# Striaght Line Charge of Finite Length (E Field)

1. Sep 20, 2011

### jegues

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,

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

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?

#### Attached Files:

• ###### EX1.8.JPG
File size:
56.5 KB
Views:
123
Last edited: Sep 20, 2011
2. Sep 21, 2011

### jegues

I'm still looking for help one this one!