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Convergent field, divergent potential?

  1. Nov 27, 2004 #1
    If you want to calculate the electric field at a distance r from a line of infinite length and uniform charge density you could one of three things:

    1. Employ symmetry and Gauss' law.
    2. Use superposition and integrate from minus to plus infinity along the rod.
    3. Integrate to find the potential and differentiate.

    1. and 2. work fine and unsurprisingly give the same result. But when I try 3., I get an integral of the form:
    [tex]\int^{\infty}_{-\infty} \frac{b ds}{\sqrt{a^2 + s^2}}[/tex]
    Equal to an inverse sinh, which diverges, surely impossible to differentiate. Why is this?
     
  2. jcsd
  3. Nov 27, 2004 #2

    Dr Transport

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    When working with infinite extent object like a charged rod, it is cutomary to not integrate over then entire length because the potential will become infinite. What you want to do is because of symmetry, leave out the 3rd integration and obtain a potential per unit length.

    Set up the equations for potential, and only do it in 2-d.
     
  4. Nov 27, 2004 #3
    What do you mean by a 'potential per unit length'? Length of the line/rod? This changes according to where this unit length is. Thanks - please clarify.
     
  5. Nov 27, 2004 #4

    Dr Transport

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    if the rod is infinte in extent, the potential is in a plane only and depends on the distance from the rod. Out of the plane contributions cancel.
     
  6. Nov 27, 2004 #5

    krab

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    The electric field is finite, but the potential, if referenced to infinity, is not finite. Since only potential differences are important anyway, just declare the potential at some distance R to be zero.
     
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