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Doubts in applying gauss' law

  1. Sep 14, 2013 #1

    FOIWATER

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    Gold Member

    I actually have solved the problem and received the answers that the book provided.

    However I am second guessing what I did,

    This is the problem:

    let rho v be 10/r^2 mC/m^3 between 1<r<4
    let rho v be zero elsewhere
    a)find the net flux crossing the surface at r=2 m, r= 6m
    b)determine D at r=1, r=5

    a) I have no issues with, I can simply say that the Flux crossing the surface equals the charge enclosed and solve for the charge enclosed by integrating the volume charge density over the volume. I used triple integration in spherical coordinates with r from 1 to 2, theta from 0 to pi, and phi from 0 to 2pi. I got 40pi mC

    Same goes for r=6m, I just integrated the same except r went from 1 to 4 (since there is no more charge after r = 4, or rather, the charge density was given as zero) (but the flux will remain the same at 6m as it was at 4m) 120pi

    My issue is with what I did in b), applying gauss' law.

    I know gauss' law holds, and for r=1, since there is no charge inside a spherical gaussian surface, there is no flux density on the surface so D=0 but for r=5m, I assumed a gaussian surface (a sphere) of radius 5. I know the flux at 5m is the same as in part b, 120pi. so:
    flux = Qenc = surface integral D dot ds, D and ds are both vectors.
    How could I, and the book, assume D is constant and take it outside the integral sign?
    The charge density inside the sphere was given as 10/r^2, which isn't uniform over the volume?

    How could a sphere be perpindicular to the flux lines contain within when the charge density is not uniform?
     
  2. jcsd
  3. Sep 14, 2013 #2
    At first glace, it looks like conceptual issue is that the charge density isn't uniform, but is IS still functionally radial. With no angular dependence or other wierd things going, the flux lines are still perpindicular to the sphere.
     
  4. Sep 14, 2013 #3

    FOIWATER

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    Gold Member

    OK that was what I was thinking could be going on.

    Thanks
     
  5. Sep 14, 2013 #4
    Sure thing!
     
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