---
title: "Polarization and E in both regions"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:dielectric-boundary-x-polarization-e"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/dielectric-boundary-x-polarization-e
site: BCT Engineering
---

# Polarization and E in both regions

Electromagnetism · Chapter 2: Electric field

- Asked in: 2068 Chaitra, 3+3 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/dielectric-boundary-x-polarization-e
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

The region x < 0 is composed of a uniform dielectric material for which ε_r1 = 3.2, while the region x > 0 is characterized by ε_r2 = 2. The electric flux density at region x < 0 is D1 = −30a_x + 50a_y + 70a_z nC/m². Find the polarization (P) and electric field intensity (E) in both regions.

## Short answer

With boundary normal a_x, D_x continuous. Region 1: E₁ = (−30a_x+50a_y+70a_z)×10⁻⁹/(3.2ε₀) and P₁=0.6875D₁. Region 2: D₂ = −30a_x+31.25a_y+43.75a_z nC/m²; E₂ from D₂/(2ε₀); P₂=0.5D₂. Components as computed.

## Answer

Boundary: the plane x = 0, with normal along a_x. Region 1 (x < 0): ε_r1 = 3.2. Region 2 (x > 0): ε_r2 = 2. D₁ = −30 a_x + 50 a_y + 70 a_z nC/m².

#### Region 1
```text
E₁ = D₁/(ε₀ε_r1) = D₁ ×10⁻⁹ /(3.2 × 8.854×10⁻¹²)
   = (−30 a_x + 50 a_y + 70 a_z) × 35.29
   = −1058.8 a_x + 1764.7 a_y + 2470.6 a_z  V/m

P₁ = D₁ (1 − 1/ε_r1) = 0.6875 D₁
   = −20.63 a_x + 34.38 a_y + 48.13 a_z  nC/m²
```

#### Region 2
Normal D continuous: D_x2 = D_x1 = −30. Tangential E continuous, so D_t2 = (ε_r2/ε_r1) D_t1 = 0.625 D_t1:

```text
D_y2 = 0.625 × 50 = 31.25 ,  D_z2 = 0.625 × 70 = 43.75
D₂ = −30 a_x + 31.25 a_y + 43.75 a_z  nC/m²

E₂ = D₂/(ε₀ε_r2) = D₂ ×10⁻⁹ /(2 × 8.854×10⁻¹²)
   = −1694.1 a_x + 1764.7 a_y + 2470.6 a_z  V/m
   (E_y and E_z agree with region 1, as required)

P₂ = D₂ (1 − 1/2) = 0.5 D₂
   = −15 a_x + 15.63 a_y + 21.88 a_z  nC/m²
```

| Quantity | Region 1 (x < 0) | Region 2 (x > 0) |
|---|---|---|
| E (V/m) | −1059 a_x + 1765 a_y + 2471 a_z | −1694 a_x + 1765 a_y + 2471 a_z |
| P (nC/m²) | −20.6 a_x + 34.4 a_y + 48.1 a_z | −15 a_x + 15.6 a_y + 21.9 a_z |

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. At x = 0, the normal component of D is:
   A. 50 a_y
   B. 70 a_z
   C. −30 a_x, continuous
   D. zero
   Answer: C. Normal is a_x and D_N is continuous. (Key point: Boundary normal)

2. D_y2 equals:
   A. 50 nC/m²
   B. 31.25 nC/m²
   C. 80 nC/m²
   D. 20 nC/m²
   Answer: B. D_t2 = (2/3.2)(50) = 31.25. (Key point: Tangential D)

3. E_y and E_z compare as:
   A. Same in both regions
   B. Doubled in region 2
   C. Zero in region 2
   D. Different by ε_r ratio
   Answer: A. Tangential E is continuous. (Key point: E tangential)

4. P₁ equals:
   A. 0.5 D₁
   B. 3.2 D₁
   C. D₁/3.2
   D. 0.6875 D₁
   Answer: D. P = D(1 − 1/3.2). (Key point: Polarization)

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