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Electric flux through a hemisphere

  1. Feb 21, 2014 #1
    During my physics lecture, the professor said that flux on a closed surface is equal to zero. What exactly is a closed surface defined as?

    This problem also requires the use of the Flux = Field * Area formula. In relation to the direction of the electric field, what is the area for Part B? I'm just having trouble understanding the importance of closed surfaces vs. non-closed surfaces and how to tell them apart. Thanks
     

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  3. Feb 21, 2014 #2

    CAF123

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    Flux through a surface is more generally defined as ##\phi_E = \int_S \mathbf{E} \cdot \mathbf{dS}##, so it is the integral of the normal component of ##\mathbf{E}## over the surface ##S##. This surface ##S## will have a boundary curve if it is an open surface (and in your example, this boundary curve is a circle.) By virtue of Stokes', the flux through any surface with the same boundary curve will be the same.
     
  4. Feb 21, 2014 #3

    collinsmark

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    That's true of electric flux, *if* there is no charge inside the closed surface.

    If you have a charge or charges completely surrounded by a closed surface, the electric flux through the closed surface is proportional to the total amount of charge contained within.

    Any shape you can imagine that completely surrounds and contains a volume.

    A closed, six-sided box is a closed surface as an example. A sphere is a closed surface, if the sphere doesn't have any holes in it.

    Any three dimensional shape you can imagine is a close surface as long as that shape completely encloses some sort of volume without having any holes in it.

    Take the box, and remove one of its sides (making it a 5-sided, open box), and it is no longer a closed surface.

    Take a sphere and poke a hole in it, or cut it in half as is done in this problem, and it is not a closed surface.

    There is a convention that the differential surface vector of a closed surface always points "outward" (as opposed to pointing inside). That's merely a convention though.

    There is no such convention for open surfaces though, and you'll have to arbitrarily choose which side of the surface is positive on a case by case basis.

    This part is slightly tricky. But the answer is easy once you develop an intuition about it. I'd like you to think about this one. So I won't give any detailed hints.

    Just choose which side of the surface is positive or negative, and once you choose it be consistent. Keep in mind a flux line passing though the surface in one direction cancels out a flux line passing through the surface in the opposite direction.

    (In terms of differentials, the differential area vector is always perpendicular to the plane of the differential surface. But you will still have to choose which side is positive and which side is negative, and then maintain consistency.)
     
    Last edited: Feb 21, 2014
  5. Feb 21, 2014 #4

    collinsmark

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    And one last clarification,

    Keep in mind that that's not a multiplication sign in the formula. That's the vector dot product! It makes a difference. :smile:

    Take an open, flat surface with area A and a uniform electric field E and the flux is
    [tex] \Phi_E = \vec E \cdot \vec A [/tex]
    Note the vector notation. Converting that to scalars,
    [tex] \Phi_E = E A \cos \theta[/tex]
    where [itex] \theta [/itex] is the angle between the surface vector (perpendicular to the surface) and the electric field direction.

    More generally, with curved surfaces (as CAF123 points out), that becomes,
    [tex] \Phi_E = \int_S \vec E \cdot \vec {dA} [/tex]
     
  6. Feb 21, 2014 #5
    So the convention of flux depends on the direction of surface vectors? I think I understand... that would mean that any flux entering the top of the sphere perpendicular to the axis is cancelled when passing through the bottom of the sphere, since each surface that the E-field lines pass through have a different convention?

    Thanks for your help by the way
     
  7. Feb 21, 2014 #6

    collinsmark

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    Yes, that's right!

    Not because of different conversions. The convention must be the same for all points on the surface.

    For example, you can choose the convention that the surface vector always points outward, from hemisphere's center of curvature. Or alternately, you can choose the different convention that the surface vectors point toward the center of curvature. Your convention must be such that the vector is perpendicular to the differential surface, but you'll always have two choices: the top side or the bottom side. Once you choose, consistency must be maintained.

    The overall flux is zero in Part B) because for every E(dA)cos(θ) passing through the surface from the outside -> in, there is another -E(dA)cos(θ) passing from the inside -> out. In this situation there are no lines of flux that only pass through the surface in one direction.
     
  8. Feb 21, 2014 #7
    Ahh ok I see now, thank you for your help!
     
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