# Infinitely uniform line charge

• Apr 7th 2009, 06:27 AM
august
Infinitely uniform line charge
Hi all,

In fact, this is a classical integral, whose answer is always given without showing the steps, in electromagnetism.

In Cylindrical coordinates

$
\int_{+\infty}^{-\infty} \! \frac{\rho r \hspace{1mm} dz}{4 \pi \epsilon_o (r^2 + z^2)^{\frac{3}{2}}} \hspace{1mm} \bf{r}
$

$
\frac{\rho}{2\pi\epsilon_{o}r} \hspace {1mm} \bf{r}
$

I don't know which integration technique I should use to solve this?

• Apr 7th 2009, 07:11 AM
chisigma
Because the variable of integration is z the integral becomes...

$\frac {\rho\cdot r}{4\cdot \pi\cdot \epsilon_{0}} \cdot \int_{-\infty}^{+\infty} \frac{dz}{(r^{2} + z^{2})^{\frac{3}{2}}}$ (1)

If you take in mind that...

$\int \frac{dz}{(r^{2} + z^{2})^{\frac{3}{2}}} = \frac{z}{r^{2}\cdot \sqrt{r^{2}+z^{2}}} + c$

... you find...

$\int_{-\infty}^{+\infty} \frac{dz}{(r^{2} + z^{2})^{\frac{3}{2}}} = \frac{2}{r^{2}}$ (2)

... so that the (1) becomes...

$\frac{\rho}{2\cdot \pi\cdot\epsilon_{0}\cdot r}$

Kind regards

$\chi$ $\sigma$
• Apr 7th 2009, 08:30 AM
august
Quote:

Originally Posted by chisigma
Because the variable of integration is z the integral becomes...

$\frac {\rho\cdot r}{4\cdot \pi\cdot \epsilon_{0}} \cdot \int_{-\infty}^{+\infty} \frac{dz}{(r^{2} + z^{2})^{\frac{3}{2}}}$ (1)

If you take in mind that...

$\int \frac{dz}{(r^{2} + z^{2})^{\frac{3}{2}}} = \frac{z}{r^{2}\cdot \sqrt{r^{2}+z^{2}}} + c$

... you find...

$\int_{-\infty}^{+\infty} \frac{dz}{(r^{2} + z^{2})^{\frac{3}{2}}} = \frac{2}{r^{2}}$ (2)

... so that the (1) becomes...

$\frac{\rho}{2\cdot \pi\cdot\epsilon_{0}\cdot r}$

Kind regards

$\chi$ $\sigma$

thanks although I gained nothing!!

My problem essentially is how to get from $\int \frac{dz}{(r^{2} + z^{2})^{\frac{3}{2}}}$ to $\frac{z}{r^{2}\cdot \sqrt{r^{2}+z^{2}}} + c$

Do I have to use partial fractions?
• Apr 7th 2009, 09:05 AM
Quote:

Originally Posted by august
thanks although I gained nothing!!

My problem essentially is how to get from $\int \frac{dz}{(r^{2} + z^{2})^{\frac{3}{2}}}$ to $\frac{z}{r^{2}\cdot \sqrt{r^{2}+z^{2}}} + c$

Do I have to use partial fractions?

z= r tan(x)

dz = r sec^2(x) dx

Denominator

= r^3 sec^3(x)

Integration becomes

$\int{\frac{r\times sec^2(x) dx}{r^3 sec^3(x)}}$

$\int{\frac{ dx}{r^2 sec(x)}}$

$\frac{1}{r^2}\int{cos(x) dx}$

if tan(x) = z/r

sin(x) = ?

that's the integration thingy!