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 dielectric constant of 2 and slab-2 has a dielectric constant of 1.5. With the area of each of the top and bottom conducting plates is much greater than a?, we can assume the the surface charge densities +o and -o on the top and bottom plates is uniform. (a) Find the electric displacement D in each slab. (b) Find the electric field E in each slab. (c) Find the potential difference between the plates.
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 dielectric constant of 2 and slab-2 has a dielectric constant of 1.5. With the area of each of the top and bottom conducting plates is much greater than a?, we can assume the the surface charge densities +o and -o on the top and bottom plates is uniform. (a) Find the electric displacement D in each slab. (b) Find the electric field E in each slab. (c) Find the potential difference between the plates.
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Transcribed Image Text:**Parallel-Plate Capacitor with Dielectric Slabs**
The space between the plates of a parallel-plate capacitor (referenced as Figure 4.24) is filled with two slabs of linear dielectric material. Each slab has a thickness \(a\), making the total distance between the plates \(2a\). Slab-1 has a dielectric constant of 2, and Slab-2 has a dielectric constant of 1.5. Given that the area of each of the top and bottom conducting plates is much greater than \(a^2\), it is assumed that the surface charge densities \(+\sigma\) and \(-\sigma\) on the top and bottom plates are uniform.
**Tasks:**
(a) **Find the electric displacement \( \mathbf{D} \) in each slab.**
(b) **Find the electric field \( \mathbf{E} \) in each slab.**
(c) **Find the potential difference between the plates.**
(d) **Find the locations and amounts of all bound charge.**
(e) **Based on the values of bound charge, recalculate \( \mathbf{E} \) and verify your answer from (b).**
(f) **Analyze how your results relate to the formula for the addition of two series capacitors.**
Graphs/Diagrams:
- None provided in the text; Figure 4.24 is referenced but not included.
Transcribed Image Text:The image illustrates a diagram labeled as "Figure 4.24," showcasing two slabs, each labeled "Slab 1." These slabs are depicted as parallel layers separated by a distance, both having a thickness denoted as "a."
Key elements in the diagram include:
1. **Charge Distribution**:
- One slab has a positive surface charge density indicated by "+σ."
- The other slab has a negative surface charge density indicated by "−σ."
2. **Dimensions**:
- The thickness of each slab is represented by the label "a."
This configuration often represents a parallel plate capacitor setup where the slabs act as plates with opposite charge densities, potentially illustrating concepts related to electric fields, capacitance, or electrostatic interactions in an educational setting.
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