Figure 4.24

4.18   The space between the plates of a parallel-plate capacitor (Fig 4.24) is filled with two slabs of linear dielectric material. Each slab has a thickness a, so that the total distance between the plates is 2a. Slab 1 has a dielectric constant of 2, and slab 2 has a dielaecric constant of 1.5. The free charge density on the top plate is s and on the bottom plate is -s

1. Find the electric displacement D in each slab.

2. Find the electric field in each slab.

3. Find the potential difference between the plates.

4. Find the location and amount of all bound charge.

5. Now that you know all the charge (free and bound), recalculate the field in each slab, and confirm your answer to (b).