Coulomb's Law - Line of Charge

[WesleyPipes]

Well, I am raging over my homework and not even going to turn it in because I can't complete enough of it. I really just need to figure this problem out so I don't feel as bad about it. I know this shouldn't be as hard as I am making it, but all examples only have answers containing a single-direction vector.

Homework Statement

A line of charge with uniform density Pl=8(uC/m) exists in air along the z-axis between z=0 and z=5 cm. Find E at (0,10cm,0).

Homework Equations

R = y(.1)-z(z) --- really Vector(R) = Yhat(.1)-Zhat(z)
Mag(R) = sqrt(.01+z^2)
Rhat' = Vector(R)/Mag(R)

Line Distribution:
E=(1/(4*Pi*Epsilon))*Integral(z=0 to z=0.05)[Rhat'*(Pl)/Mag(R)^2]dL

The Attempt at a Solution

Sorry if the equations are ridiculously hard to read.

Each time I try this, I get an answer in the +y direction when I feel like the answer vector should be in the (-z,+y) direction.

Thanks guys.

 

[SammyS]

In your final expression for E, dL is dz. Here's the Latex version of that expression incorporating all of your other results.

\vec{E}=\frac{1}{4\pi\epsilon_0}\,\int_{0}^{.05}\frac{\Pi\hat{R}}{R^2}\,dz

=\frac{1}{4\pi\epsilon_0}\,\int_{0}^{.05}\frac{\Pi(0.1\hat{y}-z\hat{z})}{R^3}\,dz=\frac{1}{4\pi\epsilon_0}\,\int_{0}^{.05}\frac{\Pi(0.1\hat{y}-z\hat{z})}{(0.01+z^2)^{3/2}}\,dz​

This splits into the difference of two integrals.

\vec{E}=\frac{\Pi}{4\pi\epsilon_0}\left(0.1\hat{y}\,\int_{0}^{.05}\frac{1}{(0.01+z^2)^{3/2}}\,dz\ -\hat{z}\,\int_{0}^{.05}\frac{z}{(0.01+z^2)^{3/2}}\,dz\right)

which gives the two components you want.

 

[WesleyPipes]

Thanks, knew I was missing something simple. I feel a lot better now haha.