1. ## 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

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

3. 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

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

5. You are right the good equation is:

ρv=ρvo*sin(PI*x/(2*a)),for -a<x<a
Sorry !

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 = (-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

7. Thank you very much topsquark for your explantion.
It is much simpler with your choice of gaussian surface.
B

8. 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..

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