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Homework Help: The Electric Field of a Continuous Distribution of Charge

  1. Jun 8, 2010 #1
    part 2 The Electric Field of a Continuous Distribution of Charge

    1. The problem statement, all variables and given/known data
    find the electric potential at point p
    29619jm.jpg



    2. Relevant equations

    v=Kq/r....v=Er.

    3. The attempt at a solution
    2m3gt1d.jpg
    using this pic above and dQ=(Q/L)(dX)
    v=(K)(dQ)/(L-(X_i)+d) then sub dq=dX(Q/L)

    we get...
    (KQ/(L))(dX/(L+d-(X_i)) then integrating

    V=(KQ/L)((-ln(d))+(ln(L+d)))

    is this correct? if not could some one tell me where i went wrong.
    thanks in advance.
     
    Last edited: Jun 8, 2010
  2. jcsd
  3. Jun 8, 2010 #2

    kuruman

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    That looks about right except that I would combine the logarithm arguments and write it as

    [tex]V=\frac{kQ}{L}ln \left(1+\frac{L}{d}\right)[/tex]

    Then you can see more clearly what happens when d is much larger than L.
     
  4. Jun 8, 2010 #3
    i see what you are doing but as d>>L it goes to zero..

    but does not tell anything about whats below.

    a) an infinitely long wire with total charge Q
    b) an infinitely long wire with total charge Qd/L
    c) a point charge of magnitude Q
    d) an electric dipole with moment QL
     
  5. Jun 8, 2010 #4

    gabbagabbahey

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    Not quite, as [itex]d\to\infty[/itex] [itex]V[/itex] will go to zero. For [itex]d\gg L[/itex] you'll want to use the Taylor series expansion of [itex]\ln(1+x)[/itex] with [itex]x=\frac{L}{d}[/itex] to get an idea of how the potential behaves.
     
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