Straight Line Charge of Finite Length

**jegues** · September 21st 2011, 04:31 PM

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,

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

Where does the ld in the denominator come from?

From trig relationships it can be shown that,

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

which accounts for the d in the denominator, but where does the l come from? Is it that,

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

And if so, why? (That doesn't make any sense to me)

Also, where does he get $\hat{R} = cos\theta \hat{i} - sin\theta \hat{k}$ ? Where does this come from?

Also, he mentions that $\theta$ ranges from $\theta_{1}$ to $\theta_{2}$ , but $\theta_{1}$ is moving counterclockwise fashion, so shouldn't we conclude $\theta_{1} > 0$ and for $\theta_{2}$ moving in a clockwise fashion, $\theta_{2} < 0$ ? In his solution he has this reversed, so what am I mixing up?

Thread: Straight Line Charge of Finite Length

LinkBack

Thread Tools

Display

Straight Line Charge of Finite Length

Similar Math Help Forum Discussions

This is a straight line?

[SOLVED] Line Integral along a straight line.

straight line

straight line

Infinitely uniform line charge

Search Tags