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Find the uniform charge density

  1. Sep 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Upload Problem 2.3 and I uploaded the question in a reply below


    2. Relevant equations
    D/2ε
    E=KQ/R^2
    E=E1+E2+E3

    3. The attempt at a solution
    I would upload my work but I am in a location where I cant right now but I can type what I did hopefully it will be clear enough. When r is outside of the sphere E=KQ/r^2 and when r is inside the sphere E=KQr/a^3. Well when r=1 it is inside and when r=2 it is inside but however at r=3 it is outside I assume then to solve the problem I need to set D/2ε=KQ/r^2 or some combination of this to get it correct, however because the equation adds the "a" term. Which is shown in the problem but is not given I dont know what to do?
     
    Last edited: Sep 13, 2013
  2. jcsd
  3. Sep 13, 2013 #2
    opps forgot to upload
     

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  4. Sep 13, 2013 #3
    Oh and the question is 2.3 sorry should have mentioned this too
     
  5. Sep 13, 2013 #4
    ok I can upload my work now
     

    Attached Files:

  6. Sep 14, 2013 #5

    rude man

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    Hard to read your work, and looks strange. At any rate my answer was quite different and negative.
     
  7. Sep 14, 2013 #6
    Yah I figured I was wrong any hints?
     
  8. Sep 14, 2013 #7

    rude man

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    Sure thing.

    Start with the basic formula relating the D flux across a closed surface and the free charge within the volume circumscribed by that surface.
     
  9. Sep 14, 2013 #8
    double integral of E dot Ds = Q/epsilon0
     
  10. Sep 14, 2013 #9

    rude man

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    No. E does not appear. Just D, area and q.
     
  11. Sep 14, 2013 #10
    Ok closed surface integral of E dot dA = Qfree
     
  12. Sep 14, 2013 #11
    Whoops not E. I meant D
     
  13. Sep 14, 2013 #12
    D dot dA=Qfree
     
  14. Sep 14, 2013 #13

    rude man

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    Excellent.

    Now, how do you determine the total free charge inside the Gaussian surface at 3m?
     
  15. Sep 14, 2013 #14
    Ok well should I guess I need to figure out limits of integration do the limits ave anything to do with (pi)r^2
     
  16. Sep 14, 2013 #15

    rude man

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    You need to determine the total free charge inside the Gaussian surface at r = 3m. No integration required.
     
  17. Sep 14, 2013 #16
    Hmm well what would D be here 2 because of 8-6 then 2x(pi)(3)^2
     
  18. Sep 14, 2013 #17

    rude man

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    The 8 and -6 are surface charge densities, not charges. You need to compute the total net charge as seen by the Gaussian spherical shell at r = 3m.

    Don't think about D until you have the net free charge. One step at a time!
     
  19. Sep 15, 2013 #18
    Well how about this 8/(2(epsilon))=Kq1/(1)^2 and 6/(2(epsilon))=Kq2/(2)^2 where q1=16(pi) and q2= -48(pi) and then q3=q1+q2 so -32(pi) for Qfree? am I close I mean U figure I ave to use the surface carge densities in dome way to find the total charge
     
  20. Sep 15, 2013 #19

    rude man

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    Yes, but not that way!

    Charge = charge density times area. k and epsilon have nothing to do with this problem.
     
  21. Sep 15, 2013 #20
    Ok so 8(pi)(1)^2-6(pi)(2)^2=-16(pi)
     
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