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Electric field in a semi-circle (non-uniformly charged)

  1. Mar 25, 2012 #1
    1. The problem statement, all variables and given/known data
    A plastic semi-circle/arc with radius R has a non-uniformly distributed charge upon it.
    The density of the charge/length is [itex]\lambda = \lambda_0 sin\theta[/itex]
    θ is 0 in the middle of the arc

    find the electric field in the point where the radius come from

    2. Relevant equations

    [itex]dE=\frac{kdQ}{R^2}[/itex]
    [itex]\phi = \int{Edr}[/itex]

    3. The attempt at a solution

    charge differential to angle differential is :
    [itex]dQ=\lambda R = \lambda_0 sin\theta d\theta[/itex]

    for the field I separated for x and y
    [itex]E_x =\frac{k\lambda_0}{R} \int_{0}^{\pi}sin^2\theta d\theta=\frac{k\lambda_0}{R}=\frac{k\lambda_0\pi}{2R} [/itex]

    the same way I get
    [itex]E_y= \frac{k\lambda_0}{R}\int_{0}^{\pi}sin\theta cos\theta d\theta=0 [/itex]

    this is counter-intuitive for me, I thought the electric field should be pointing in the y direction.
    I guess I need to have my answers inverted, but would be glad to understand why, and maybe shed some light on my mistakes...


    edit: I think I've found my mistake... I should integrate once from 0 to [itex]\frac{\pi}{2}[/itex] and once from 0 to [itex]-\frac{\pi}{2}[/itex]
    and add what I get
     
    Last edited: Mar 26, 2012
  2. jcsd
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