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How do i find electric field of a sphere?

  1. Mar 20, 2009 #1
    we have solid ball with a radius of R=5cm and a charge density of "rho"=-3C/m3,
    inside this ball, we make a hollow ball shaped space with a radius of R/3 with its centre at 2R/3 from the centre of the big ball.

    diagram below
    http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5314956645280048530 [Broken]

    what is the electric field at point:

    A-on the leftmost point of the hollow
    B-on the top point of the hollow
    C-at the centre of the big ball

    ==============================================================
    these are all my calculations, but i am only stuck with C, so if you dont fell like going through the whole story,skip to the last step (4)
    ===============================================================

    what i did was the following:
    my workings:
    http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5315181847071965298 [Broken]

    i say that my field is the total of 2 fields, one of the original ball, and one of the "imaginary hollowed out ball" (which has a density equal to the negative of the density of the big ball)

    1) i chose a gauss surface as a sphere, after my calculations i get : E=([tex]\rho[/tex]r/3[tex]E[/tex]0)
    i will use this expression throughout my calculation changing the radius every time

    2) for point A, i can say that the field in the y and z axis is 0 because of the symetry of the point.
    as for the axis:
    Ex=([tex]\rho[/tex]R/3[tex]E[/tex]0) - ([tex]\rho[/tex](R/3)/3[tex]E[/tex]0)

    EAx=(2[tex]\rho[/tex]r/9[tex]E[/tex]0)
    -----------------------------------------------------------------------------------
    3) for point B, z axis is still 0 because of symetry
    Ey=([tex]\rho[/tex](sqrt(5)R/3)/3[tex]E[/tex]0*sin(26.565)) -([tex]\rho[/tex](R/3)/3[tex]E[/tex]0) = 0
    Ex=([tex]\rho[/tex](sqrt(5)R/3)/3[tex]E[/tex]0*cos(26.565)) - 0 =

    EBx=(2[tex]\rho[/tex]r/9[tex]E[/tex]0)
    ------------------------------------------------------------------------------------
    4) for point C i say that since it is exactly in the centre of the big ball, the big ball's field at C is equal to 0, because of perfect symetry and because R=0, so the total field at point C is only the field of my hollowed out imaginary ball,
    because of the symetry, the imaginary ball too has 0 field on y and z axis so i only care about x

    ECx=([tex]\rho[/tex](2R/3)/3[tex]E[/tex]0)
    ECx=(2[tex]\rho[/tex]r/9[tex]E[/tex]0)
    ---------------------------------------------------------------------------------------
    HOW CAN THIS BE??
    for A, B i understand that it is possible that the field is the same, but for C i do not, the correct answer is supposed to be

    ECx=(2[tex]\rho[/tex]r/36[tex]E[/tex]0)
    but i cannot get to it, where am i going wrong??
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 20, 2009 #2

    alphysicist

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    Homework Helper

    Hi Dell,

    This expression is true, but remember that it applies when the point you are looking at is inside the sphere. (Notice that if you plot it out, it gets larger and larger as you move away from the center; this is true as long as you are inside the sphere, but once you move outside the sphere the electric field begins decreasing.)

    But the field point for part c is outside the imaginary sphere. So what you need for part c is to find the electric field outside of a sphere (but with the formula still including the charge density of the sphere). Does this help?
     
    Last edited by a moderator: May 4, 2017
  4. Mar 20, 2009 #3
    i think so, can i take the sphere as if all its charge were in its centre and use Kq/r^2 for my field?
     
  5. Mar 20, 2009 #4

    alphysicist

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    That sounds right (but when you rewrite it with the charge density you'll have to be sure to keep track of what the different R's mean). Do you get the answer?
     
  6. Mar 20, 2009 #5
    yes thanks, worked perfectly
     
  7. Mar 20, 2009 #6

    alphysicist

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    Glad to help!
     
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