The space between the plates of a parallel-plate capacitor

Problem 18P

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 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.

Reference fig 4.24

(a) Find the electric displacement D in each slab.

(b) Find the electric field E in each slab.

(c) Find the polarization P in each slab.

(d) Find the potential difference between the plates.

(e) Find the location and amount of all bound charge.

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

Solution 18P

Step 1 of 8:

In this question, we have parallel plate capacitor and the space between these plates is filled with two slabs of linear dielectric material, thickness of each slab is  ,

In part a, we need to find electric displacement of each slab

In part b, we need to find the electric field E on each slab

In part c, we need to find polarization on each slab

In part d, we need to find potential difference between the plates

In part e, we need to find location and amount of total bound charges

In part f, since we have that total charges we need to recalculate the electric field and confirm the solution of part b

Data given

Thickness of each slab

Distance between the plates

Dielectric of slab 1

Dielectric of slab 2

Charge density of top plate

Charge density on lower plate