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Math Help - Gauss Law

  1. #1
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    Gauss Law

    Hey, I need help to use the Gauss law in this problem:

    We have A planar slab of charge with a charge density ρv=ρvo*sin(2*pi/(2*a)),for -a<x<a
    the thickness of the slab is 2*a.
    the horizontal y axis passes through the middle of the stab.
    the x axis is vertical
    a) Find the electric field vector in the region -a<x<a.


    I know I need to use the Gauss law.
    But I have some problems to establish the integral.
    What surface should I take?

    I tried the cylinder as surface of integration.
    I got:
    FLUX=2(pi)x*L*Dx=Qenclosed
    Then I am stuck.
    The Qenclosed integration give me some problem.
    Please can someone help me to establish the Gauss law to solve the problem?

    Thank you
    Last edited by braddy; February 20th 2007 at 11:47 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by braddy View Post
    Hey, I need help to use the Gauss law in this problem:

    We have A planar slab of charge with a charge density ρv=ρvo*sin(2*pi/(2*a)),for -a<x<a
    the thickness of the slab is 2*a.
    the horizontal y axis passes through the middle of the stab.
    the x axis is vertical
    a) Find the electric field vector in the region -a<x<a.


    I know I need to use the Gauss law.
    But I have some problems to establish the integral.
    What surface should I take?

    I tried the cylinder as surface of integration.
    I got:
    FLUX=2(pi)x*L*Dx=Qenclosed
    Then I am stuck.
    The Qenclosed integration give me some problem.
    Please can someone help me to establish the Gauss law to solve the problem?

    Thank you
    I need both a visual assist as well as an equation assist. I'm not seeing just how the axes are here. The x is perpendicular to the slab whereas y is parallel to it? And your equation for charge density doesn't contain a variable.

    -Dan
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  3. #3
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    For the picture since I could not upload something with microsoft paint I send you an image from a website:
    General Physics II (picture at the botom of the page)
    For the equation, I just have the gauss law.
    My problem is to find the surface enclosed.
    Also the volume to find the total charge.
    Thank you
    B
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by braddy View Post
    For the picture since I could not upload something with microsoft paint I send you an image from a website:
    General Physics II (picture at the botom of the page)
    For the equation, I just have the gauss law.
    My problem is to find the surface enclosed.
    Also the volume to find the total charge.
    Thank you
    B
    Okay for the picture, but your equation for the charge density has no variable in it!! What is the correct equation for the charge density?

    -Dan
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  5. #5
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    You are right the good equation is:

    ρv=ρvo*sin(PI*x/(2*a)),for -a<x<a
    Sorry !
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by braddy View Post
    You are right the good equation is:

    ρv=ρvo*sin(PI*x/(2*a)),for -a<x<a
    Sorry !
    Now we're in business. Actually, any Gaussian surface that has an axis of symmetry along the x-axis will do. A cylinder obviously will work, but a rectangular prism will also do. The rectangular prism will likely solve your integration problem for you.

    So let's do that. I have a Gaussian surface that is a rectangular prism of dimensions -b to b in the x direction, -c to c in the y direction, and -d to d in the z direction. (Rho is a volume density so we need that third dimension.) So the charge Q in this volume will be the integral over y and z (which have constant components since the charge density only varies in x) giving:
    Q = (4cd)*Int(p(x)dx) = (4cd)*Int(p0*sin(pi*x/(2a))dx) <-- Calling "rho" p.

    Q = (4cdp0)*Int(sin(pi*x/(2a))dx)

    Change the integration variable to y = (pi/(2a))*x ==> dy = (pi/(2a))dx and changing the integration limits from -b to b into -pi*b/(2a) to pi*b/(2a):
    Q = (4dcp0)*(2a/pi)*Int(sin(y)dy) = (8adcp0)/(pi)*(-cos(y))

    Q = (-8adcp0)/(pi)*(cos([pi*b/(2a)]) - cos([-pi*b/(2a)]))

    Q = (-8adcp0)/(pi)*(cos([pi*b/(2a)]) - cos([pi*b/(2a)])) <-- cosine is an even function so cos(-x) = cos(x)

    Thus Q = 0.

    No matter what Gaussian surface you choose, the enclosed charge will be 0, thus the flux of the Electric field will be 0 N/C. Thus E inside the material will be uniformly 0 N/C.

    -Dan
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  7. #7
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    Thank you very much topsquark for your explantion.
    It is much simpler with your choice of gaussian surface.
    B
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  8. #8
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    Hey,
    I still have a question for my problem.

    The planar slab is then surrounded by two planes of charge, one carrying ρso at the top (in the positive direction of x=b)
    and the other -ρso at x=-b.
    What should be ρso to make the field vector be zero at x=0?

    Since for a plane the electric field is E=ρso/(2*epsilon 0), then the value of ρso would be zero!
    But something seems odd...

    please can someone check my reasonning?

    EDIT: I just change some things since I realized I wrote the wrong formula for the electric field for a plane..
    Last edited by braddy; February 22nd 2007 at 06:36 AM.
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by braddy View Post
    Hey,
    I still have a question for my problem.

    The planar slab is then surrounded by two planes of charge, one carrying ρso at the top (in the positive direction of x=b)
    and the other -ρso at x=-b.
    What should be ρso to make the field vector be zero at x=0?

    Since for a plane the electric field is E=ρso/(2*epsilon 0), then the value of ρso would be zero!
    But something seems odd...

    please can someone check my reasonning?

    EDIT: I just change some things since I realized I wrote the wrong formula for the electric field for a plane..
    Your reasoning matches mine. Something IS fishy. Perhaps I erred somewhere in my original solution? (If so I can't find it.)

    -Dan
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