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Electric field at center of semi-circular bent rod of charge

  1. Jul 19, 2008 #1
    I have the correct answer just the wrong direction

    1. The problem statement, all variables and given/known data
    A uniformly charged insulating rod of length 14.0cm is bent into the shape of a semicircle as shown in figure XXX. The rod has a total charge of 7.5 micro C. Find the magnitude and direction of the electric field at P, the center of the semicircle.


    2. Relevant equations

    dE = k.dq.1/r^2


    3. The attempt at a solution
    I set things up like this
    [​IMG]

    Since the vertical components would cancel I just looked at the x components.

    [tex]
    dE_{x} = \frac{k_{e}\cdot dq\cdot\sin\theta}{r^2}
    [/tex]

    [tex]
    = \frac{k_{e}\cdot dx\cdot\lambda\cdot\sin\theta}{r^2} \ \ \mbox{as dq = dx\cdot\lambda}
    [/tex]

    [tex]
    = \frac{k_{e}\cdot r\cdot d\theta\cdot\lambda\cdot\sin\theta}{r^2} \ \ \mbox{as dx = ds = r\cdot d\theta}
    [/tex]

    [tex]
    = \frac{k_{e}\cdot d\theta\cdot\lambda\cdot\sin\theta}{r} \ \ \mbox{as dx = ds = r\cdot d\theta}
    [/tex]

    So
    [tex]
    E = \frac{k_{e}\cdot \lambda}{r} \int d\theta\cdot\sin\theta}
    [/tex]
    (integral is from zero to pi)

    [tex]
    E = \frac{k_{e}\cdot \lambda}{r} ( -1 - 1)
    [/tex]

    [tex]
    E = \frac{-2\cdot k_{e}\cdot Q}{L\cdot r} \ \ \mbox{as \lambda = \frac{Q}{L}}
    [/tex]
    When I substitute the values is I get the correct answer but it's negative. The bent rod is positive so the resultant electric field should be AWAY from the semicircle and therefor positve?!?

    What have I done wrong?
     
  2. jcsd
  3. Jul 19, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The integral of sin from 0 to pi is 2. How did you get (-1-1)? BTW the antiderivative of sin(x) is -cos(x).
     
  4. Feb 10, 2009 #3
    Sorry for reactivating this old thread, but I figured better that than starting a whole new thread.

    I have gotten to the point of [tex]

    E = \frac{2\cdot k_{e}\cdot Q}{L\cdot r} \

    [/tex]

    but now my question is what is the r value? Of what my reasoning is telling me, it should be [tex]

    E = \frac{2\cdot k_{e}\cdot Q}{L^{2}} \
    [/tex]

    Is this right? Can any one assist me?
     
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