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Two homogeneous dielectric regions 1 (p ≤ 4 cm) and 2 (p ≥ 4 cm) have dielectric constants 3.5 and 1.5, respectively. If D₂ = 12a, 6a+ 9a, nC/m², calculate (a) E₁ and D₁, (b) P₂ and pp2, (c) the energy density for each region.

Two homogeneous dielectric regions 1 (p ≤ 4 cm) and 2 (p ≥ 4 cm) have dielectric constants 3.5 and 1.5, respectively. If D₂ = 12a, 6a+ 9a, nC/m², calculate (a) E₁ and D₁, (b) P₂ and pp2, (c) the energy density for each region.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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**Problem Statement:** Two homogeneous dielectric regions, denoted as region 1 (\( \rho \leq 4 \, \text{cm} \)) and region 2 (\( \rho \geq 4 \, \text{cm} \)), have dielectric constants of 3.5 and 1.5, respectively. Given the electric displacement vector for region 2, \( \mathbf{D_2} = 12a_\rho - 6a_\phi + 9a_z \, \text{nC/m}^2 \), calculate the following: (a) The electric field \( \mathbf{E_1} \) and displacement \( \mathbf{D_1} \) in region 1. (b) The polarization \( \mathbf{P_2} \) and the susceptibility \( \chi_{p,\nu_2} \) for region 2. (c) The energy density for each region. **Explanation:** 1. **Regions and Dielectric Constants:** - Region 1 (\( \rho \leq 4 \, \text{cm} \)), Dielectric constant = 3.5 - Region 2 (\( \rho \geq 4 \, \text{cm} \)), Dielectric constant = 1.5 2. **Given Data:** - \( \mathbf{D_2} = 12a_\rho - 6a_\phi + 9a_z \, \text{nC/m}^2 \) 3. **Required Calculations:** (a) Determine \( \mathbf{E_1} \) and \( \mathbf{D_1} \): - Use the relationship between the displacement vector \( \mathbf{D} \) and electric field \( \mathbf{E} \) with the dielectric constant. (b) Compute \( \mathbf{P_2} \) and \( \chi_{p,\nu_2} \): - Determine the polarization using \( \mathbf{P} = \mathbf{D} - \epsilon_0 \mathbf{E} \). - Find the susceptibility from \( \chi_p = \epsilon_r - 1 \). (c) Energy Density: - Use the formula \( u = \frac{1}{2} \mathbf{E} \cdot \
Transcribed Image Text:**Problem Statement:** Two homogeneous dielectric regions, denoted as region 1 (\( \rho \leq 4 \, \text{cm} \)) and region 2 (\( \rho \geq 4 \, \text{cm} \)), have dielectric constants of 3.5 and 1.5, respectively. Given the electric displacement vector for region 2, \( \mathbf{D_2} = 12a_\rho - 6a_\phi + 9a_z \, \text{nC/m}^2 \), calculate the following: (a) The electric field \( \mathbf{E_1} \) and displacement \( \mathbf{D_1} \) in region 1. (b) The polarization \( \mathbf{P_2} \) and the susceptibility \( \chi_{p,\nu_2} \) for region 2. (c) The energy density for each region. **Explanation:** 1. **Regions and Dielectric Constants:** - Region 1 (\( \rho \leq 4 \, \text{cm} \)), Dielectric constant = 3.5 - Region 2 (\( \rho \geq 4 \, \text{cm} \)), Dielectric constant = 1.5 2. **Given Data:** - \( \mathbf{D_2} = 12a_\rho - 6a_\phi + 9a_z \, \text{nC/m}^2 \) 3. **Required Calculations:** (a) Determine \( \mathbf{E_1} \) and \( \mathbf{D_1} \): - Use the relationship between the displacement vector \( \mathbf{D} \) and electric field \( \mathbf{E} \) with the dielectric constant. (b) Compute \( \mathbf{P_2} \) and \( \chi_{p,\nu_2} \): - Determine the polarization using \( \mathbf{P} = \mathbf{D} - \epsilon_0 \mathbf{E} \). - Find the susceptibility from \( \chi_p = \epsilon_r - 1 \). (c) Energy Density: - Use the formula \( u = \frac{1}{2} \mathbf{E} \cdot \
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