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Positively charged plastic rod of length L

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data
    25-36.jpg

    The figure shows a positively charged plastic rod of length L = 2.40 m and uniform linear charge density 8.00×10-3 C/m. Setting V= 0 at infinity, find the electric potential at point P for d = 1.536 m.


    2. Relevant equations
    asfd.jpg


    3. The attempt at a solution
    Alright, I plugged in the values to this formula. It should be setup as:
    (2)*(8.99e9)*(8e-3)*ln[((2.4/4) + sqrt((2.4)^2/4 + (1.536)^2))/1.536]

    All of this churns out to be 72811647.46708032. However, this is not correct. Any ideas?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 25, 2010 #2

    Delphi51

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    Homework Helper

    I've been away from this for a long time, but I think there may be an error in your formula. It should have the form of the difference between an upper and a lower value of an integral. Check out this thread
    https://www.physicsforums.com/showthread.php?t=296367
    and note that it uses x in place of your d.
     
  4. Feb 26, 2010 #3

    ehild

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    Homework Helper
    Gold Member



    You have a mistake when plugging in the data, this is the correct formula for the figure on the left:

    (2)*(8.99e9)*(8e-3)*ln[((2.4/2) + sqrt((2.4)^2/4+ (1.536)^2))/1.536]

    ehild
     
  5. Feb 26, 2010 #4

    ehild

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    Gold Member

    Delphi, the formula is OK, the integral is taken between x=0 and x=L/2, and doubled.

    ehild
     
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