Striaght Line Charge of Finite Length (E Field)

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Straight Line Charge of Finite Length (E Field)

Homework Statement



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.

Homework Equations





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} \left( \int_{\theta_{1}} ^{\theta_{2}} cos\theta d\theta \hat{i} - \int_{\theta_{1}} ^{\theta_{2}} sin\theta d\theta \hat{k} \right)

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

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

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

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

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

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

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