How to Calculate Electric Potential and Field for a Charged Line Segment?

[dingo_d]

Homework Statement

Find the potential \phi and electric field \vec{E} for homogenous charged line segment with the length 2a, that lies between a and -a along the z-axis, if the total charge on the segment is q

Homework Equations

\phi(\vec{r})=\int\frac{\rho(\vec{r}')}{|\vec{r}-\vec{r}'|}d\tau '
\vec{E}=-\nabla\phi

The Attempt at a Solution

So since I have line segment on z-axes I set:

\vec{r}'=z'\hat{z},\ z\in[-a,a] so the distance between the point where I look the potential and the charged segment is:
|\vec{r}-\vec{r}'|=\sqrt{x^2+y^2+(z-z')^2}.

I'm dealing with line segment so my charge density is:

\rho(\vec{r}')d\tau'=\lambda dz', where \lambda=\frac{q}{2a}.

So after putting that all in integral I get:

\phi(\vec{r})=\frac{q}{2a}\int_{-a}^a\frac{dz'}{\sqrt{x^2+y^2+(z-z')^2}}

and the result (by checking Bronstein and Semendyayev, even Mathematica) is:

\phi(\vec{r})=\frac{q}{2a}\ln\left(\frac{z+a+\sqrt{x^2+y^2+(z+a)^2}}{z-a+\sqrt{x^2+y^2+(z-a)^2}}\right).

Now I got the solved problems hand written from a guy who finished this course years ago and in his notes it says that the solution is:

\phi(\vec{r})=\frac{q}{2a}\ln\left(\frac{z-a+\sqrt{x^2+y^2+(z-a)^2}}{z+a+\sqrt{x^2+y^2+(z+a)^2}}\right).

And he wrote that that's the same result as from the book where he got it (didn't mention the name -.-)...

So what am I doing wrong?

 

[Mindscrape]

Yours looks like the right one. I got the same integral as you at least. Both answers are missing a k (for permittivity) though. :p

 

[dingo_d]

Yours looks like the right one. I got the same integral as you at least. Both answers are missing a k (for permittivity) though. :p

Oh I'm working in cgs so I just say its 1 :D

Hmmm that's interesting...

 

[Mindscrape]

I just looked into this a bit further since you still had your doubts. My mathematica license ran out, so I google searched and stumbled upon this, which gives something completely different.
http://www.physics.princeton.edu/~mcdonald/examples/EM/rowley_ajp_74_1120_06.pdf

The integral is right and that is the most important thing, in my opinion.

 

[dingo_d]

Yeah, but why the different result when I'm just putting limits on my integration :\

 

[Mindscrape]

I plugged the integral into Matlab's symbolic integration, and it gave me what the was in the paper I cited.

-log(-z-a+(x^2+y^2+z^2+2*z*a+a^2)^(1/2))+log(-z+a+(x^2+y^2+z^2-2*z*a+a^2)^(1/2))

I'm not really sure what mathematica or the book gives because I don't have either right now. :)