Why Is Choosing the Right Gaussian Surface Crucial in Gauss' Law Calculations?

[Neophyte]

Homework Statement

I know

D . n(hat)dS =Qencl

I do not really have problems with the dS or Qencl but I do not really know why they choose the gaussian surfaces. Like on the second page where z> pi/2 It seems to me that one would go from (pi/2) to z but won't z effectively just be (pi/2)

Does the DzA*(1) because normal is positive z(hat) come from the top unshaded portion on the right gaussian surface? And Thus the bottom unshaded portion is -Dz because below origin *(-1) from the -z(hat) what is the significance of the shaded portion as oppose to the unshaded? I guess shaded portion is where the charge is coming from?

It seems that you get the limits from the shaded region which would make sense for the charge enclosed

But |z|< (pi/2)
the second surface does not make much sense to me because obviously I am thinking about it incorrectly because I would assume

-Dz(-1)from z(comp) = Qencl

The flux coming out should be on the outside of the slab so the shaded region should not have much to do with the left side of the equation or the gaussian surface I suppose since A will just cancel out.

I mean the answer was right I am not sure if the gaussian surfaces are drawn correctly but I think they are, but obviously I had no idea what was happenin

Wheres does the (-1)*(A) come from in |z| < (pi/2)?

Any help would be appreciated.
Thanks for your time.

 

[jbsteele]

Homework Equations D . n(hat)dS =Qencl The Attempt at a Solution The Gaussian surfaces are used to calculate the electric flux through the surface. The flux is calculated by summing up the electric field over the surface. In this case, the two Gaussian surfaces are used to calculate the total electric field inside the slab. The first Gaussian surface is used to calculate the electric field in the region where z>π/2 and the second Gaussian surface is used to calculate the electric field in the region where |z|<π/2. The shaded portion of the Gaussian surfaces indicate the regions where the electric field is coming from. The unshaded portions indicate the regions where the electric field is going out. The -Dz(-1)*A in the |z|<π/2 region comes from the fact that the electric field is pointing in the negative z-direction in this region. Therefore, we have to multiply the electric field with -1*A to get the correct electric flux.