Problems and puzzles in electric fields 9789811532962, 9789811532979

Table of contents :
Preface......Page 5
Acknowledgements......Page 6
Contents......Page 7
1.1 Purpose of This Book......Page 13
1.2 Composition of the Book......Page 14
2.2 Other Notes......Page 16
Question A1: Simple Solution for Two-Dimensional Fields?......Page 18
Question A3: Magnitude of the Image Charge......Page 19
Answer A1......Page 20
Detailed Explanation......Page 21
Answer A3......Page 22
Detailed Explanation......Page 23
Question B1: Lines of Electric Force on Rotation Axis in an Axisymmetric Configuration......Page 25
Question B2: Two Conductors at Different Potentials......Page 26
Question B3: Two Conductors at the Same Potential......Page 27
Question B5: Mutual Capacitance Between Two Conductors......Page 28
Answer B1......Page 29
Detailed Explanation: Role of Electrostatic Capacitances......Page 30
Answer B4......Page 31
Application......Page 32
Question C1: Maximum Electric Field in Two Configurations......Page 33
Question C2: Maximum Electric Field in Three Configurations......Page 34
Question C3: Sphere-to-Ground Versus Sphere-to-Sphere Configuration......Page 35
Detailed Explanation......Page 36
Application 2: Utilization Factor......Page 37
Detailed Explanation......Page 38
Application 1: Breakdown (Sparkover) Voltage of a Non-uniform Gap in Air......Page 39
Application 3: Designing Discharge Gaps......Page 40
Answer C3......Page 41
Detailed Explanation......Page 42
Question D2: Zero Field at a Conductor Wedge Tip......Page 43
Answer D1......Page 44
Answer D2......Page 45
Detailed Explanation: Numerical Field Calculation......Page 46
Comment 1......Page 47
Detailed Explanation......Page 48
Question E2: Thin Foil Standing on a Grounded Plane Under a Uniform Field......Page 49
Detailed Explanation......Page 50
Detailed Explanation......Page 51
Question F2: Uniform Field-Forming Electrodes 1: Square Electrodes......Page 53
Answer F1......Page 54
Answer F2......Page 55
Detailed Explanation......Page 56
Application......Page 57
Answer F3......Page 58
Comment: Borda profile......Page 59
Question G1: Electric Field on a Single Sheet of Charge......Page 60
Answer G2......Page 61
Detailed Explanation......Page 62
Application: Space Charge in Atmospheric Air......Page 63
Question H2: Ion Current in a Non-uniform Field......Page 64
Question H3: Current Based on the Energy Balance......Page 65
Detailed Explanation......Page 66
Answer H2......Page 67
Detailed Explanation......Page 68
3.4 Composite Dielectrics......Page 69
Question I2: The λ Function in Probe Measurement......Page 70
Question I3: Solid Supporting Dielectric 1......Page 71
Question I4: Solid Supporting Dielectric 2......Page 72
Detailed Explanation......Page 73
Application: Field Calculation Including a Floating Conductor (Ref. I1-1)......Page 74
Answer I3......Page 75
Application: Field Strength Meter......Page 76
Question J2: Sharp Wedge-Shaped Tip of a Solid Dielectric......Page 77
Answer J1......Page 78
Detailed Explanation......Page 79
Application: Significance of the Takagi Effect......Page 80
Answer J2......Page 81
Comment: Cone-Shaped Interface of a Solid Dielectric......Page 82
Detailed Explanation......Page 83
Detailed Explanation......Page 84
Comment 2: Application of an Equivalent Circuit......Page 86
Application 2: Electrorheological Fluid......Page 87
Question K2: What Is Surface Conduction?......Page 88
Question K3: Proposal for Field Calculations Including Surface Conduction......Page 89
Answer K1......Page 90
Detailed Explanation......Page 91
Detailed Explanation......Page 92
Answer K3......Page 93
Detailed Explanation......Page 94
3.5 Other Topics......Page 96
Question L2: Significant Force Caused by Balanced Forces?......Page 97
Question L3: Electric Field and Force on a Charged Spherical Dielectric Shell......Page 98
Question L5: Holding a Charge with Electrostatic Force 2: A Cage with Uniform Surface Charge......Page 99
Answer L1......Page 100
Application 1: Effect of Induced Charge in Gas Discharge......Page 101
Detailed Explanation......Page 102
Answer L3......Page 103
Answer L4......Page 104
Detailed Explanation......Page 106
Comment: Problem Setting in the Journal of the Institute of Electrical Engineers of Japan......Page 107
Question M1: Shaking Hands Under Atmospheric Electricity......Page 109
Question M5: Boundary Conditions for Calculating Induced Current......Page 110
Comment......Page 111
Answer M2......Page 112
Detailed Explanation......Page 113
Application 1: Calculation of Induced Current Inside a Human Body......Page 114
Application: Current Induced by the Magnetic Field of a DC Transmission Line......Page 115
Detailed Explanation......Page 116
4 Three Simple Questions Not Related to Electric Fields......Page 118
5.1 Governing Equations......Page 119
5.2 Uniqueness Theorem in Electrostatic Fields......Page 120
5.3 Boundary Conditions for Composite Dielectric Fields......Page 121
5.4 Basic Equations for Capacitive–Resistive (Mixed) Fields......Page 122
5.5 Boundary Conditions for Mixed Fields......Page 123
5.6 Summary of Equations in Mixed Fields......Page 125
6.1 Image Charge Method (in Answer€A3)......Page 126
6.2 Critical Comment on Answer€C2......Page 127
6.3 Variable Separation Method Applied to Question€D1......Page 130
6.5 Ellipsoidal Coordinates Applied to Question€E1......Page 131
6.6 Ellipsoidal Coordinates Applied to Question€E2......Page 132
6.7 Electric Field for a Semi-infinite Plane Electrode Above a Grounded Plane (in Answer€F2)......Page 133
6.8 Relationship Between the Surface Shape of an Electrode and the Electric Field (in Answer€F3)......Page 134
6.9 Comparison of Electric Fields Between Two-Dimensional and Axisymmetric Configurations (in Answer€F3)......Page 135
6.10 Axisymmetric Uniform Field-Forming Electrodes in Practice (in Answer€F3)......Page 136
6.11 Application of Gauss’s Theorem (in Answer G1)......Page 137
6.12 Electric Field in a Uniform Ion Flow Field (in Answer€G3)......Page 138
6.13 Estimation of Threshold Space Charge in a Non-uniform Field Condition (in Answer€G3)......Page 140
6.14 Green’s Reciprocity Theorem (in Answers H1, H3 and I2)......Page 141
6.15 Ion Current in Concentric-Sphere Configuration (in Answer€H2)......Page 142
6.17 Refraction Law for a Line of Electric Force (in Answer€J1)......Page 143
6.18 Analysis of the Field in Fig.€J1-1 by the Variable Separation Method......Page 144
6.19 Contact of Three Dielectrics with Straight Interfaces (in Answer€J1)......Page 146
6.20 Analysis of the Configuration in Fig.€J2-1 Using the Variable Separation Method......Page 147
6.21 Analysis of the Field for a Cone-Shaped Dielectric Interface (in Answer€J2)......Page 149
6.22 Analysis of the Axisymmetric Configuration in Question€J4 Using the Successive Image Charge Method......Page 150
6.23 Analysis of the Two-Dimensional Case in Question€J4 Using the Successive Image Charge Method......Page 151
6.24 Contact-Point Electric Fields (in Answer€J4)......Page 152
6.25 Hemispherical Solid Dielectric on a Grounded Plane Under a Uniform Field (in Answers€K5 and M5)......Page 153
6.26 Examples of Induced Charge and Floating Force (in Answer€L1)......Page 154
6.27 Induced Charge and Floating Force for a Hemispherical Conductor on a Grounded Plane (in Sect.€6.26)......Page 155
6.28 Force on a Charged Spherical Shell Above a Grounded Plane (in Answer€L3)......Page 156
6.29 Force or Electric Field Exerted on a Charged Particle in the Two Configurations Shown in Fig.€L4-1......Page 158
References......Page 160
Index......Page 163

