Yes, thank you!
The I actually got the ln(r) from the unit vector, since
\hat{r} = \frac{\vec{r}}{r} and when integrating 1/r you get ln(r).
You are right though, my calculations are a mess, too many mistakes. Sorry about that. :/
At least I now know that this configuration does not...
I am not sure I understand what you mean.
Though I think I have made a mistake.
I wanted to integrate using the surface area for cylindrical coordinates, i.e.:
\int \vec{E} \; d\vec{A} = \int \vec{E} \; \hat{r} \; dr \; d\phi \; dx = \vec{E} \; \vec{r} \; ln(r) \; 2 \pi \cdot 2 L...
Hi,
on page 63 of David J. Griffiths' "Introduction to Electrodynamics" he calculates the electric field at a point z above a line charge (with a finite length L) using the electric field in integral form.
E_z = \frac{1}{4 \pi \epsilon_0} \int_{0}^{L} \frac{2 \lambda z}{\sqrt{(z^2 + x^2)^3}}...