(adsbygoogle = window.adsbygoogle || []).push({}); Straight Line Charge of Finite Length (E Field)

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

Find the expression for theEfield 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?

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# Striaght Line Charge of Finite Length (E Field)

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