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Homework Help: Flux through a disk

  1. Mar 23, 2009 #1
    in this electrostatics question i have a disk with a radius of R, and i have a charge q placed at a distance of R/2 perpendicular to the disk from its centre. i am asked to find the flux through the disk

    WHAT I DID:-- here are my calculations and i diagram
    http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5312348851145303922 [Broken]

    i take an imaginary "container" as my gauss surface, which is a half sphere "bowl" on one side and the "lid" is my disk. now the flux in through the bowl is my flux out through the lid since there is no charge inside my container,

    i took a spherical surface with a radius of sqrt(5/4)R which is flat on one side(the disk -green in my diagram)since there is no internal charge, all the ingoing flux, (through the disk) is equal to the outgoing flux( through the spherical part)

    now the area of the shere is 2pi*r*h which comes to [(5-sqrt5)/2]pi*R^2


    flux=0.27(q/epsilon) ===>which is the fluc throught the bowl part, so the flux throught the disk is

    flux=(-0.27(q/epsilon)) right??

    but the correct answer in my textbook is exactly half of this??
    can anyone see where i have gone wrong?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 23, 2009 #2


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    That looks like an OK approach.

    But isn't the integral over the outside surface of your spherical section about the same to calculate as just doing the rings that make up the surface of the disk using the more straightforward ∫EcosθdA ?
  4. Mar 24, 2009 #3
    but E is not constant for the surface of the disk, that is why i prefer the spere. can you not see how i can get an answer half what i got, because according to my book that is the correct one
  5. Mar 24, 2009 #4


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    To calculate the surface area of the spherical section in an integral still involves an integration in θ. You don't show enough of your work to tempt me to want to understand where you may have made an error.
  6. Mar 25, 2009 #5
    i showed all my work in the atteched link,
    but anyway i dont see how integrating the sphered part involves θ, since E is constant for a shpere with q in its centre, symetry, . the normal to the area and the field are ALWAYS at θ=0/180 between them
    my work
    http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5312348851145303922 [Broken]
    Last edited by a moderator: May 4, 2017
  7. Mar 25, 2009 #6


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    Sorry. I see now that you did. I forgot that it was in the original image.

    In reviewing your work, I approached it as taking the surface area of the spherical cap which is as you used 2πrh and I agree that it looks to come to

    (5 - √5)/2*πR2

    Noting then that the total surface area is 4πr2

    And taking that radius as √5/2*R that yields a total surface area of 5πR2

    Since Total Φ = Q/εo

    Then through the spherical cap you get

    Φ = Ac/AT * Q/εo = (5 - √5)/2*πR2 / 5πR2 * Q/εo

    And since as you observed the disk projects onto the spherical cap ...

    Basically I don't see from the variables that you have calculated it incorrectly. If your math is OK ... then I don't know what to say.
    Last edited by a moderator: May 4, 2017
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