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Flux through a spherical surface

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculate the flux of vector field F = -3r through sphere radius 5 at the origin.

    2. Relevant equations



    3. The attempt at a solution

    Since the orientation is always exactly opposite of the orientation of the surface, I expect a negative answer.

    Also, since they are along the same direction, and n is a unit vector the dot product must reduce to a multiplication of magnitudes.

    (I can't find the vector arrow in itex..! F is a vector and has an arrow above!)

    [itex]F \bullet \widehat{n} \Delta A = || F ||\Delta A [/itex]

    Which is 300pi over the entire surface area, and since it's in the opposite direction, -300pi.

    What am I doing wrong?
     
  2. jcsd
  3. Apr 19, 2012 #2
    Pardon the bump, but, any ideas? Should I go back to square 1?
     
  4. Apr 19, 2012 #3

    HallsofIvy

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    Why do you think you are doing anything wrong?
     
    Last edited: Apr 19, 2012
  5. Apr 19, 2012 #4
    Well, because my answer was marked incorrect.
     
  6. Apr 19, 2012 #5

    sharks

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    F = -3r? I'm not familiar with this format for the vector field. Did you type it correctly? For a vector field, you normally need to specify if -3r is in which direction?
    For example: [itex]\vec F = -3r \vec i[/itex]

    Normally, i would use the Gauss Divergence theorem to solve this easily.
    [tex]\iiint\ div\vec F \,.dV[/tex]
     
  7. Apr 19, 2012 #6
    No, r is in the direction of radial increase. These aren't Cartesian vectors.

    Gauss Divergence is in the next chapter, the professor said it would cut the work down considerably but I think I should learn the basics first.
     
  8. Apr 19, 2012 #7

    sharks

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    OK, so i think what you need to learn is how to use surface integral to find flux through a surface, S.

    [tex]\iint_S \vec F. \hat n \,.d \sigma[/tex]
    where [itex]d\sigma[/itex] is the differential area and [itex]\hat n[/itex] is the unit normal vector outwards from the surface.

    I would suggest finding only the upper half of the sphere, where [itex]\hat n[/itex] is positive, and then you multiply your final result (flux) by 2 to get the total flux. But maybe there's a simpler way.

    However, according to what i know, you normally first need to have (or in this case, derive) the vector field [itex]\vec F[/itex] in terms of its unit direction vectors.

    In your case, the equation of the sphere would be: [itex]x^2+y^2+z^2=25[/itex]
     
    Last edited: Apr 19, 2012
  9. Apr 19, 2012 #8
    Well, yes I know that. That dot product is just a multiplication of magnitudes. F and the sphere's orientation are the same, but opposite, at all points.

    And since the vector field is constant magnitude, I can pull it out of integration.

    Which leaves me with the integral of differential area multiplied by the constant vector field magnitude, which is just the total surface area multiplied by the constant vector field magnitude. Which is what I've done.

    So what's wrong with that thinking?
     
  10. Apr 19, 2012 #9
    Why convert this to cartesian coordinates? It's already in unit vector notation, and I'm dealing with a spherical surface. Wouldn't this only complicate the problem?
     
  11. Apr 19, 2012 #10

    sharks

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    Then, since F=-15, you can find out [itex]\hat n=\frac{∇ \phi }{\left| ∇ \phi \right|}[/itex] and [itex]d\sigma[/itex], from its surface area formula.
    Let [itex]\phi (x,y,z)=x^2+y^2+z^2-25[/itex]

    I think [itex]\vec F=\left[ 0\,, 0\,, -15 \right]^T[/itex]
     
    Last edited: Apr 19, 2012
  12. Apr 19, 2012 #11
    I don't really understand why I'd need any equation for this sphere at all here.
     
  13. Apr 19, 2012 #12

    tiny-tim

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    area = 4π(5)2, flux = field x area (since it's normal to the area), = -300π …

    looks ok to me :confused:
     
  14. Apr 19, 2012 #13
    Wiley Plus strikes again, it seems. :grumpy:

    Thank you very much.
     
  15. Apr 19, 2012 #14
    *Sigh*

    I think I see where we erred.

    The field is 3r, which actually has a magnitude of 15 at all points of the surface, not 3, because of the sphere's radius of 5.

    So the answer should be -1500π
     
  16. Apr 20, 2012 #15

    tiny-tim

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    oh!! :rolleyes:

    [itex]-3\mathbf{r}[/itex], not [itex]-3\mathbf{\hat{r}}[/itex] :biggrin:
     
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