Citation preview

Tadasu Takuma Takatoshi Shindo

Problems and Puzzles in Electric Fields

Problems and Puzzles in Electric Fields

Tadasu Takuma Takatoshi Shindo •

Problems and Puzzles in Electric Fields

123

Tadasu Takuma Kawasaki, Kanagawa, Japan

Takatoshi Shindo Central Research Institute of Electric Power Industry Yokosuka, Kanagawa, Japan

ISBN 978-981-15-3296-2 ISBN 978-981-15-3297-9 https://doi.org/10.1007/978-981-15-3297-9

(eBook)

© Springer Nature Singapore Pte Ltd. 2020 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Singapore Pte Ltd. The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721, Singapore

Preface

Electric fields play key roles in many practical problems and in various environmental phenomena. This book aims to help readers develop a better understanding of electrostatic fields using the form of problems and puzzles (presented as questions) and answers, instead of the tedious explanations often found in textbooks. In that sense, this book on electric fields is unique. The authors hope that with this approach, the book will attract the interest of students and engineers who study electric fields and of general readers who are interested in the subject. The book is filled with questions that have unexpected answers and with questions that are often misunderstood, or rarely completely understood. Most of these questions are original. At a glance, the questions look simple and very easy to answer; nevertheless, once readers try to solve them, they will find that the questions can be really tough nuts to crack. Lecturers could also use the questions in this book in their classes. This book will be useful not only from an academic or an educational point of view, but also for engineers working in such fields as electrical discharges and their applications, high-voltage equipment in DC and AC circuits, and electrostatic devices. This is because the book introduces many varied practical applications related to electrostatic fields. Readers need not approach the contents of the book in order from beginning to end. Rather, the authors recommend that some of the following questions are attempted first: Question B2, C2, D2, E1, L2 and L5. This will give a feeling for the scope and style of the book and reveal the interesting insights that can be gained from solving problems. Kawasaki, Japan Yokosuka, Japan

Tadasu Takuma Takatoshi Shindo

v

Acknowledgements

The authors wish to express their appreciation to Prof. Masao Kitano (Kyoto University) and the late Prof. Teruya Kouno (formerly, University of Tokyo) for their useful discussions. They also would like to thank Prof. Boonchai Techaumnat (Chulalongkorn University), Dr. Akiyoshi Tatematsu (CRIEPI, Central Research Institute of Electric Power Industry, Japan), and the late Dr. Tadashi Kawamoto (CRIEPI) who carried out some of the numerical calculations cited in the book. They are also grateful to Dr. Kenichi Yamazaki (CRIEPI) for checking a part of Subject M, and Mr. David Smallbones for his copyediting of the manuscripts.

vii

Contents

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Purpose of This Book . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Composition of the Book . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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2 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Electrostatic and Quasi-electrostatic Fields . . . . . . . . . . . . . . . . . 2.2 Other Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3 Questions and Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Basic Concepts in Electric Fields . . . . . . . . . . . . . . . . . Subject A: Basics of Electric Fields . . . . . . . . . . . . . . . Question A1: Simple Solution for Two-Dimensional Fields? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question A2: Superposition of Electric Fields . . . . . . Question A3: Magnitude of the Image Charge . . . . . Question A4: Application of the Neumann Boundary Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer A1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer A2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer A3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer A4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subject B: Lines of Electric Force and Electrostatic Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question B1: Lines of Electric Force on Rotation Axis in an Axisymmetric Configuration . . . . . . . . . . Question B2: Two Conductors at Different Potentials Question B3: Two Conductors at the Same Potential Question B4: Capacitance of an Isolated Conductor . Question B5: Mutual Capacitance Between Two Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Answer B1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer B2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer B3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer B4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer B5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Mono-dielectric Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subject C: Spherical Conductors . . . . . . . . . . . . . . . . . . . . . . Question C1: Maximum Electric Field in Two Configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question C2: Maximum Electric Field in Three Configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question C3: Sphere-to-Ground Versus Sphere-to-Sphere Configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer C1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer C2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer C3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subject D: Electric Field at a Conductor Wedge Tip . . . . . . . . Question D1: Electric Field at a Sharp Conductor Wedge Tip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question D2: Zero Field at a Conductor Wedge Tip . . . . . . Question D3: Infinitely High Field at a Conductor Pit . . . . . Answer D1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer D2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer D3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subject E: Electric Field for a Conducting Needle or a Foil on a Grounded Plane . . . . . . . . . . . . . . . . . . . . . . . Question E1: Needle on a Grounded Plane Under a Uniform Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question E2: Thin Foil Standing on a Grounded Plane Under a Uniform Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer E1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer E2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subject F: Electrode Configurations and Electric Fields . . . . . Question F1: Electric Field Decreasing Radially on a Plane . Question F2: Uniform Field-Forming Electrodes 1: Square Electrodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question F3: Uniform Field-Forming Electrodes 2: Why Is the Borda Profile Not Utilized in Practice? . . . . . . . Answer F1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer F2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer F3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3.3

