How to Calculate the Resultant Electric Field from a Line Charge?

[diredragon]

1. Homework Statement
Calculate the resultant electric field acting on some point $x$. The electric field is generated by a long thin rod and the line charge density is given. $p_l=\frac{dQ}{dL}$

Homework Equations

3. The Attempt at a Solution [/B]
I have uploaded two images, one of the problem and one of my solution. My problem is the following. When i calculate the $dL$ part my solutions differs from the one given in the book. Can someone take a look at the?
just a little edit: in the picture you see $pldl$it is actually $p_ldl$

 

[Simon Bridge]

You seem to be defining dl to be the length of the line segment between $z$ and $z+dz$ (top diagram) ... but surely that is just $dz$? Why change the name?
Looking at your dl then ... actually I think that may work ... you'd get $dl = x\;d\alpha / \cos^2\alpha$ or something...
Some of the other expressions in the first pic do not work... like $r\; d\alpha \cos\alpha = dl$ is not true, and I don't see how you did $r=\sqrt{\alpha_0^2+z^2}$

Why are you trying to do the integration wrt $\alpha$ anyway?

Go back a bit: at position x, the radial component of the field due to charge $dq$ between $z$ and $z+dz$ is:
$dE_x = (dq/r^2)\cos\alpha$ (using your diagrams).

Notice that $\cos\alpha = x/r$, $r^2=x^2+z^2$, and $dq = \lambda\; dz$
... where $\lambda$ is the linear charge density. now the calculation is easy to set up.

 

[diredragon]

You seem to be defining dl to be the length of the line segment between $z$ and $z+dz$ (top diagram) ... but surely that is just $dz$? Why change the name?
Looking at your dl then ... actually I think that may work ... you'd get $dl = x\;d\alpha / \cos^2\alpha$ or something...
Some of the other expressions in the first pic do not work... like $r\; d\alpha \cos\alpha = dl$ is not true, and I don't see how you did $r=\sqrt{\alpha_0^2+z^2}$

Why are you trying to do the integration wrt $\alpha$ anyway?

Go back a bit: at position x, the radial component of the field due to charge $dq$ between $z$ and $z+dz$ is:
$dE_x = (dq/r^2)\cos\alpha$ (using your diagrams).

Notice that $\cos\alpha = x/r$, $r^2=x^2+z^2$, and $dq = \lambda\; dz$
... where $\lambda$ is the linear charge density. now the calculation is easy to set up.

Oh yeah i see..i was looking at the small segment wrong and that's $x_0$ in the square root, I just wrote it funny, sorry bout that and thanks! :)