Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook