Calculating Electric Displacement in Dielectric-Filled Capacitor Slabs

[BearY]

Homework Statement

Introduction to Electrodynamics (4th Edition) By J Griffth Ch.4
Problem 4.18
The space between the plates of a parallel-plate capacitor is filled with two slabs of linear dielectric material. Each slab has thickness a, so the total distance between the plates is 2a. Slab 1 has a dielectric constant of 2, and slab 2 has a dielectric constant of 1.5. The free charge density on the top plate is σ and on the bottom plate −σ.

(a) Find the electric displacement D in each slab

Homework Equations

$$\epsilon_r = \frac{\epsilon}{\epsilon_0}$$
$$D = \epsilon_r\epsilon_0 E $$
And for parallel plate capacitor.
$$E_{vac} = \frac{\sigma}{\epsilon_0}$$

The Attempt at a Solution

How do I relate $E$ to $E_{vac}$? I see in the text when you don't need to worry about boundary condition, $$D = \epsilon_0 E_{vac}$$. But I am not very sure what that means in this setup. There are 3 boundaries in this problem. Solution online and the solution given by my professor to a similar question seem to have$$\epsilon= \epsilon_0 $$ somehow. The link seems trivial(?) so that neither of these solutions explained this.

 

[Charles Link]

Using the electric displacement $D$, Gauss' law reads $\nabla \cdot D=\rho_{free}$. $\\$ Applying Gauss' with one face of the pillbox between the plates is all you need for this problem.