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Electric Charge, Space Charge, and Drifting Charge . . . . . . . Subject G: Charges and Electric Fields . . . . . . . . . . . . . . . . Question G1: Electric Field on a Single Sheet of Charge . . Question G2: Electric Field on a Conductor Surface . . . . . Question G3: Laplace’s Equation and Poisson’s Equation . Answer G1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer G2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer G3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subject H: Electric Current Caused by Drifting Charge . . . . . Question H1: Ion Current in a Uniform Field . . . . . . . . . . Question H2: Ion Current in a Non-uniform Field . . . . . . . Question H3: Current Based on the Energy Balance . . . . . Answer H1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer H2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer H3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Composite Dielectrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subject I: Polarization Charge and Accumulated Charge on a Dielectric Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question I1: What Is Polarization Charge? . . . . . . . . . . . . Question I2: The k Function in Probe Measurement . . . . . Question I3: Solid Supporting Dielectric 1 . . . . . . . . . . . . Question I4: Solid Supporting Dielectric 2 . . . . . . . . . . . . Answer I1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer I2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer I3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer I4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subject J: Triple-Junction Effect . . . . . . . . . . . . . . . . . . . . . Question J1: Triple Junction 1: Finite Contact Angle (Takagi Effect) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question J2: Sharp Wedge-Shaped Tip of a Solid Dielectric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question J3: Sharp Dielectric Pit . . . . . . . . . . . . . . . . . . . Question J4: Triple-Junction Effect 2: Zero Contact Angle Answer J1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer J2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer J3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer J4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subject K: Electric Fields Including Conduction . . . . . . . . . . Question K1: Ohmic Conduction and Ion Flow . . . . . . . . Question K2: What Is Surface Conduction? . . . . . . . . . . . Question K3: Proposal for Field Calculations Including Surface Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3.5

Question K4: Comparison of Electric Fields for DC and AC Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question K5: Electric Fields with and Without Conduction . Answer K1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer K2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer K3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer K4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer K5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Other Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subject L: Electric Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question L1: Force on a Charge in an Electric Field . . . . . . Question L2: Significant Force Caused by Balanced Forces? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question L3: Electric Field and Force on a Charged Spherical Dielectric Shell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question L4: Holding a Charge with Electrostatic Force 1: Surrounding Point or Line Charges . . . . . . . . . . . . . . . . . . Question L5: Holding a Charge with Electrostatic Force 2: A Cage with Uniform Surface Charge . . . . . . . . . . . . . . . . Answer L1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer L2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer L3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer L4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer L5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Subject M: Induction on the Human Body . . . . . . . . . . . . . . . Question M1: Shaking Hands Under Atmospheric Electricity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question M2: Induction on a Falling Body . . . . . . . . . . . . . Question M3: Induction from a Power-Frequency High-Voltage Source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question M4: Induction from a High-Voltage DC Source . . Question M5: Boundary Conditions for Calculating Induced Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer M1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer M2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer M3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer M4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answer M5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4 Three Simple Questions Not Related to Electric Fields . . . . . . . . . . . 107 5 Fundamentals of Electrostatic and Quasi-electrostatic Fields . . . . . . 109 5.1 Governing Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 5.2 Uniqueness Theorem in Electrostatic Fields . . . . . . . . . . . . . . . . 110

Contents

5.3 5.4 5.5 5.6

xiii

Boundary Conditions for Composite Dielectric Fields . . . Basic Equations for Capacitive–Resistive (Mixed) Fields Boundary Conditions for Mixed Fields . . . . . . . . . . . . . Summary of Equations in Mixed Fields . . . . . . . . . . . . .

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111 112 113 115

6 Supplementary Explanations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Image Charge Method (in Answer A3) . . . . . . . . . . . . . . . . . 6.2 Critical Comment on Answer C2 . . . . . . . . . . . . . . . . . . . . . . 6.3 Variable Separation Method Applied to Question D1 . . . . . . . 6.4 Field Behavior Near a Sharp Conical Tip (in Sect. 6.3) . . . . . 6.5 Ellipsoidal Coordinates Applied to Question E1 . . . . . . . . . . . 6.6 Ellipsoidal Coordinates Applied to Question E2 . . . . . . . . . . . 6.7 Electric Field for a Semi-infinite Plane Electrode Above a Grounded Plane (in Answer F2) . . . . . . . . . . . . . . . . . . . . . 6.8 Relationship Between the Surface Shape of an Electrode and the Electric Field (in Answer F3) . . . . . . . . . . . . . . . . . . 6.9 Comparison of Electric Fields Between Two-Dimensional and Axisymmetric Configurations (in Answer F3) . . . . . . . . . 6.10 Axisymmetric Uniform Field-Forming Electrodes in Practice (in Answer F3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.11 Application of Gauss’s Theorem (in Answer G1) . . . . . . . . . . 6.12 Electric Field in a Uniform Ion Flow Field (in Answer G3) . . 6.13 Estimation of Threshold Space Charge in a Non-uniform Field Condition (in Answer G3) . . . . . . . . . . . . . . . . . . . . . . 6.14 Green’s Reciprocity Theorem (in Answers H1, H3 and I2) . . . 6.15 Ion Current in Concentric-Sphere Configuration (in Answer H2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.16 Electrostatic Capacitances in Fig. I2-1 . . . . . . . . . . . . . . . . . . 6.17 Refraction Law for a Line of Electric Force (in Answer J1) . . 6.18 Analysis of the Field in Fig. J1-1 by the Variable Separation Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.19 Contact of Three Dielectrics with Straight Interfaces (in Answer J1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.20 Analysis of the Configuration in Fig. J2-1 Using the Variable Separation Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.21 Analysis of the Field for a Cone-Shaped Dielectric Interface (in Answer J2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.22 Analysis of the Axisymmetric Configuration in Question J4 Using the Successive Image Charge Method . . . . . . . . . . . . . 6.23 Analysis of the Two-Dimensional Case in Question J4 Using the Successive Image Charge Method . . . . . . . . . . . . . . . . . . 6.24 Contact-Point Electric Fields (in Answer J4) . . . . . . . . . . . . . 6.25 Hemispherical Solid Dielectric on a Grounded Plane Under a Uniform Field (in Answers K5 and M5) . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . .

117 117 118 121 122 122 123

. . . .

. . . .

. . . .

. . 124 . . 125 . . 126 . . 127 . . 128 . . 129 . . 131 . . 132 . . 133 . . 134 . . 134 . . 135 . . 137 . . 138 . . 140 . . 141 . . 142 . . 143 . . 144

xiv

Contents

6.26 Examples of Induced Charge and Floating Force (in Answer L1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.27 Induced Charge and Floating Force for a Hemispherical Conductor on a Grounded Plane (in Sect. 6.26) . . . . . . . . . . . 6.28 Force on a Charged Spherical Shell Above a Grounded Plane (in Answer L3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.29 Force or Electric Field Exerted on a Charged Particle in the Two Configurations Shown in Fig. L4-1 . . . . . . . . . . .

