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Potential of parallel infinite charged wires

  1. Nov 19, 2008 #1
    1. Two infinitely long wires running parallel to the x-axis carry uniform charge densities [tex] +\lambda [/tex] and [tex] -\lambda [/tex]. Find the potential at any point. Show that the equipotential surfaces are circular cylinders, and locate the axis and radius of the cylinder corresponding to a given potential [tex] V_0 [/tex]

    2. Relevant equations.

    Potential of one wire charge density [tex] \lambda [/tex] I calculated to be :

    [tex] V_a(b) = \frac{-\lambda}{2\pi\epsilon_0} \ln (b/a) [/tex] where a,b are shortest distances to the wire, and distance a is the reference.

    I let the wires be equidistant from the x-axis, denoting the distance d. I put the [tex] -\lambda [/tex] wire at (-d,0) and the [tex] +\lambda [/tex] at (d,0).

    3. The attempt at a solution

    Setting the reference at the origin, the reference distance for both wires is [tex] a = d [/tex]. Letting [tex] r_+, r_- [/tex] be the distance from a point (x,y) to the positive and negative wires, respectively, the potential at (x,y) is then

    [tex] V(x,y) = \frac{\lambda}{2\pi\epsilon_0}( -\ln(r_+/d) + ln(r_-/d) ) [/tex]
    [tex] = \frac{\lambda}{2\pi\epsilon_0}\ln(\frac{r_-}{r_+}) [/tex].

    The distances are

    [tex] r_+ = \sqrt{(x-d)^2 + y^2} [/tex]
    [tex] r_- = \sqrt{(x+d)^2 + y^2} [/tex].

    So the potential can be written

    [tex] V(x,y) = \frac{\lambda}{4\pi\epsilon_0}\ln(\frac{(x+d)^2 + y^2}{(x-d)^2 + y^2}) [/tex].

    If so, then the eqn. of a [tex] V_0 [/tex] equipotential surface is

    [tex] k = \frac{(x+d)^2 + y^2}{(x-d)^2 + y^2}) [/tex] where [tex] k = \exp( \frac{4\pi\epsilon_0 V_0}{\lambda} ) [/tex].

    This simplifies to

    [tex] x^2 + y^2 + d^2 - 2xd\frac{k+1}{k-1} = 0 [/tex].

    and I don't see how this eqn. could define a cylinder, and moreover, on intuitive grounds, I don't see how a cylinder could possibly be the geometry of the equipotential surfaces. By symmetry, an equipotential cylinder would have to be centered at the origin -- right? -- but that would means that the potential at [tex] (d + \varepsilon, 0 ) [/tex] a bit to the right of the positive wire would be the same as the potential at [tex] (-d - \varepsilon, 0 ) [/tex] a bit to the left of the negative wire!
  2. jcsd
  3. Nov 19, 2008 #2


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    Strictly speaking, [tex]x^2 + y^2 + d^2 - 2xd\frac{k+1}{k-1} = 0 [/tex] defines a circle, not a cylinder. However, the potential is independent of z, so these circles are the same for a given [itex]V_0[/itex] all the way along the z-axis, and together this infinite collection of circles (of the same radius and center) define a cylindrical surface.

    To see that your equation represents a circle, just "complete the square" in 'x', an you can rewrite your equation into the form [tex](x-a)^2+y^2=R^2[/tex] which of course, represents a circle centered at [itex](a,0)[/itex] with radius [itex]R[/itex]... What do you get for your [itex]a[/itex] and [itex]R[/itex] values?....Clearly, your cylinder is not centered at the origin, but rather runs parallel to the z-axis and is centered at [itex](a,0)[/itex].
  4. Nov 19, 2008 #3
    Now that you mention it, that was the plain-as-day thing to do. Thanks!
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