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Math Help - Infinitely uniform line charge

  1. #1
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    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

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

    the answer is:
    <br />
\frac{\rho}{2\pi\epsilon_{o}r} \hspace {1mm} \bf{r}<br />

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

    Thanks in advance
    Last edited by august; April 7th 2009 at 06:28 AM. Reason: Clarification
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  2. #2
    MHF Contributor chisigma's Avatar
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    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
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  3. #3
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    Quote Originally Posted by chisigma View Post
    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?
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by august View Post
    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!
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