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Equipotential cylinder

  1. Mar 26, 2008 #1
    An infinite string with linear charge density [tex]\lambda[/tex] is put parallel to the axis of an infinite conducting grounded cylinder (V=0) of radius [tex]R[/tex], the distance between the string and the center of the cylinder is [tex]l[/tex]. Find the potential outside the cylinder.

    Is it possible to solve this problem using the method of images, i.e. placing an image string with some charge density [tex]\lambda_{1}[/tex] inside the cylinder so that the cylinder will be an equipotential surface? If V weren't 0, I could place a string of equal charge (i.e. [tex]-\lambda[/tex]) somwehere inside and it should work, but for V=0, the charges can't be equal.
     
  2. jcsd
  3. Mar 26, 2008 #2

    clem

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    Science Advisor

    Yes, it is the 2D example of a charge outside a sphere.
     
  4. Mar 26, 2008 #3
    Ok, but how do I calculate this? If I place an image string with charge [tex]\lambda_{1}[/tex] inside the cylinder, at distance [tex]a[/tex] from the center,I get the following equation for equipotential surfaces:
    [tex]\lambda ln(r)-\lambda_{1} ln(r_{1})=C[/tex], where [tex]r, r_{1}[/tex] are distances form resp. 1st and 2nd string. Now if I put C=0 (since on the surface of the cylinder V=0, and I want the cylinder to be an equipotential surface), I get:
    [tex]\lambda ln(r)=\lambda_{1} ln(r_{1})[/tex]. Now I need to calculate [tex]\lambda_{1}[/tex] and [tex]a[/tex] from this. Any help?
     
  5. Mar 27, 2008 #4

    pam

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    It's done like the 3D problem.
    Write [tex](R\hat{r}-{\vec L})e^\lambda=(R\hat{r}-{\vec L}')e^{-\lambda'}[/tex].
    Then let L'=R^2/L.
     
    Last edited: Mar 27, 2008
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