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Infinite array of charged wires

  1. May 19, 2010 #1
    In the Feynman Lectures, Volume II, Chapter 7, the final two pages (7-10 and 7-11), Feynman describes an infinite array of parallel charged wires and uses a fourier series to solve for the field above them. He shows that the series can be expressed entirely in terms of cosine terms and that the coefficients have the value F(n)*e^(-z/k) where n is the order of the term, z is the distance from the plane of the wires, and k is a constant. What I don't understand is how one would then find the function F(n) (which he does not show). Later (in chapter 12 of the same volume) says that F(1) is twice the average field strength and that solving for the function F(n) is straightforward (in relation to another, related problem).
    How would you go about solving for the function?
    I've found the field via the coulomb interaction at several points where symmetry makes it easy but I'm not sure how this helps.
    Any help would be appreciated.
     
  2. jcsd
  3. May 20, 2010 #2
    The parallel charged wires can be represented as an array of delta functions spaced [itex]a[/itex] apart (times [itex]\lambda[/itex], the linear charge density of the wires). Then it would just be the Fourier Series of a delta function.

    [tex]
    F_n = \frac{2}{a} \int_{-a/2}^{a/2} \lambda\delta(x) \cos \frac{2n\pi x}{a} \, dx = 2\lambda
    [/tex]
     
  4. May 21, 2010 #3
    I just solved it last night. The delta function approach doesn't produce the correct answer; F(n)=lambda/(2pi*n*epsilon_0) (sorry I'm not sure how to embed latex into the forums). I found this by manipulating the infinite series for the field a distance a directly above one of the wires into a form which, when expanded, gave a function along the lines of e^(-n). (The field is related to the hyperbolic cotangent of pi*z/a).
     
  5. May 21, 2010 #4
    You're right. My delta function approach doesn't work because the Fourier coefficients of the potential are what's needed, not the charge density.
     
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