Finding E Field at Point from non uniform charge density

[fmpak93]

1. Portion of z-axis for which |z| < 2 carries a non uniform charge density of 10|z| (nC/m). Using cylindrical coordinates, determine E in free space at P(0,0,4). Explicitly show your integration.

Homework Equations

E = (1/4πε₀) ∫ dQ*a_R/(R²)

The Attempt at a Solution

https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11004892_945246908827650_417954135_n.jpg?oh=d538ef93ebc53d426c29d8a2f0116327&oe=54ECD699&__gda__=1424795396_496617364f01c6d9bac875a291ef70af
I don't know what I'm doing wrong. the linear charge lies on the z axis and the point of interest lies on (0,0,4) so only the z unit vector applies right? As shown in the paper, the real answer is E = 34.20 a_z (V/m).
I

 

[TSny]

Welcome to PF!

Everything looks good except for your expression for $\vec{R}$. Try evaluating your expression for z = +2 and z = -2. Does it make sense?

 

[fmpak93]

Welcome to PF!

Everything looks good except for your expression for $\vec{R}$. Try evaluating your expression for z = +2 and z = -2. Does it make sense?

I would assume the R vector we can find as the difference in points between the source and destination. So our destination point is P(0,0,4), and our source charge point would have to be (0,0,|z|). Thus making our vector R = (4-|z|)a_z. Is this correct?

 

[TSny]

Thus making our vector R = (4-|z|)a_z. Is this correct?

No. Should R be the same for z = +2 and z = -2?

 

[fmpak93]

No. Should R be the same for z = +2 and z = -2?

Oh I see, it shouldn't. The absolute factor only applies to the charge density, the length of the position vector will still vary from top to bottom like always. So then just R = (4-z)?

 

[fmpak93]

No. Should R be the same for z = +2 and z = -2?

Never-mind, you were right. I got the answer. THANKS! :)