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Help with Maxwell’s equation in integral form

  1. Mar 12, 2005 #1
    Electric field due to a cylindrical charge distribution using Gauss' law.
    Charge is distributed with density [tex]\rho_{0}e^{-r^{2}}[/tex] C/m^3 in cylindrical region r < 1. Find D (displacement flux density vector) everywhere.

    I did used this equation
    [tex]\int_{s}D\cdot\,dS=\int_{V}\rho\*d\upsilon[/tex]

    Since this is a cylindrical charge distribiution I used Gaussian surface in the shape of a cylinder.

    [tex]\int_{s}D\cdot\,dS=\rho\*l[/tex]

    So if I understand good the D=0 inside cylinder. therefore r>R is valid.
    The surface area is [tex]2\pi\*rL[/tex].

    I'm having a problem to set up the the equation or I'm doing everything wrong?
    Thanks for any help and recommendation.
     
  2. jcsd
  3. Mar 12, 2005 #2

    dextercioby

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    1.U need to evaluate the charge...What's the charge in the cylinder...?


    Daniel.
     
  4. Mar 12, 2005 #3
    [tex]\int_{s}D\cdot\,dS=\int_{V}\rho\*d\upsilon[/tex]
    So fom this I need to evaluate volume integral?

    [tex]\int_{s}D\cdot\,dS=Q_{V}[/tex]

    [tex]Q_{V}=\int_{V}\rho_{0}e^{-r^{2}}[/tex]

    But what limits should I used for x,y,z (0 and 1 for all three)?
    Thanks
     
  5. Mar 12, 2005 #4

    dextercioby

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    Nope,you need the cilindrical coordinates [tex] r,\varphi,z [/tex].What's the volume element in cilindrical coordinates...?

    Daniel.
     
  6. Mar 12, 2005 #5
    [tex]Q_{V}=\int_{V}\rho_{0}e^{-r^{2}}[/tex]

    [tex]Q_{V}=\int_{r=0}^{r}\int_{\phi=0}^{2\pi}\int_{z=0}^{l}\rho_{0}e^{-r^{2}}dr\,d\phi\,dz[/tex]

    Is that correct?
     
  7. Mar 12, 2005 #6

    dextercioby

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    U need another "r" in the volume element.And the cilinder has radius 1 (see text of the problem)...

    Daniel.
     
  8. Mar 12, 2005 #7
    [tex]Q_{V}=\int_{r=0}^{1}\int_{\phi=0}^{2\pi}\int_{z=0}^{l}\rho_{0}e^{-r^{2}}r dr\,d\phi\,dz[/tex]

    That is not easy integration, but I think I got it.

    [tex]Q_{V}=l\pi\rho_{0}(e-1)e^{-1}}[/tex]

    Is this is correct what would be next step?
     
  9. Mar 12, 2005 #8

    dextercioby

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    It is correct.Now u have to apply Guass' theorem which gives the flux of the induction [itex] \vec{D} [/itex]...

    Daniel.
     
  10. Mar 12, 2005 #9
    if I understand well D depend on r only, so:

    [tex]\int_{s}D\cdot\,dS=\int_{\phi=0}^{2\pi}\int_{z=0}^{l}D_{r}(r)a_{r}\cdot{r}\,d\phi\,dz\,a_{r}=[/tex]

    [tex]=2\pi\,rlD_{r}(r)[/tex]

    And this I should compare to Qv and find D, right?
     
  11. Mar 12, 2005 #10

    dextercioby

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    It's the same cyclinder of radius unity...That "r" is 1...

    Daniel.
     
  12. Mar 12, 2005 #11
    So this is my D

    [tex]D=\frac{\rho_{0}(e-1)e^{-1}}{2r} a_{r}[/tex]

    I really appreciate your help
     
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