫ o ln( r 2 /r 1 ) . Note that I have done this calculation in SI units. The conversion from SI to esu is 1 4 πǫ o ( SI ) = 1( esu ). This factor converts the above result to the answer in the book. The Bfield for r 1 < r < r 2 (in SI units) is vector B = μ o I 2 πr ˆ φ . This may be derived using Ampere’s law. Then the flux is (integrate over a rectangular region from r 1 to r 2 , and length a): Φ = integraldisplay vector B · dvectora = μ o I 2 π integraldisplay a integraldisplay r 2 r 1 1 r drdz = μ o I 2 π a ln( r 2 /r 1 ) Then the inductance per length is: L a = Φ Ia = μ o 2 π ln( r 2 /r 1 ) Again this result is in SI units. The conversion factor in this case is: μ o 4 π ( SI ) = 1 c 2 ( esu ). This yields the result in the book. These conversion factors may be found in the appendix of the third edition of Jackson’s book on classical electrodynamics....
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 Spring '09
 DUDKO
 mechanics, International System of Units, Conversion of units, ESU, Ph12a Solutions

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