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Math Help - Straight Line Charge of Finite Length

  1. #1
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    Question Straight Line Charge of Finite Length

    1. The problem statement, all variables and given/known data

    Find the expression for the E field at an arbitrary point in space due to a straight line of length l uniformly charged with total charge Q. The ambient medium is air.

    2. Relevant equations



    3. The attempt at a solution

    I am following this example with the solution given in my textbook and I am confused about one part.

    He somehow makes the following transition,

    \vec{E} = \frac{1}{4 \pi \epsilon_{0}} \int_{l} \frac{Q^{'}dl}{R^{2}}\hat{R} = \frac{Q}{4\pi \epsilon_{0} ld} \int_{\theta_{1}} ^{\theta_{2}} \left( cos\theta \hat{i} - sin\theta \hat{k} \right)d\theta

    Where does the ld in the denominator come from?

    From trig relationships it can be shown that,

    \frac{1}{R^{2}} = \frac{d\theta}{dz} \frac{1}{d}

    which accounts for the d in the denominator, but where does the l come from? Is it that,

    l = \frac{dz}{d\theta}?

    And if so, why? (That doesn't make any sense to me)

    Also, where does he get \hat{R} = cos\theta \hat{i} - sin\theta \hat{k}? Where does this come from?

    Also, he mentions that \theta ranges from \theta_{1} to \theta_{2}, but \theta_{1} is moving counterclockwise fashion, so shouldn't we conclude \theta_{1} > 0 and for \theta_{2} moving in a clockwise fashion, \theta_{2} < 0? In his solution he has this reversed, so what am I mixing up?
    Attached Thumbnails Attached Thumbnails Straight Line Charge of Finite Length-ex1.8.jpg  
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