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Finding the Electric potential

  1. Apr 12, 2010 #1
    1. The problem statement, all variables and given/known data

    The figure shows a thin plastic rod of length L = 11.8 cm and uniform positive charge Q = 58.9 fC lying on an x axis. With V = 0 at infinity, find the electric potential at point P1 on the axis, at distance d = 3.45 cm from one end of the rod.

    http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c24/qu_24_30.gif

    2. Relevant equations

    dq = [tex]\lambda[/tex]dx

    [tex]\lambda[/tex] = Q/L

    [tex]\int[/tex]dV = dq/r

    3. The attempt at a solution

    so after doing some substitutions i get

    [tex]\int[/tex]dV = k [tex]\int[/tex] [tex]\lambda[/tex]/ (d+L) dL

    Simplifying it I get:

    [tex]\int[/tex]dV = k[tex]\lambda[/tex][tex]\int[/tex]1/(d+L) dL

    After U-Sub I get:

    k[tex]\lambda[/tex]ln(d+L)

    Plugging all the values in I would get -0.008439 V. But it seems to be wrong. I'm sure it's somewhere around my integration that I messed up on. Any help would be great.

    Thanks!
     
  2. jcsd
  3. Apr 12, 2010 #2

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    Hello seraphimhouse,

    The limits of integration are important on this one. It seems you've evaluated the indefinite integral, but you need to evaluate the definite integral over the correct limits.

    I'm guessing you are new to [tex] \LaTeX [/tex], so you can use my equation below as a template if you're unfamiliar on how to write the limits on the integral sign (if you'd like).

    [tex] V = \int _{l=a} ^b \frac{k \lambda}{(d+l)}dl [/tex]

    Of course, you need to choose appropriate values for a and b. And then evaluate the expression, once you have solved the indefinite integral (through substitution, like you've already done).

    (Hint: the resulting expression, when evaluated with the bottom limit, is not zero! :wink:)
     
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