. . 145 . . 146 . . 147 . . 149

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

Chapter 1

Introduction

1.1

Purpose of This Book

Electromagnetism is a major component of the physics education curricula of high schools, colleges, and universities. However, it is more difficult to understand than the other major component, mechanics. One reason for this difficulty is that the principal variables in electromagnetism, i.e., potentials and fields, cannot be seen directly nor experienced easily, whereas those in mechanics, such as displacement and velocity, can easily be seen and perceived. Compared to other electromagnetic fields, electrostatic fields are considered easier to understand because they are principally based on a single scalar potential. However, the study of electrostatic fields involves many concepts which are confusing or easily misunderstood. The authors have often observed that many students have an inaccurate or shallow understanding of electrostatic fields, even at the end of a lecture. One of the reasons for this poor understanding is that academic learning is built on a sophisticated system of highly abstract theories. This book is a collection of problems and puzzles (presented as Questions) intended to enable readers to develop a better understanding of electrostatic fields. Most questions in this book are not like those provided in other textbooks, which mainly consist of explanations and calculations of electric fields and capacitances, among others. Instead, the authors have attempted to compose problems and puzzles under the following categories: (1) Questions with unexpected answers, (2) Questions that are either often misunderstood or rarely completely understood by students, and (3) Questions related to practical applications. Moreover, the authors have attempted to select configurations that are as simple as possible.

© Springer Nature Singapore Pte Ltd. 2020 T. Takuma and T. Shindo, Problems and Puzzles in Electric Fields, https://doi.org/10.1007/978-981-15-3297-9_1

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2

1

Introduction

Electrostatic fields are associated with many practical phenomena and applications. Some typical examples are electrical discharges and their applications, high-voltage or high-field equipment in DC and AC circuits (such as transmission and distribution lines), and electrostatic devices (such as those used in high-field-strength emission or electron beam generation). Electric fields also play a key role in various environmental phenomena. The authors hope that this book will contribute to the development of a better understanding of such phenomena and their practical applications.

1.2

Composition of the Book

Chapter 2 supplies some general notes on the scope of the book and the idealizations used. Section 2.1 explains the electrostatic and quasi-electrostatic fields that are the core subjects of the book. In contrast, Sect. 2.2 exemplifies the simplified or idealized concepts and/or conditions relating to charge and the configurations and shapes of conductors and solid dielectrics. Chapter 3 is the main part of this book and contains the questions (the problems and puzzles referred to in the book title) and answers. The questions and answers are organized into five sections (Sects. 3.1–3.5) and thirteen subjects (Subjects A– M) as follows: Section 3.1: Basic Concepts in Electric Fields (two Subjects with nine Questions) Section 3.2: Mono-dielectric Fields (four Subjects with eleven Questions) Section 3.3: Electric Charge, Space Charge, and Drifting Charge (two Subjects with six Questions) Section 3.4: Composite Dielectrics (three Subjects with thirteen Questions) Section 3.5: Other Topics (two Subjects with ten Questions). The Questions (and Answers) are arranged sequentially across the thirteen Subjects. The titles of the Subjects and the 49 Questions (and Answers) are given in the Table of Contents. In Chap. 3, most Answers contain additional items such as Detailed Explanation, Comments, and (related) Applications. Chapter 4 (Coffee Break) is intended for the enjoyment of readers and offers three simple questions not related to electric fields but pertinent to other topics in physics or technology. Chapter 5 summarizes the fundamental characteristics of the electrostatic and quasi-electrostatic fields that this book covers; these topics are usually studied in colleges and universities. The chapter deals with basic equations, boundary conditions, and the effects of conduction, among others. The “uniqueness theorem” in electric fields is also explained. Readers who are familiar with the content of this chapter may freely skip this part.

1.2 Composition of the Book

3

Chapter 6 gives deeper or somewhat more detailed explanations than those provided in the individual answers in Chap. 3. After Chap. 6, a list of references is given in the order of the Questions and Answers described above.

Chapter 2

Notes

2.1

Electrostatic and Quasi-electrostatic Fields

An electric field is created either by an electric charge or by a magnetic field changing in time. The former corresponds to the electrostatic and quasi-electrostatic fields that are addressed in this book. The latter field that results from a changing magnetic field is beyond the scope of this book. Specifically, the word “static” implies that a field is invariant in time, i.e., it is in the DC state. The correct description of a power-frequency AC field is quasi-static because the field changes with time at a low frequency. If a field involves only dielectrics with zero conductivity and perfect conductors (as materials) and only a single type of voltage waveform (source), its field behavior at an instantaneous voltage is identical to that of a DC source. This remains true provided the applied voltage does not change very rapidly compared to the field’s traveling time. In other words, when a field changes slowly (as in power-frequency sources), the instantaneous potential and field strength are always similar throughout the region under consideration (i.e., they are merely proportional to the applied voltage at a given instant). When several voltage sources exist and/or the dielectrics possess conductivity, we conveniently resort to expressions involving complex variables (the so-called “phasor notation”). Chapter 5 explains these treatments in more detail.

2.2

Other Notes

It is to be noted that this book deals with simplified or idealized concepts and configurations that do not exist in practice. These concepts are explained simply here, although most are commonly encountered in any textbook on electromagnetism.

© Springer Nature Singapore Pte Ltd. 2020 T. Takuma and T. Shindo, Problems and Puzzles in Electric Fields, https://doi.org/10.1007/978-981-15-3297-9_2

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2

Notes

The basic concepts are summarized as follows: A. Charge (1) Uniform surface charge (2) Point charge. All real charges are discrete entities at the infinitesimally fine, microscopical level of elementary particles. Therefore, a uniform, continuous charge is a hypothetical concept. In contrast, a point charge is theoretically possible, but its infinitesimally small size would create an infinitely high field strength, resulting in discharge. This fact is not taken into consideration when addressing common electric field behaviors. B. Field distribution or electrode configuration (1) One-dimensional and two-dimensional configurations (2) Uniform field. Any field distribution or electrode configuration is three-dimensional in practice, and the field does not extend to infinity without being modified. Moreover, a field that is uniform everywhere does not exist in practice. Practical uniform fields and their corresponding electrodes are discussed in Questions F2 and F3. C. Shape or structure of a solid conductor or solid dielectric (1) (2) (3) (4)

Extremely sharp tip or edge Extremely fine or thin conductor Perfectly smooth surface Uniform and constant material properties.

Real materials are neither perfectly sharp nor perfectly smooth. They are either somewhat round or with irregularities at the microscopical level. Concerning material properties, the dielectric constant e and conductivity r are treated as constant and uniform throughout each medium in this book. In practice, however, they change to a greater or lesser extent depending on such factors as the applied electric field and temperature, among others.

Chapter 3

Questions and Answers

3.1

Basic Concepts in Electric Fields

As explained later in Chap. 5, there are two governing equations for electrostatic fields: Laplace’s equation and Poisson’s equation. However, many useful additional concepts supplement these equations. This section collects questions concerning basic concepts, lines of electric force, image charges, and electrostatic capacitance.

