How Does Gauss' Law Apply to Infinite Charge Slabs?

[shanty]

Homework Statement

An infinite slab of uniform charge density ρ occupies the region -∞<x< ∞, -∞<y<∞, -d/2<z<d/2.

Homework Equations

Use Gauss' Law to calculate the electric field density for -∞<z<∞

The Attempt at a Solution

D=εE.
Using that, I just need to find D = $\frac{1}{4*Pi}$∫$\frac{ρ}{r^2}$dv.

But I don't know what to use as a radius...

 

[E_M_C]

The equation you wrote is not Gauss' Law. Also, r² is not a "radius." It is the squared distance between the location of the charged slab and the point at which you wish to sample the electric flux density.

Gauss' Law is:

$\oint \vec{E} \cdot d\vec{a} = \frac{Q_{Enclosed}}{ε_0}$

Start by choosing a closed Gaussian surface: a cylinder, a pillbox, etc. Then write down what you would get on the LHS of the equation, which is the total electric flux. Then, on the RHS, write the total charge enclosed Q_Enclosed in terms of the charge density. What do you get?

 

[shanty]

D = (d*ρ)/2.

∫∫D(dot)ds = ∫∫∫ρdv

Where bounds are -x < x < x and -y < y < y. z bounds for the volume are -d/2 < z < d/2.

I see. Thanks.