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Striaght Line Charge of Finite Length (E Field)

  1. Sep 20, 2011 #1
    Straight Line Charge of Finite Length (E Field)

    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,

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

    Where does the ld in the denominator come from? We can conclude from the trig relationships that,

    [itex]\frac{1}{R^{2}} = \frac{d\theta}{dz} \frac{1}{d}[/itex]

    which can account for the d in the denonminator, is it supposed to be that,

    [itex] l = \frac{dz}{d\theta}[/itex] ?

    Also, how did he devise, [itex]\hat{R} = cos\theta \hat{i} - sin\theta \hat{k}[/itex] in the first place? Where does he get this from?

    He also mentions that [itex]\theta[/itex] ranges from [itex]\theta_{1}[/itex] to [itex]\theta_{2}[/itex], but [itex]\theta_{1}[/itex] is moving in the counterclockwise fashion, thus shouldn't we conclude [itex]\theta_{1} > 0[/itex] and for [itex]\theta_{2}[/itex] moving in a clockwise fashion, [itex]\theta_{2} < 0[/itex]? He's got these reversed(like in the figure attached), so what am I mixing up?
     

    Attached Files:

    Last edited: Sep 20, 2011
  2. jcsd
  3. Sep 21, 2011 #2
    I'm still looking for help one this one!
     
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