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Electric Flux depedent on position inside a surface, yes?

  1. Jul 12, 2009 #1
    Sample Problems to calculate Electric Flux

    01. A point charge of 1.8mC is at the centre of a cubical Gaussain surface having each side 50cm. What is the net electric flux through the surface?

    Suggested answer:
    According to Gauss' theorem, flux = q / sigma sub0 or 1.8mC/8.854 x 10^-12

    I would question if this is true no matter WHERE inside the cubical Gaussian surface the charge is. If it does matter, why is there no varible for the position inside a given surface????
     
  2. jcsd
  3. Jul 12, 2009 #2

    jtbell

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    Staff: Mentor

    As long as the amount of charge inside the surface stays the same, the total electric flux through the surface does not depend on where the charge is located. It also doesn't depend on the size or shape of the surface, so long as the surface is closed.
     
    Last edited: Jul 12, 2009
  4. Jul 12, 2009 #3
    I will take your word on that but it doesn't seem logical given the following.

    Here's the equation for flux. Phi=q/sigma sub0

    If this is true, flux has nothing at all to do with the radiating vectors' angle with respect to the surface they are passing through, since angle is not a part of the equation for flux.

    So that sounds like these arrows can exit the surface at any angle as long as they pass through. As an analogy, a light blub in a glass sphere. The amount of light leaving the glass sphere is independent of the position of the bulb in the sphere.

    So why is it stated that flux varies with how the surface faces the flow???

    My question in a nutshell: Does flux vary wiith respect to the angle the field makes with the surface? If so, the why is the angle not included in the equation for flux?
    Phi=q/sigma sub0
     
  5. Jul 12, 2009 #4

    Doc Al

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    Note that Gauss's law only talks about the total flux; it makes no comment about the flux through any portion of the surface.

    As you move the charge around within the surface, the field and thus the flux through some portion of the surface may certainly change, yet the total flux remains the same.
     
  6. Jul 12, 2009 #5
    Oh, so the angle is only important with dA, flux thru a small portion of the overall surface.
    So I can throw ANYTHING whatsoever in side the sphere and it's always zero?
     
  7. Jul 12, 2009 #6
    Woops. I made a mistake. I mean whatever is inside the sphere is
    Phi=q/epsilon sub0
    but what is in a small portion is
    Surface integral of vector E dot producted with vector dA
    Is that better?
     
  8. Jul 12, 2009 #7
    Another mistake. Rats.

    I mean whatever is inside the sphere is
    Phi=q/epsilon sub0
    but what is in a small portion is
    Surface E field times dA times cos theta =Vector E dot producted with dA
    Is that better?
     
  9. Jul 12, 2009 #8

    diazona

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    Homework Helper

    Yeah, that sounds right.
     
  10. Jul 12, 2009 #9
    Or to be even more exact.

    I mean whatever is inside the sphere is
    Phi(total)=q/epsilon sub0
    but what is in a small portion is
    Phi(little chunk)=E field times dA times cos theta =Vector E dot producted with dA
    Is that better?
     
  11. Jul 12, 2009 #10

    Doc Al

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    Staff: Mentor

    Sounds good.
     
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