Electric Field due to Continuous Line Charge

[darkpsi]

Homework Statement

Find the electric field a distance z above the midpoint of a straight line segment of length
2L, which carries a uniform line charge λ.

Homework Equations

1) my textbook says :
E(r) = 1/4πε_o ∫_V λ(r')/r² r dl'


2) and this also works? :
E(r) = 1/4πε_o ∫_V λ(r')/r³ r dl'

both where r is the unit vector from the charge to the point we are caluclating the field at

The Attempt at a Solution

I can integrate once I get the setup down just fine. I just wanted to know when and why this two are the same. For example, in my book it says for the setup for this problem :
E(r) = 2 * 1/4πε_o ∫_V(λdx/r²)cosθz

For one thing, how did that cos appear and since it equals z/r couldn't 2) have just been used from the start? And if 2) can be used could I apply it to surface and volume integrals?

 

[ApexOfDE]

General formula for E is:

dE = 1/4πεo (dq/r^2)$$\hat{r}$$

However, r = r.$$\hat{r}$$ (r_hat is unit vector)
So you can write"

dE = 1/4πεo (dq/r^3).r

--
Electric field of a point above distance Z from a line is a cumulative Electric field caused by every charge in straight line.

Lets say an angle formed by E caused by one charge and a vertical axis is theta. From there, you can get:
E_x = E * sin(theta)
E_y = E * cos(theta)

And because E in x-axis will automatically blow up, you just need to find E in y-axis.

Hope this helps

 

[darkpsi]

Oh now I see. I wasn't even thinking that because r_hat wasn't pointing vertically that the z component wasn't all of the electric field. I was thinking the x-fields canceled so the field in the z direction was exactly equal to the electric field, but its actually only cos(theta) of it. Thanks a bunch