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*D_{x}=Q_{enclosed}
Then I am stuck.
The Q_{enclosed} integration give me some problem.
Please can someone help me to establish the Gauss law to solve the problem?
Thank you
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
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
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..