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The flux through the hemispherical surface

  1. Sep 20, 2015 #1
    1. The problem statement, all variables and given/known data
    The flux through the hemispherical surface in the figure shown below is
    HEMI.png

    2. Relevant equations
    25px-OiintLaTeX.svg.png 658ce88d8e6ff0b3aa4dc97ee27b244c.png

    3. The attempt at a solution
    But the above formula is for closed surface .But hemisphere is not a closed surface to make it closed surface we will need another hemisphere .
    then the flux through the hemispherical surface should be =Q/2ε0
    but this is wrong it should be πR^2E
     
  2. jcsd
  3. Sep 20, 2015 #2

    Hesch

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    The flux = π*R2 * E.

    Think about what π*R2 actually is.
     
  4. Sep 21, 2015 #3
    area of a circle.
     
  5. Sep 21, 2015 #4

    haruspex

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    The answer must be expressed in terms of the facts given. Your diagram indicates a (uniform?) field and a radius. It does not show a charge.
    To answer the question in terms of E, imagine closing the hemisphere with a flat disc radius R. What is the flux of E through the disk? Do any flux lines pass through the hemispherical shell but not through the disk, or vice versa?
     
  6. Sep 21, 2015 #5
    No.
     
  7. Sep 21, 2015 #6
    If we go for finding flux for disk
    It will be E cos 180 degrees 4pir^2
    =-4pir^2E
    which is not the answer
     
  8. Sep 21, 2015 #7

    BvU

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    "which is not the answer" is not correct: Haru asks for the flux through the disk and that is ##-4\pi r^2\,E\ ##.

    1. [edit] sorry, overlooked the 4. flux through the disk and that is ##\ -\pi r^2\,E\ ##.

    You did very well letting the area vector for the disk point to the left (the outside of the hemisphere).

    Now your relevant equation comes in very handy: Let S be the hemisphere. There is no charge inside, so ##\oint \vec E \cdot d\vec S = 0##. For the disk part you have a value and the bowl has to have the equal and opposite value to get that zero.
     
    Last edited: Sep 21, 2015
  9. Sep 21, 2015 #8

    Hesch

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    The flux through the disk = πR2*E.

    The E-field is perpendicular to the disk ( not to the all over hemisphere ).

    When you close a volume by the disk+hemisphere, that does not include any charge, the sum of the fluxes: ψdisk + ψhemis = 0 →
    ψhemis = -ψdisk
     
  10. Sep 21, 2015 #9
    what's that?
     
  11. Sep 21, 2015 #10

    BvU

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    The curved part of the hemisphere
     
  12. Sep 21, 2015 #11
    How?Can you please elaborate.
     
  13. Sep 21, 2015 #12

    BvU

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    You said so yourself in #6 !
     
  14. Sep 21, 2015 #13
    I wrote
     
  15. Sep 21, 2015 #14

    BvU

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    Agrees nicely, I would say !

    [edit] Wow, now I see a 4. Where did you get that from
     
  16. Sep 21, 2015 #15

    Hesch

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    Ψ = Area*Ψdens*sin φ = πR2*E*1.

    How does 4πR2 get into it ?
     
  17. Sep 21, 2015 #16

    haruspex

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    How do you get ##4\pi R^2##? What's the area of a disk?
     
  18. Sep 21, 2015 #17
    It was always there!
     
  19. Sep 21, 2015 #18
    I thought we use surface area in gauss's formula,not area.
     
  20. Sep 21, 2015 #19
    Ok.Let me ask one thing
    surface area is always of three dimensional figures?it can not be there in two dimensional figures such as circle and square?
     
  21. Sep 21, 2015 #20

    haruspex

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    You have a uniform field E at right angles to a plane surface (disk) of area ##\pi R^2##. What flux does that give?
     
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