---
title: "Dielectric boundary at z=0, region 1 data"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:dielectric-boundary-z-3-2"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/dielectric-boundary-z-3-2
site: BCT Engineering
---

# Dielectric boundary at z=0, region 1 data

Electromagnetism · Chapter 2: Electric field

- Asked in: 2075 Chaitra, 7 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-z-3-2
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Let the region z < 0 be composed of a uniform dielectric material for which ε_R1 = 3.2, while the region z > 0 is characterized by ε_R2 = 2. Let D1 = −30a_x + 50a_y + 70a_z nC/m² and find: (a) D_t1 (tangential component of D in region 1); (b) polarization (P1); (c) E_n2 (normal component of E in region 2); (d) E_t2 (tangential component of E in region 2).

## Short answer

Tangential D₁ = −30 a_x+50 a_y nC/m². P₁=0.6875 D₁ = −20.6 a_x+34.4 a_y+48.1 a_z nC/m². Normal E₂ = 3.95 a_z kV/m. Tangential E₂ = −1.06 a_x+1.76 a_y kV/m.

## Answer

**Given:** Boundary at z = 0. Region 1 (z < 0): ε_R1 = 3.2. Region 2 (z > 0): ε_R2 = 2. D₁ = −30 a_x + 50 a_y + 70 a_z nC/m². The normal to the boundary is a_z.

**Boundary conditions (no free surface charge on a dielectric interface):**
- Normal D is continuous: D_N1 = D_N2.
- Tangential E is continuous: E_t1 = E_t2, i.e. D_t1/ε₁ = D_t2/ε₂.

#### (a) Tangential component of D in region 1

```text
D_t1 = −30 a_x + 50 a_y   nC/m²        (|D_t1| = 58.3 nC/m²)
```

#### (b) Polarization P₁

```text
P₁ = D₁ − ε₀E₁ = D₁ − D₁/ε_R1 = D₁ (1 − 1/3.2) = 0.6875 D₁
P₁ = −20.6 a_x + 34.4 a_y + 48.1 a_z   nC/m²
```

#### (c) Normal component of E in region 2

D_N2 = D_N1 = 70 nC/m²:

```text
E_N2 = D_N2 / (ε_R2 ε₀) = 70 × 10⁻⁹ / (2 × 8.854 × 10⁻¹²) = 3953 V/m
E_N2 = 3.95 a_z kV/m
```

#### (d) Tangential component of E in region 2

E_t2 = E_t1 = D_t1 / (ε_R1 ε₀):

```text
E_t2 = (−30 a_x + 50 a_y) × 10⁻⁹ / (3.2 × 8.854 × 10⁻¹²)
     = (−30 a_x + 50 a_y) × 10⁻⁹ / (2.833 × 10⁻¹¹)
     = −1059 a_x + 1765 a_y   V/m
```

**Answers:**
- (a) D_t1 = −30 a_x + 50 a_y nC/m²
- (b) P₁ = −20.6 a_x + 34.4 a_y + 48.1 a_z nC/m²
- (c) E_N2 = 3.95 kV/m (along a_z)
- (d) E_t2 = −1.06 a_x + 1.76 a_y kV/m

## Check yourself

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

1. At the boundary z = 0, which components of D1 are tangential?
   A. The a_x and a_y components
   B. The a_z component only
   C. All components
   D. None
   Answer: A. The normal is a_z, so a_x and a_y are tangential. (Key point: Boundary orientation)

2. P₁ = D₁ − ε₀E₁ simplifies to:
   A. D₁/ε_R1
   B. D₁(ε_R1 − 1)
   C. ε₀ D₁
   D. D₁(1 − 1/ε_R1) = 0.6875 D₁
   Answer: D. With E₁ = D₁/(ε₀ε_R1), P₁ = D₁(1 − 1/3.2). (Key point: Polarization)

3. Why is D_N2 = 70 nC/m²?
   A. Normal E is continuous
   B. Tangential D is continuous
   C. Normal D is continuous across a boundary with no free surface charge
   D. D is the same on both sides entirely
   Answer: C. With no free charge, D_N1 = D_N2. (Key point: Normal D)

4. How is E_t2 found?
   A. E_t2 = D_t1/(ε_R2 ε₀)
   B. E_t2 = E_t1 = D_t1/(ε_R1 ε₀)
   C. E_t2 = E_N2
   D. E_t2 = 0
   Answer: B. Tangential E is continuous, so E_t2 = D_t1/(ε_R1ε₀) = -1059 a_x + 1765 a_y V/m. (Key point: Tangential E)

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