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Gauss' Law (Why the opposite sign?)

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data

    4. In Fig. 23-28, a butterfly net is in a uniform electric field of magnitude [itex]{E = 3.0}[/itex] mN/C. The rim, a circle of radius [itex]a[/itex] = 11 cm, is aligned perpendicular to the field. The net contains no net charge. Find the electric flux through the netting.

    [​IMG]

    [Fig 23-28]

    2. Relevant equations

    Gauss' Law

    [tex]
    {{\Phi}_{E}} = {{\oint}{\vec{E}}{\cdot}{d{\vec{A}}}}
    [/tex]

    3. The attempt at a solution

    At first I thought this was a simple problem. I reasoned that the net electric flux ([itex]{{\Phi}_{E}}[/itex]) was zero because all the electric field lines that enter leave also.

    However, the book answer was not zero and so proved that answer wrong.

    Then I thought to just evaluate the electric flux noting that the the [itex]{\vec{E}}[/itex] and [itex]{\vec{A}}[/itex] point in the same direction so that,

    [tex]
    {{\Phi}_{E}} = {{\oint}{\vec{E}}{\cdot}{d{\vec{A}}}}
    [/tex]

    [tex]
    {{\Phi}_{E}} = {{\oint}{|\vec{E}|}{|d{\vec{A}}}|}{{cos}{\theta}}
    [/tex]

    Where,

    [tex]
    {\theta} = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}degrees
    [/tex]

    Since, [itex]{\vec{E}}[/itex] and [itex]{\vec{A}}[/itex] point in the same direction.

    So,

    [tex]
    {{\Phi}_{E}} = {|\vec{E}|}{{\oint}{|d{\vec{A}}}|}
    [/tex]

    [tex]
    {{\Phi}_{E}} = {|\vec{E}|}{|{\vec{A}}}|}
    [/tex]

    Where [itex]A > 0[/itex], so letting, [itex]{|{\vec{A}}}|} = A[/itex]. Noting that [itex]A = {{\pi}{{a}^{2}}}[/itex].

    [tex]
    {{\Phi}_{E}} = {|\vec{E}|}{A}
    [/tex]


    [tex]
    {{\Phi}_{E}} = {|\vec{E}|}{{\pi}{{a}^{2}}}
    [/tex]

    However the book answer is the negative of the above.

    So my question is, why the negative answer?

    Thanks,

    -PFStudent
     
    Last edited: Oct 2, 2007
  2. jcsd
  3. Oct 2, 2007 #2

    learningphysics

    User Avatar
    Homework Helper

    I suppose they want the flux through the other side of the net...

    You're right that the net flux through the net (ie adding the flux through both sides) is zero...

    But I guess they just want the flux through one side of the net.
     
  4. Oct 2, 2007 #3
    Well, from what the problem says how would you know which side of the net (left or right)?

    If you take the [itex]{{\Phi}_{E}}[/itex] of the right side the angle between [itex]{\vec{E}}[/itex] and [itex]{\vec{A}}[/itex] is 180 degrees and so the [itex]{{\Phi}_{E}}[/itex] is negative.

    However,

    If you take the [itex]{{\Phi}_{E}}[/itex] of the left side the angle between [itex]{\vec{E}}[/itex] and [itex]{\vec{A}}[/itex] is 0 degrees and so the [itex]{{\Phi}_{E}}[/itex] is positive.

    So how do you know which one to take?

    Thanks for the reply learningphysics.

    Thanks,

    -PFStudent
     
  5. Oct 2, 2007 #4

    learningphysics

    User Avatar
    Homework Helper

    Yeah, you can't really tell which side they want... I guess the flux through the left is what they want, because that is the direction in which the butterflies go into the net... Not sure.
     
  6. Oct 2, 2007 #5
    Ah, "through the netting" as in the direction through which butterfiles travel in to the net, which is (coming from the right side) the left side.

    Ok, now that makes sense, because looking at the left side the [itex]{\vec{A}}[/itex] points to the left and the [itex]{\vec{E}}[/itex] points to the right. So, the angle between the two is 180 degrees and that is where the book answer got the negative sign.

    Yea, now it makes sense. But I still think that was a badly worded problem. :tongue2:

    Thanks for the help learningphysics.

    Thanks,

    -PFStudent
     
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