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A question on flux, and on field integration

  1. Nov 9, 2012 #1
    hey all,
    i've recently started studying electrostatics, and i have couple of question about things that i did not fully understand, and would very much appreciate if someone could set me straight.

    1) how can a cube, with a single charge in the middle of it have a flux? don't the field lines cancel each other, thus achieving 0 electric field for each plane? i mean, you have a field vector going up on the Y axis, and down on the same axis (and so on for the others), so how come the flux [itex]\Phi=A\overline{E}[/itex] isn't zero?

    2) if i have a flat disk on a plane, with uniform charge density [itex]\sigma[/itex], why when i integrate, i do so for small rings, and not for very small circles? why is it: [itex]E=\int\frac{kσ2∏rdr}{(...)}[/itex] instead of [itex]E=\int\frac{kσ2∏dr}{(...)}[/itex]?

    thank you very much for your help
  2. jcsd
  3. Nov 9, 2012 #2
    For your first question I would say that flux is a scalar and a unit outward normal is defined before calculating it.For example-if field line along +Y axis and normal is outward then flux is positive.For -Y ,normal is again outward(down in most preferred),so again it is positive because of two - signs.
    For second one,you will have to cover the whole area of disk(for integration) and a circle does not have any thickness so you must have to select a ring.Also note that it should have the units of area which is only in the first case,not the second.
  4. Nov 9, 2012 #3
    so what you are saying (in regards to question no. 1) is that when i get E=0, it's because im adding vectors for which the total sum is zero (opposite directions) while the flux does not "consider" that case, and is a "per case" thing - each face has it's own flux, even though total E is zero?

    and another one:
    how can i find the field exerted by infinite plane on a point along the Z axis, but without using Gauss law? do i just say that a dq part of the charge is equal to kσdxdy and then
    [itex]\overline{E}[/itex]=[itex]\int\int kσdxdy\cdot\frac{z}{\sqrt{x^{2}+y^{2}}}[/itex] ?

    thank you!
    Last edited: Nov 9, 2012
  5. Nov 9, 2012 #4


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    It doesn't make any sense to talk about the "total E" in this situation. You can only add E's that are located at the same point, produced by different charges or other sources. In your example, there's no meaningful way to add the E at the top of the cube to the E at the bottom of the cube.

    Not quite. E is a vector, so when you add the contributions to the total E at a given point, you have to do each component separately, in general:

    $$E_z = \int {\int {dE_z}}$$

    and similarly for ##E_x## and ##E_y##. In this example, you can argue from symmetry that the total ##E_x## and ##E_y## are both zero. For ##E_z## you have to take into account the angle between the z-direction and the line that defines r. And remember, Coulomb's Law has an r2 in the denominator. :wink:

    $$E_z = \int {\int {\frac {k \sigma dx dy} {r^2} \cos \theta}}$$
  6. Nov 9, 2012 #5
    but doesn't the E that go in the +Y direction "cancels" the one going in the -Y direction?
  7. Nov 9, 2012 #6


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    Staff: Mentor

    Yes, that's what I mean by "argue from symmetry..." :smile:
  8. Nov 9, 2012 #7
    so, if i want to treat E as a vector, i must relate it to a certain point? i thought the field was a vector anyhow, so i can add it however i wanted.
    i'm adding a picture that might make things clear. the green arrows are E exerted by a single charge at (0,0,0). so, from symmetry, you argue that E in that axis cancels each other, but can't you say as well, that if you were to add them up (0,y,0)+(0,-y,0) you'd get the same answer? -note that i refer to E outside the box, as i understand that inside (from gauss law) the field is zero in any case

    http://img842.imageshack.us/img842/3222/cubeq.jpg [Broken]

    thank you!

    edit: i think i'm getting it: i can treat E in that manner if, like you say, i have a point on which E is exerted, but if i'm talking about a sphere / cube / other as a shape, there is no point for which i can add the vectors?

    thank for your patience, when i get things slow, i understand them fast :)
    Last edited by a moderator: May 6, 2017
  9. Nov 9, 2012 #8

    Meir Achuz

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    What book are you using. Your questions have simple answers in most texts.
  10. Nov 11, 2012 #9
    Flux arises by scalar product,so in the situation where E is non zero flux can be zero because of a cosine term which may cancel all the flux.Think when electric field is passing through a region(closed) with no charge inside of it
  11. Nov 11, 2012 #10
    For a closed surface around a charge, all the E-field lines are "poking" in or out, so there IS a net flux.

    If you replace the charge with a bar magnet, some of the field lines poke out and some poke in, resulting in no net flux.

    Gauss' Laws.
  12. Nov 12, 2012 #11
    i have principles of physics, 9th edition, by Halidy & Resnick.

    @greswd: if i put a bar inside, wouldn't i have 0 E at the top and bottom?
  13. Nov 15, 2012 #12
    and another question please: I understand Gauss's law in regards to plains (thanks to you). but how does it work when dealing with a sphere? i know how to integrate the whole thing, but i'm trying to get it in a more intuitive manner. when i calculate the flux , I'm looking at infinitesimal area dA, which is the perpendicular vector to the surface. so far, so good. but when I'm dealing with spheres, I integrate over infinitesimal spheres - or more accurate, over their surface area which is 3D.

    what I think is, that because the flux goes through the entire sphere - say we have a 1C charge inside - and is different between these little spheres (the flux is proportional to electric field lines) so you need to sum the flux through all these little spheres in order to get the total flux, and that's why dA is the entire 3D surface area of an infinitesimal sphere.
    Am I correct? If not, how do these dA look like? if they are 2D (r^2), then how can i get just the flux for one such dA?

    Don't laugh if my question seems stupid, but I always thought of area as a 2D thing
    Last edited: Nov 15, 2012
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