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A question about line images(electrostatics)

  1. Apr 29, 2012 #1

    ShayanJ

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    Gold Member

    In Nayfeh's electricity and magnetism,the equation of equipotential surfaces of two lines of charges are found and they are infinite cyliners with
    radius and center(center of the circle which is in the plane,we're solving problem in) given by the following equations:

    [itex] x_0=\frac{m^2+1}{m^2-1} a [/itex]
    [itex] R^2=x_0^2-a^2=\left(\frac{2ma}{m^2-1}\right)^2 [/itex]

    with m a constant and 2a the distance between lines of charge.

    Then the results above are used to find the potential of a configuration like an infinite line of charge near an infinite conducting cylinder.
    [itex] x_0 [/itex](the center of circle which is part of cylinder) and a(which is half the distance of charge line and its image) are found and using them,m
    is found which determines the potential function.
    My problem is this.When we see the configuration,as soon as we setup the coordinate system and choose the origine,[itex] x_0 [/itex] is determined.
    It shouldn't depend on other parameters!

    thanks
     
  2. jcsd
  3. Apr 30, 2012 #2
    I don't have that book on hand. Can you describe what each of these symbols means.
     
  4. May 2, 2012 #3

    ShayanJ

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    Gold Member

    At first the potential of two line charges with charge densities [itex] \lambda [/itex] and [itex] -\lambda [/itex] is found to be:

    [itex] \phi=\frac{\lambda}{2 \pi \varepsilon_0}\ln {\frac{\rho_-}{\rho_+}} [/itex]

    Where [itex] \rho_- [/itex] and [itex] \rho_+ [/itex] are the distace of the observation point to the line with minus and plus charge,respectively,in
    cylindrical coordinates.

    The equipotential surfaces are found by equating [itex]\frac{\rho_-}{\rho_+}[/itex] to a constant, m.
    They are cylinders with radius R.
    And [itex] x_0 [/itex] is the center of the circle which is the intersection of the cylinder and the xy plane.
     
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