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Potential of an infinite rod using Green's function

  1. Mar 24, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Same problem as in https://www.physicsforums.com/showthread.php?t=589704 but instead of a spherical shape, consider an infinite line of constant charge density [itex]\lambda _0[/itex].


    2. Relevant equations
    Given in the link.


    3. The attempt at a solution
    I assume Phi will be the same along any parallel line to the charge distribution. So I can calculate [itex]\Phi (y)[/itex] in an x-y plane where x is the line direction and y=0 is where the line is.
    I get that [itex]|\vec x - \vec x'|=\sqrt {x^2+y^2}[/itex] so that [itex]\Phi (y)=\int _{-\infty }^{\infty } \frac{\lambda _0 dx}{\sqrt {y^2+x^2}}=\lambda _0 \ln ( \sqrt {y^2+x^2 } ) \big | _{-\infty}^{\infty}[/itex] but this diverge.
    By intuition I know that the equipotentials must be parallel to the charge distribution but here I get that the potential is infinite everywhere.
    I don't know what's going on.
     
  2. jcsd
  3. Mar 24, 2012 #2
    Again this time you'll need to subtract an infinite constant to make your potential finite and meaningful. The green's function you used assumes the sources are localized, i.e., zero boundary condition at infinity, whiles your source extend to infinity, therefore it won't work.
     
  4. Mar 24, 2012 #3

    fluidistic

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    Hmm so I cannot solve the problem via the given formula?
     
  5. Mar 28, 2012 #4

    fluidistic

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    My professor said today that the problem cannot be solved via the integral formula given because rho does not satisfy some properties. The problem was ill posed. :smile:
     
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