---
title: "Dielectric boundary, E1 given, find E2 D2 P2"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:dielectric-boundary-find-e2-d2"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/dielectric-boundary-find-e2-d2
site: BCT Engineering
---

# Dielectric boundary, E1 given, find E2 D2 P2

Electromagnetism · Chapter 2: Electric field

- Asked in: 2073 Shrawan, 2+2+2+1+1 marks
- Video: [Dielectric boundary condition with examples by Prof. Niraj Kumar VIT Chennai](https://www.youtube.com/watch?v=IsPbg_F1DXk) — RF Design Basics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/dielectric-boundary-find-e2-d2
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Define electric dipole and polarization. The region z < 0 contains a dielectric material for which ε_r1 = 2.5, while the region z > 0 is characterized by ε_r2 = 4. Let E1 = −30a_x + 50a_y + 70a_z V/m. Find: (a) E2 (b) D2 (c) polarization in region 2 (P2).

## Short answer

Electric dipole: equal opposite charges separated, moment p = Qd. Polarization: dipole moment per unit volume. Using tangential E continuous and normal D continuous: E₂ = −30 a_x + 50 a_y + 43.75 a_z V/m; D₂ = −1.06 a_x + 1.77 a_y + 1.55 a_z nC/m²; P₂ = −0.80 a_x + 1.33 a_y + 1.16 a_z nC/m².

## Answer

#### Electric dipole
An **electric dipole** is a pair of equal and opposite point charges +Q and −Q separated by a small distance d. Its dipole moment is p = Q d (vector from −Q to +Q), in C·m.

#### Polarization
**Polarization** P is the dipole moment per unit volume of a dielectric: P = lim (Σ pᵢ)/Δv, in C/m². When an external field is applied, bound charges in the dielectric shift and form dipoles. For a linear isotropic material P = ε₀χ_e E = ε₀(ε_r − 1)E, and D = ε₀E + P.

#### Boundary: z = 0, normal along a_z
Region 1 (z < 0): ε_r1 = 2.5. Region 2 (z > 0): ε_r2 = 4. E₁ = −30 a_x + 50 a_y + 70 a_z V/m.

Conditions: tangential E is continuous; normal D is continuous (no free surface charge), so ε_r1 E_z1 = ε_r2 E_z2.

#### (a) E₂
```text
E₂ tangential = E₁ tangential : −30 a_x + 50 a_y
E_z2 = (ε_r1/ε_r2) E_z1 = (2.5/4)(70) = 43.75
E₂ = −30 a_x + 50 a_y + 43.75 a_z  V/m
```

#### (b) D₂
```text
D₂ = ε₀ε_r2 E₂ = (8.854×10⁻¹²)(4) E₂
   = 35.42×10⁻¹² (−30 a_x + 50 a_y + 43.75 a_z)
   = −1.062 a_x + 1.771 a_y + 1.549 a_z  nC/m²
```
(Equivalently D₂ = ε₀(−120 a_x + 200 a_y + 175 a_z).)

#### (c) P₂
```text
P₂ = ε₀(ε_r2 − 1)E₂ = 3 ε₀ E₂ = 26.56×10⁻¹² (−30, 50, 43.75)
   = −0.797 a_x + 1.328 a_y + 1.162 a_z  nC/m²
```

**Answers:** E₂ = −30 a_x + 50 a_y + 43.75 a_z V/m; D₂ = −1.06 a_x + 1.77 a_y + 1.55 a_z nC/m²; P₂ = −0.80 a_x + 1.33 a_y + 1.16 a_z nC/m².

## Check yourself

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

1. The dipole moment of charges +Q and −Q separated by d is:
   A. p = Q/d
   B. p = Qd, directed from +Q to −Q
   C. p = 2Q
   D. p = Qd, directed from −Q to +Q
   Answer: D. p = Qd points from the negative to the positive charge. (Key point: Dipole moment)

2. Polarization P is:
   A. Total charge in a dielectric
   B. Dipole moment per charge
   C. Dipole moment per unit volume, in C/m²
   D. Energy per unit volume
   Answer: C. P = lim Σp/Δv. (Key point: Polarization)

3. Which conditions give E₂?
   A. Normal E continuous; tangential D continuous
   B. Tangential E continuous; normal D continuous
   C. Both D components continuous
   D. Both E components continuous
   Answer: B. With no free charge: E_t1 = E_t2 and ε_r1E_z1 = ε_r2E_z2. (Key point: Boundary conditions)

4. E_z2 equals:
   A. (2.5/4)(70) = 43.75 V/m
   B. (4/2.5)(70) = 112 V/m
   C. 70 V/m
   D. 0
   Answer: A. ε_r1 E_z1 = ε_r2 E_z2 gives 43.75. (Key point: E₂ normal)

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