Subject A: Basics of Electric Fields Question A1: Simple Solution for Two-Dimensional Fields? For simplicity, we consider two-dimensional fields with coordinates x and y. Laplace’s equation for this configuration is @2U @2U þ 2 ¼0 @x2 @y

ðA1-1Þ

The following expression describes the electric potential U U ¼ ax2 þ by2 þ cxy þ dx þ ey þ f;

ðA1-2Þ

where a, b, …, f are constants. This formula satisfies Laplace’s equation if constant b is equal to −a, which suggests that any two-dimensional field can be expressed as U ¼ ax2  ay2 þ cxy þ dx þ ey þ f:

ðA1-3Þ

Is this correct? If it is incorrect, show examples that are not compatible with Eq. (A1-3) and explain the reason why it is incorrect. © Springer Nature Singapore Pte Ltd. 2020 T. Takuma and T. Shindo, Problems and Puzzles in Electric Fields, https://doi.org/10.1007/978-981-15-3297-9_3

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3

Questions and Answers

Question A2: Superposition of Electric Fields We consider a spherical conductor located in the gap between parallel-plane electrodes, as shown in Fig. A2-1. In Fig. A2-1(a), the upper plane electrode is at V and the lower plane is grounded. If the sphere has no net charge, negative and positive charges −Q and Q are induced, respectively, on the upper and lower surfaces of the sphere, as shown in Fig. A2-1(a). Neither of these surfaces is hemispherical unless the sphere is located exactly in the middle of the gap. In Fig. A2-1(b), both parallel-plane electrodes are grounded, and the sphere is charged with 2Q. Figure A2-1(c) shows a sphere with charge 2Q located in the same position between an upper electrode at V and a grounded lower plane. In short, the charge in Fig. A2-1(c) is the superposition of Fig. A2-1(a) and (b), but the superposition of the electric fields is not obvious. Is it correct that the electric field of Fig. A2-1(c) is the superposed fields of Fig. A2-1(a) and (b)?

Question A3: Magnitude of the Image Charge The image charge method is explained in Sect. 6.1. As mentioned there, when a point charge exists above a grounded plane conductor, the magnitude of its image charge is equal to the whole charge induced on the plane surface. Here, we consider a spherical conductor and a point charge. When a point charge Q1 exists outside a grounded spherical conductor, Q1 induces a charge on the surface of the sphere. The electric field outside the sphere is given by Q1 and its image charge, Q2, which is situated at a point inside the spherical conductor. Q2 is an imaginary charge which replaces the original spherical conductor for electric field calculations. When a point charge Q3 exists inside a hollow spherical conducting shell, Q3 induces a charge on the inner surface of the conducting shell. The electric field

V

V -Q Q

2Q

22Q

(a)

(b)

(c)

d

0

Fig. A2-1 Spherical conductor between parallel-plane electrodes

3.1 Basic Concepts in Electric Fields

9

inside the shell is created by Q3 and its image charge, Q4, which is situated at a point outside the conducting shell. Is it correct that the magnitude of the image charge is equal to the total amount of charge induced on the surface of the sphere or spherical shell? If not, explain the reason.

Question A4: Application of the Neumann Boundary Condition In solving electrostatic fields, we apply Laplace’s equation and boundary conditions. Textbooks say that two kinds of boundary conditions exist: the Dirichlet and the Neumann boundary conditions. In a mono-dielectric medium with only conductors present, however, only one boundary condition seems necessary, i.e., the Dirichlet boundary condition in which a constant voltage appears on the surface of each conductor. The Neumann boundary condition is @U=@n ¼ En ¼ 0;

ðA4-1Þ

where U is the electric potential and n is a normal vector. This equation means that the normal component En of an electric field is zero at the boundary. Where or in what surface is the Neumann boundary condition applied? Also explain why the Neumann boundary condition is necessary.

Answer A1 It is not correct. There are many fields not compatible with Eq. (A1-3). One typical example is the field for coaxial conducting cylinders of radius R1 and R2 (R1 < R2). When the inner cylinder has voltage V and the outer cylinder is grounded, the potential Ф and field E are expressed as U ¼ V lnðR2 =r Þ= lnðR2 =R1 Þ

and E ¼ V=fr lnðR1 =R2 Þg;

ðA1-4Þ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi where r ¼ x2 þ y2 . This field cannot be expressed by Eq. (A1-3). It also should be noted that if the potential Ф is constant, then it always satisfies Laplace’s equation. However, it is evident that a constant potential is not a general expression of an electric field. It is incorrect because satisfying Laplace’s equation is only a necessary condition, not a necessary and sufficient condition for a correct electrostatic field distribution. A correct solution must satisfy the boundary conditions in addition to Laplace’s equation.

10

3

Questions and Answers

Comment This question is not “puzzling” at all and seems quite easy to answer. However, the authors have observed that most students cannot answer this question properly. Even if they answer that it is incorrect, many cannot provide the reason. This is because Laplace’s equation has an authoritative aura as the foundation for solving electrostatic fields.

Answer A2 It is correct. The uniqueness theorem in electrostatic fields guarantees that if a solution satisfies both Laplace’s equation and the relevant boundary conditions, then it is the only correct solution. This theorem is explained in Sect. 5.2. Consequently, we have only to examine the governing equation and the boundary conditions for the three configurations. It is evident that the three electric fields in Fig. A2-1(a–c) all satisfy Laplace’s equation. Concerning the boundary conditions, the potential of the parallel-plane conductors (electrodes) in Fig. A2-1(c) is the superposition of Fig. A2-1(a) and (b). We can furthermore confirm that the potential of the central conductor is the superposition of those in Fig. A2-1(a) and (b), as explained below.

Detailed Explanation We consider the equivalent circuit with electrostatic capacitances for the central conductor with charge Q and voltage Vs as Q ¼ C1 ðVs  U Þ þ C2 ðVs  W Þ

ðA2-1Þ

where C1 and C2 are the capacitances of the central conductor, respectively, with the upper and lower plane conductors. These capacitances do not depend on the potentials. U and W are the potentials of the upper and lower conductors, respectively. From this equation, in Fig. A2-1(a), C1 ðVa  V Þ þ C2 Va ¼ 0;

ðA2-2Þ

C1 Vb þ C2 Vb ¼ 2Q;

ðA2-3Þ

in Fig. A2-1(b),

3.1 Basic Concepts in Electric Fields

11

Fig. A2-2 Equivalent circuit of capacitances for Fig. A2-1

U C1 Vs

C2

W

and in Fig. A2-1(c), C1 ðVc  V Þ þ C2 Vc ¼ 2Q:

ðA2-4Þ

In these equations, Va, Vb, and Vc refer to the potentials of the central conductor in each case. Adding Eqs. (A2-2) and (A2-3), and subtracting Eq. (A2-4) from the sum, we can readily obtain Vc ¼ Va þ Vb :

ðA2-5Þ

Furthermore, we can extend the same explanation to a more general case in which the charge of the central conductor is arbitrary, but only on the condition that the charge in Fig. A2-1(c) is the superposition of those in Fig. A2-1(b) and (c). This can be easily confirmed using the equivalent circuit shown in Fig. A2-2.

