Calculating Electric Displacement in Dielectric-Filled Capacitor Slabs

In summary, the problem involves two slabs of linear dielectric material with different dielectric constants between parallel plates. The electric displacement D in each slab can be found using the equations ε_r = ε/ε_0 and D = ε_rε_0E, where E_vac represents the electric field in a vacuum. Gauss' law can be applied to find the electric displacement using one face of a pillbox between the plates.
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BearY
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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.
 
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  • #2
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.
 
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Related to Calculating Electric Displacement in Dielectric-Filled Capacitor Slabs

1. What is the electric displacement problem?

The electric displacement problem, also known as the dielectric displacement problem, is a fundamental issue in electromagnetism that arises when trying to explain the behavior of electric fields in materials with varying degrees of conductivity.

2. How is the electric displacement problem related to Maxwell's equations?

The electric displacement problem is related to Maxwell's equations through the introduction of a new variable, the electric displacement field, which accounts for the effects of polarization in materials. This allows for a more complete understanding of the behavior of electric fields in materials.

3. Why is the electric displacement problem important?

The electric displacement problem is important because it allows us to better understand and predict the behavior of electric fields in materials, which has numerous practical applications in fields such as electronics, telecommunications, and materials science.

4. How is the electric displacement problem solved?

The electric displacement problem is solved by using Maxwell's equations, along with the appropriate boundary conditions, to determine the electric displacement field in a given material. This can then be used to calculate other properties of the electric field, such as the electric flux density.

5. What are some real-world examples of the electric displacement problem?

The electric displacement problem can be observed in many everyday objects, such as capacitors, where the behavior of the electric field is affected by the polarization of the dielectric material between the plates. It is also important in the design of electronic devices and in understanding the behavior of electromagnetic waves in different materials.

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