1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric Flux through a hemisphere

  1. Aug 30, 2007 #1
    1. The problem statement, all variables and given/known data
    "What is the flux through the hemispherical open surface of radius R? The uniform field has magnitude E. Hint: Don't use a messy integral!"

    [tex]\mid \vec{E} \mid= E[/tex]
    radius = [itex]R[/itex]

    2. Relevant equations
    Electric Flux over a surface (in general)
    [tex]\Phi = \int \vec{E} \cdot \,dA = \int E \cdot \,dA \cos\theta[/tex]

    Surface area of a hemisphere
    [tex]A = 2\pi r^2[/tex]


    3. The attempt at a solution
    If it were a point charge at the center (the origin of the radius, [itex]R[/itex]), all of the [itex]\cos \theta[/itex] values would be 1, making this as simple as multiplication by the surface area. The only thing that comes to mind for this, however, is somehow integrating in terms of [itex]d\theta[/itex] and using the angle values on both axes of the hemisphere:
    [tex]\left( \frac{\pi}{2}, \frac{-\pi}{2}\right)[/tex]

    But i can't just stick an integral in there like that... can I? I'm really lost on this one...
     
  2. jcsd
  3. Aug 30, 2007 #2

    learningphysics

    User Avatar
    Homework Helper

    You don't need integrals... think of the flux lines... think of another area through which all the flux lines go, that are going through the hemisphere... flux = number of flux lines...

    so if the flux lines that go through the hemisphere are the same flux lines that go through this other area, then the flux through that other area is the same as the flux through the hemisphere.
     
  4. Aug 30, 2007 #3
    so... you're basically saying that shape doesn't matter and the answer is:

    [tex]2\pi R^2 E[/tex]

    and was ridiculously simple, and my real flaw is i was trying to overcomplicate??? which i tend to do a lot...

    and as suggested, i still want to think it has to be more complicated than this since it's not an enclosed surface, and it's not dealing with a point charge... and the flux at the center of the hemisphere's surface would be [itex]E \cdot \,dA[/itex] where the cosine would equal 1... whereas at the edge the cosine would be 0... and i would still have to account for everything in between... right? which is where i start thinking i would need to integrate. but maybe i'm missing the point here...

    or maybe that doesn't matter because since it's a hemisphere and the electric field is going straight into it, meaning the y values cancel out in pairs so i shouldn't worry about cosine... but it's hitting the inside of the surface... blah, i just rambled a bunch, feel free to disregard.
     
    Last edited: Aug 30, 2007
  5. Aug 30, 2007 #4

    learningphysics

    User Avatar
    Homework Helper

    No that isn't the answer... visualize all the flux lines going through the hemisphere... now imagine a flat surface through which those same flux lines go (and only those same flux lines)... what is the shape of this flat surface? What is its area? What is the flux through this new flat surface?
     
  6. Aug 30, 2007 #5

    learningphysics

    User Avatar
    Homework Helper

    I'm referring to the base of the hemisphere. :wink:

    Assuming the field is oriented vertically, and the hemisphere is oriented in the same direction... the flux through the base (which is just a circle of radius R) is the flux through the hemisphere itself.

    But is the field oriented in the same direction as the height of the hemisphere? The question doesn't say. I'm assuming it is... if it isn't then the problem is more complicated.

    So it's [tex]\pi{R}^2E[/tex]
     
    Last edited: Aug 30, 2007
  7. Aug 30, 2007 #6
    it seems to me to be (2)E(pi)R^2 IF the field lines are directed spherically

    if the field lines are all vertical, then yes, the flux is E(pi)R^2
     
  8. Aug 30, 2007 #7
    Uniform electric field usually means a field that does not vary with position.
     
  9. Aug 30, 2007 #8
    ah, i got it now... i didn't understand the concept of flux. thanks much, everyone:)
     
  10. Nov 15, 2010 #9
    hello, im antimatt3er as you can see, well i understood is that you are trying to calculate the flux of a point charge through a hemisphere and the point charge is at O the center of the hemisphere.

    1.i doon't wannat apply gauss law because im not sure that i have symetrie case so i will use classsic integrale method
    2.[tex]\phi[/tex]=[tex]\phi(disc)[/tex]+[tex]\phi(spher)[/tex]
    3.now claculating first flux trought the disc
    [tex]\phi(disc)[/tex]=[tex]\int[/tex][tex]\vec{E}[/tex][tex]\vec{ds}[/tex]
    in all the surface the unit vector [tex]\vec{u}[/tex]that hold[tex]\vec{E}[/tex] is perpenducular to the unit vetor [tex]\vec{n}[/tex] so the cos([tex]\pi/2[/tex])=0 then the [tex]\phi(disc)[/tex]=0
    4.now going to calculate the sencond flux equal q devide by epsilon zero.
    so in the end he have symitri case and voila this is the flux i will right down the demonstartion latter on
    but my reall problem is that what is the flux of a positiv point q placed somewhere in the axsis of the hemisphers
    not just the point O i might a point R or R/2 or whatever?? plz i needd the answer soon and thanks.
     
  11. Nov 15, 2010 #10
    im soory about 7th line it should be a double integrale or surface integrale .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Electric Flux through a hemisphere
Loading...