Answer A3 It is not always correct. When a point charge Q1 is outside a sphere, as in Fig. A3-1(a), its image charge Q2 is situated at a distance R2/D from the sphere center, where D is the distance of Q1 from the sphere center. The magnitude of Q2 is Q2 ¼ RQ1 =D

ðA3-1Þ

In contrast, when a point charge Q3 is inside a hollow conductor sphere (spherical shell), as shown in Fig. A3-1(b), its image charge Q4 is situated at a distance R2/S from the sphere center, where S is the distance of Q3 from the sphere center. The magnitude of Q4 is

12

3

Questions and Answers

Image charge

R

Q1

Q2

D

(a) Point charge (Q1) outside a sphere

Image charge

R

Q3

Q4

S

(b) Point charge (Q3) inside a hollow conducting sphere Fig. A3-1 Point charge and a spherical conductor

Q4 ¼ RQ3 =S

ðA3-2Þ

It is evident that Q2 is smaller than Q1, whereas Q4 is larger than Q3. The magnitude of the image charge Q2 is the same as the total induced charge on the sphere surface. However, the charge on the inner surface induced by Q3 should be equal to Q3, but its image charge Q4 is larger than Q3. The difference in the charges (Q4 − Q3) extends to infinity as lines of electric force. Answer A4 The Neumann boundary condition is applied to a line or a plane of symmetry as well as to hypothetical or artificial boundaries. Imposing the Neumann boundary condition restricts the space or domain for analysis. This point is very important in such numerical field calculation methods as the finite difference and finite element methods. Detailed Explanation In typical mono-dielectric fields with only conductors in the domain, the only true boundaries are the conductor surfaces. In such cases, the Dirichlet boundary conditions should, in principle, be sufficient.

3.1 Basic Concepts in Electric Fields

13

As a simple example, Fig. A4-1 shows an axisymmetric configuration where upper and lower spherical conductors are facing each other. When voltage V is applied to the upper sphere and −V to the lower sphere, as in Fig. A4-1(a), the middle plane (z = 0 plane) becomes Ф = 0 and, thereby, becomes the Dirichlet boundary condition. When the same voltage V is applied to both spheres, as in Fig. A4-1(b), the middle plane becomes En = 0, thereby fulfilling the requirement for the Neumann boundary condition. It is to be noted that the Neumann condition of En = 0 holds also on the z–axis, as a line of symmetry. In both Fig. A4-1(a) and (b), the region to be analyzed is reduced to one-quarter of the whole domain. Another example is the case in which the analyzed domain extends to infinity. Numerical field calculation methods such as the finite difference and finite element methods deal with only restricted domains. In these methods, hypothetical or artificial boundaries are set up at sufficiently remote locations. A simple example is shown in Fig. A4-2, where the Neumann boundary condition is imposed on a remote vertical line, together with the Dirichlet boundary condition on a sufficiently high horizontal line.

z

z

V

V Φ= 0

En = 0

r

r V

V

(a) Two spherical conductors at opposite voltages

(b) Two spherical conductors at a same voltage

Fig. A4-1 Configurations symmetrical about a middle plane

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Questions and Answers

Φ= V

z En = 0 V Φ= 0

r V

Fig. A4-2 Examples of artificial boundaries

Subject B: Lines of Electric Force and Electrostatic Capacitance Question B1: Lines of Electric Force on Rotation Axis in an Axisymmetric Configuration Two identical elliptically tipped electrodes are facing each other separated by distance 2D, as shown in Fig. B1-1. Applying the same voltage to each electrode causes an axisymmetric electric field distribution. It is easily understood that the middle plane separated by D from each electrode tip is a symmetric plane to which the Neumann boundary condition applies, i.e., where the normal component of the electric field is zero. Two lines of electric force starting from the central tips of both electrodes must collide at the center P on the middle plane. However, the electric field must be zero at P, as described above, and it follows that on the rotation axis, no line of electric force exists and, as a result, the electric field must be zero everywhere on this axis. Is this correct?

3.1 Basic Concepts in Electric Fields

15

V

D

P D

V Fig. B1-1 Two electrodes facing each other

Question B2: Two Conductors at Different Potentials For two stressed conductors, lines of electric force usually run from the conductor with the higher potential to the conductor with the lower potential. When two conductors are at the same potential, there are no lines of electric force connecting them. Lines also extend from each conductor to a grounded conductor plane (ground). Figure B2-1(a) exemplifies a case with two spherical conductors, A and B. This relationship can be expressed as an equivalent circuit of electrostatic capacitances, as shown in Fig. B2-1(b). Are there any cases in which no lines of electric force connect two conductors at different potentials?

VA>VB CAB A

B

A

B

CA0

(a) Schematic of lines of electric force Fig. B2-1 Two conductors above a grounded plane

CB0

(b) Equivalent circuit

16

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Questions and Answers

Question B3: Two Conductors at the Same Potential It is assumed here that the reader has tried Question B2 and understood Answer B2. Figure B3-1 shows two spheres, a and b, with the same radius existing in plane symmetry, i.e., at the same height above a grounded plane. If the two spheres are at the same potential, V1, there are no lines of electric force running from one sphere to the other. The electric field distribution does not change when an equipotential surface V2 is replaced with a conductor at V2, as shown in Fig. B3-2, where V1 > V2. Answer B2 explains that because the field outside the electrodes here is identical to that in Fig. B3-1, there are still no lines of electric force connecting the two conductors.

a

b

Fig. B3-1 Equipotential surfaces formed by two identical conductors at the same potential (Dotted lines indicate equipotential surfaces.)

