# Electric flux density

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 QEnclosed in terms of the charge density. What do you get?

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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.