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Electric potential again.

  1. Mar 2, 2008 #1
    electric potential....again.

    So, I've still got my two infinite lines of charge that run parallel to the x axis running along the lines x = a and x = -a and with charge density + and - lambda respectively.

    I've figured out that using the origin as my reference point allows me to integrate the electric field of my lines from a to the point in question (r). However, because the field goes like 1/r, we end up with a ln of something in the potential. For the line that's positive a away from the x axis, there's no problem. But, for the line that runs along x = -a, how can I integrate? should I be integrating from a to r again, or from -a to r. The problem with the latter is that you can't have the natural logarithm of a negative number.

    I would REALLY appreciate any advice. I'm a little desperate....

    W. =)
  2. jcsd
  3. Mar 2, 2008 #2
    Remember, the r in 1/r is the distance to the charge so it should always be positive. If your point is (x, y) then the distance to the line x = -a is |x + a| and the distance to the line x=a is |x-a|
  4. Mar 2, 2008 #3
    that makes sense....I figured it was a that I needed to use, but I just wanted to be sure.

    Thanks for the help. =)
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