Fig. B3-2 Configuration where no line of electric force connects two conductors at potentials V1 and V2 (V1 > V2)

V2 V1 A B

3.1 Basic Concepts in Electric Fields

17

V1 V2 0

A B

Fig. B3-3 Conductor A is at zero potential and conductor B is at potential (V1 – V2)

To clarify, this configuration is shown in Fig. B3-2 with two conductors A and B at potentials V1 and V2, respectively. The field distribution is identical to that in Fig. B3-1. Then, the potential on the two conductors is changed from V1 to 0 on A and from V2 to (V1 − V2) on B, as shown in Fig. B3-3. In this configuration, lines of electric force run from the stressed conductor B to A at the grounded state. When Figs. B3-2 and B3-3 are superposed, the two conductors A and B have the same potential V1, but lines of electric force in Fig. B3-3 should remain between A and B. How can this contradiction be explained? Question B4: Capacitance of an Isolated Conductor Electrostatic capacitance C of a parallel-plate capacitor (condenser) is given by C ¼ eS=D

ðB4-1Þ

where S and D are the area and the separation of the electrode plates, respectively, and e is the dielectric constant of the medium between the electrodes. If the medium is air, e is almost equal to e0. This formula shows that when distance D becomes sufficiently large, the electrostatic capacitance C is zero. Does this mean that the electrostatic capacitance of an isolated conductor is zero? If not, explain the reason. Question B5: Mutual Capacitance Between Two Conductors We consider two configurations, X and Y, as shown in Fig. B5-1. In Configuration X, two spherical conductors, a and b, are located above a grounded conductor plane. In Configuration Y, the two upper conductors (designated A and B) are the same as in Configuration X. Configuration Y has no grounded plane, but in addition to the upper conductors, two identical spherical conductors C and D are present at their respective image locations.

18

3

a

b

Configuration X

Questions and Answers

A

B

C

D

Configuration Y

Fig. B5-1 Two configurations with multiple conductors

From the equivalent circuit of electrostatic capacitances, Qa ¼ Cao Va þ Cab ðVa  Vb Þ

ðB5-1Þ

in Configuration X, where Qa is the charge on conductor a, Va and Vb are the potentials (voltages) of the two conductors, and Cao is the capacitance of conductor a to the ground. Cab is the mutual capacitance between a and b, which determines the contribution to the charge of the potential difference (Va − Vb). In Configuration Y, QA ¼ CAB ðVA  VB Þ þ CAC ðVA  VC Þ þ CAD ðVA  VD Þ

ðB5-2Þ

The meanings of the charge, voltages, and capacitances are self-evident from the definitions given for Eq. (B5-1). When VA and VB are the same as, respectively, Va and Vb, and furthermore, VC and VD are the same as −Va and −Vb, respectively, the lower half of Configuration Y is the mirror image of Configuration X. In this condition, the electric field distribution in Configuration X is identical to that in the upper half of Configuration Y. Is the mutual capacitance CAB in Configuration Y the same as Cab in Configuration X?

Answer B1 It is not correct. Easy counterevidence is found in the case where the two electrodes are sufficiently separated. When distance 2D is sufficiently large, each electrode becomes isolated in space. Because lines of electric force extend from a stressed rod to infinity, the electric field is not zero on the tip of an isolated electrode.

3.1 Basic Concepts in Electric Fields

19

In Fig. B1-1, lines starting from parts of the rod surface other than the tip turn toward infinity before reaching the middle plane. As is well known, the density of lines of electric force at each point is proportional to the electric field. A line starting from a rod tip decays as it approaches P and finally disappears or seems to disappear at P where the electric field is zero.

Answer B2 A limitless number of cases in which no lines of electric force connect two conductors at different potentials can be shown. One example is illustrated in Fig. B2-2. Originally, two spherical conductors of equal size are located at the same height above a grounded plane. The smaller circles in the figure indicate the original conductors at voltage Va. As their potential is the same, there are no lines of electric force connecting the two conductors. An equipotential surface (which is not perfectly spherical) at Vb is replaced by a conductor with the same potential, Vb. This conductor is represented by a solid line on the right side of Fig. B2-2. The field distribution does not change when one equipotential surface is replaced with a conductor at the same potential. Therefore, the field distribution is identical to the original distribution for the two identical spheres, and, consequently, there are still no lines of electric force connecting the sphere at Va and the newly introduced conductor at Vb.

Detailed Explanation: Role of Electrostatic Capacitances As explained in Question B2, for two conductors, A and B, their electrical relationship is given by an equivalent circuit of electrostatic capacitances, as shown in Fig. B2-1(b). The potential V and charge Q on each conductor are expressed as

Fig. B2-2 Equipotential surfaces and lines of electric force

y

Va

Vb

S

x

20

3

QA ¼ CA0 VA þ CAB ðVA  VB Þ QB ¼ CAB ðVB  VA Þ þ CB0 VB

Questions and Answers

ðB2-1Þ

where CA0 and CB0 are capacitances between each conductor and the grounded plane, and CAB is the mutual capacitance between the two conductors. When VA is not equal to VB, the two conductors always have charges dependent on mutual capacitance CAB, i.e., CAB(VA – VB) and CAB(VB – VA), respectively. The potential difference is normally obtained by integrating electric fields along a line of electric force between two conductors. There is a tendency to conclude that lines of electric force should always exist between the two conductors at different potentials. However, this conclusion is not correct, as explained in Answer B2. Equation (B2-1) is of course correct, and the charge CAB(VA–VB) expresses its dependence on the potential difference between the two conductors, but this has nothing to do with lines of electric force.

Answer B3 The reason for the apparent contradiction is that we cannot superpose lines of electric force.

Detailed Explanation In the configuration of Fig. B3-3, lines of electric force start from conductor B and reach conductor A. Superposing Figs. B3-2 and B3-3 produces the condition in which both A and B are at the same potential, V1. Under this condition, any lines of electric force starting from A should not reach B at the same potential. In conclusion, we can superpose electric fields but we cannot superpose lines of electric force.

Answer B4 It is wrong. Some conditions are necessary in order for Eq. (B4-1) to hold true. In short, the electric field must be uniform ubiquitously between the electrode plates. This means that edge effects must be negligibly small and all the lines of electric force must be confined between the plates. If partial lines of electric force emerge from the plates, Eq. (B4-1) is not valid. Any charged isolated conductor has lines of electric force extending to infinity. As a result, the conductor bears electrostatic capacitance C to infinity and it has voltage V = Q/C with Q representing its charge. Values for electrostatic capacitances of some isolated conductors are given in the Application below.

3.1 Basic Concepts in Electric Fields

21

Fig. B4-1 Division of a square to calculate its capacitance

A B C C B A

B D E E D B

C E F F E C

C E F F E C

B D E E D B

A B C C B A

Comment No analytical method is available to compute the electrostatic capacitance of an isolated square conductor; however, the capacitance can be numerically obtained by dividing the square into a large number of subareas. J. C. Maxwell was the first to apply the surface charge method to field calculation. He computed the capacitance of an isolated square conductor to confirm measurements obtained by H. Cavendish. Figure B4-1 shows the 36 subareas (small squares) that he used. As shown in this figure, the number of unknowns is six. He solved by hand six simultaneous linear equations resulting from this division because no computers were available at the time. Application As mentioned in Answer B4, any isolated conductor has electrostatic capacitance. Some examples of conductors in air or in a vacuum are as follows. (a) Spherical conductor of radius R The electrostatic capacitance C can be obtained by solving Laplace’s equation for polar coordinates C ¼ 4pe0 R

ðB4-2Þ

For a sphere diameter of 1 m (R = 0.5 m), C is 55.6 pF. (b) Disc conductor of radius R The capacitance can be obtained from the ellipsoidal coordinates C ¼ 8e0 R

ðB4-3Þ

For a disc diameter of 1 m (R = 0.5 m), C is 35.4 pF. (c) Square conductor of side length L As mentioned above, no analytical method is available. The capacitance can be obtained only by using a numerical method, as also described in the comments for Answer B4. For a side length of 1 m, C is 40.8 pF (Ref. B4-1).

