Consider an infinite slab of linear dielectric with thickness 2a
containing a uniform free charge density (rho sub 'f').  Find the
electric field everywhere and the bound charge densities.

*Note a is defined to be a number > 0.

Extra Notes:  This is a rather simple problem iff there was no
thickness to the dielectric.  Unfortunately, the thickness forces us
to find the field inside the material.

Perhaps we must use the Gauss's Law equivalent for the Electric Displacement.

I'd be really nice to speak to someone who has a clue about electrostatics.

You are exactly right.  In terms of the electric displacement, D, Gauss's Law is

div(D) = (rho sub 'f')

or, in integral form

Int(D dot da) = (Q sub 'f'),

where the integral is over a closed surface which contains a total
amount (Q sub 'f') of free charge, dot is the dot product, and da is
an infinitesimal area element.

Let the y axis be perpendicular to the slab, with y=0 at the middle of
the slab.  Using the integral form of Gauss's law for electric
displacement, draw a pillbox with one surface coinciding with the
center of the slab and the other surface a distance y above it, still
within the slab.  There is no flux through the bottom surface (or the
sides, obviously) and the flux through the top is just D*A, where A is
the area of the top (or bottom) surface of the pillbox.  The enclosed
free charge is A*y*(rho sub 'f'), so:

D = (rho sub 'f')*y.

This is valid within the slab.  For a linear dielectric, 

E = D/(K * epsilon naught), 

where K is the dielectric constant, and epsilon naught is the
permittivity of free space.  So the electric field inside the slab is

E = (rho sub 'f')*y / (K * epsilon naught)      [ -a < y < a ]

There is a discontinutity in the electric field across the boundary of
the slab because of the surface bound charge.  You can find the field
outside the slab easily (it does not depend on the dielectric
constant).  It is

E = (rho sub 'f') * a / (epsilon naught)        [ abs(y) > a ]

In the last line, of course, the direction of the field is upward
(positive) above the slab and downward (negative) below it.  Assuming
(rho sub 'f') is positive.

The expression for the E-field outside of the dielectric, as you have
stated, is a  function of a.

Shouldn't the E-field outside the material be a function of the
distance from the dielectric material?


And the E-field inside the dielectric be a constant rather than dependent on y?

No, the electric field produced by an infinite sheet of charge is
independent of distance from the sheet.  See here, for example:

www.physlink.com/education/AskExperts/ae544.cfm

So, again, the E-field outside the dielectric material is constant,
independent of position.

If the field inside the dielectric were constant, it would have to be
zero everywhere (this is the limit where K goes to infinity).  You can
see this by symmetry: if the E-field had some nonzero value, how could
it decide whether to point up or down?