# Finding electric displacement $\mathbf D$ for a parallel-plate capacitor filled with two slabs of linear dielectric material

This problem refers to Griffiths Problem 4.18:  According to my solutions, my lecturer wrote:

Only free charge is on plates.

$$\mathbf D = -\sigma \hat z \ \text{between the plates and 0 everywhere else}$$

However, I don't know where this came from, especially given the "Only free charge is on plates" part.

I tried this:

Drawing a Gaussian pillbox on the top plate renders after applying Gauss' Law:

$$|\mathbf D| 2 * A = \sigma A$$

$$\implies |\mathbf D| = \sigma / 2 \ \ \text{for top slab}$$

Similarly, for the bottom slab,

$$\implies |\mathbf D| = -\sigma / 2 \ \ \text{for bottom slab}$$

Why is the magnitude of $$\mathbf D$$ in fact $$\sigma$$ and why does it go in the $$-\hat z$$ direction?

You're pretty much there. The displacement field created by each plate will have a magnitude $$|\mathbf D| = \sigma / 2$$. Directionality is important here. The upper plate is positively charged, so the $$\mathbf D$$ vector will point away from it. For the bottom, negatively charged plate, the $$\mathbf D$$ vector will point towards it. Summing up these two vectors in the middle region will give you $$\mathbf D = -\sigma \hat z$$. Note that the displacement field outside the plates is zero because the displacement fields from the positive and negative plates cancel out.