22

3

Questions and Answers

Answer B5 CAB is not the same as Cab.

Detailed Explanation In Configurations X and Y, QA is equal to Qa because the field distribution is identical in the two Configurations. When each voltage in Configuration X is substituted for the value in Y, the following equation is obtained from Eq. (B5-2) Qa ¼ CAB ðVa  Vb Þ þ 2CAC Va þ CAD ðVa þ Vb Þ ¼ 2ðCAC þ CAD ÞVa þ ðCAB  CAD ÞðVa  Vb Þ

ðB5-3Þ

The values of the capacitances do not depend on the potential distributions. In order that Eq. (B5-3) holds true for any combination of Va and Vb, the following relationship is determined CAB ¼ Cab þ CAD

ðB5-4Þ

This relationship shows that electrostatic capacitances may be different even when their electric fields are the same and that they do not depend uniquely on an electric field distribution.

3.2

Mono-dielectric Fields

This section deals with electric fields formed by conductors in a vacuum or in a single dielectric. The electric field in a single dielectric is the same as that in a vacuum if the dielectric constant is fixed and uniform in the dielectric space. A large variety of electric fields can be produced, depending on the configurations of the electrodes.

Subject C: Spherical Conductors Question C1: Maximum Electric Field in Two Configurations Figure C1-1(a) and (b) show two configurations of two spherical electrodes facing each other. In Fig. C1-1(a), voltage V is applied to the upper sphere, while the lower sphere is grounded. In Fig. C1-1(b), both spheres are insulated, and the voltages have opposite polarities. V/2 is applied to the upper sphere and –V/2 to the lower sphere.

3.2 Mono-dielectric Fields

23

Fig. C1-1 Sphere-to-sphere gaps for different applied voltages

V

V/2

d

d 0

–V/2

(a)

(b)

When the gap length (electrode separation) d is the same for both configurations, the average field strength is the same, i.e., V/d in both Fig. C1-1(a) and (b). But the electric field distribution is not the same. The maximum electric field, Em, appears on the lowest point of the upper sphere. Are the maximum values Em the same in the two configurations? If not, which is higher?

Question C2: Maximum Electric Field in Three Configurations Figure C2-1 shows three electrode configurations for a stressed sphere facing a grounded electrode: (a) sphere-to-sphere, (b) sphere-to-ground (grounded conductor plane), and (c) concentric spheres. If the applied voltage V and the gap length (electrode separation) d are the same, then the same average field (V/d) is present in all three configurations.

r

r r

r

(a)

(b)

Fig. C2-1 Gap configurations with spherical conductors of gap length d

(c)

24

3

Questions and Answers

The maximum electric field Em appears on the lowest point of the upper sphere in Fig. C2-1(a) and (b), whereas the electric field is Em on the whole surface of the central sphere in Fig. C2-1(c). Which configuration has the highest Em and which has the lowest Em? (Ref. C2-1).

Question C3: Sphere-to-Ground Versus Sphere-to-Sphere Configuration Figure C3-1(a) shows a spherical conductor at voltage V facing a grounded conductor plane. As is well known, the electric field or potential Фa in this figure is identical to the value Фb in the upper half of its mirror-image configuration shown in Fig. C3-1(b). A student inferred the electric field in this configuration as follows: (1) Adding a constant potential V uniformly to potential Фb in Fig. C3-1(b) leads to Fig. C3-1(c). This potential Фc is given by Uc ¼ Ub þ V ¼ Ua þ V

ðC3-1Þ

(2) Both Фa and Фb satisfy Laplace’s equation, so Фc also satisfies the equation because V is a constant. It is evident that Фa and Фb satisfy the boundary condition on the conductor surfaces, which also holds for Фc. (3) The uniqueness theorem dictates that a potential satisfying Laplace’s equation and all the boundary conditions is the only correct solution. This leads to the conclusion that Фc in Eq. C3-1 is the correct potential for Fig. C3-1(c).

V

V

Φa

2V

Φc

Φb d

(a)

–V

0

(b)

(c)

Fig. C3-1 Configurations of a sphere-to-sphere gap

3.2 Mono-dielectric Fields

25

(4) The electric field is given by differentiating the space potential. The field in the sphere-to-ground configuration [Fig. C3-1(a)] is identical to that in a sphere-to-sphere configuration [Fig. C3-1(c)] with twice the separation and voltage applied. Is the student’s inference correct? If not, explain the reason. Answer C1 The maximum field Em is higher in Fig. C1-1(a) than in Fig. C1-1(b).

Detailed Explanation A simple way to answer this question is to consider the case where the two spheres are adequately separated. For an isolated sphere, the electric field on the sphere surface is the applied voltage divided by the sphere radius. This means that when the two spheres are adequately separated, the field on the surface of the upper sphere in Fig. C1-1(a) is twice that in Fig. C1-1(b). If we apply the principle of superposition, this question can be more easily answered. Superposing the field Ec in Fig. C1-2, where the two spheres are both at voltage V/2, to the field Eb in Fig. C1-1(b) gives the field Ea in Fig. C1-1(a). Eb þ Ec ¼ Ea :

ðC1-1Þ

These fields are vector values, but the direction of the field on the upper sphere is always perpendicular to the sphere surface. Therefore, the electric field in Fig. C1-1 (a) is simply the sum of those in Figs. C1-1(b) and C1-2. The lines of electric force run from the surface of the spheres to infinity in Fig. C1-2, and this value is added to the field in Fig. C1-1(a). The actual values of Em in the two configurations are as follows. The electric field depends only on the ratio of the gap length d to the sphere radius r. When d = r, Em is 1.52 V/d in Fig. C1-1(a) and 1.36 V/d in Fig. C1-1(b). When d = 2r, Em is 2.34 V/d in Fig. C1-1(a) and 1.77 V/d in Fig. C1-1(b). Fig. C1-2 Sphere-to-sphere gap with both spheres at the same voltage

26

3

Questions and Answers

When d approaches zero, Em approaches V/d in both configurations. When d becomes infinitely large, Em approaches V/R in Fig. C1-1(a) and V/(2R) in (b), as explained above.

Comment Students often make the following mistakes. One is a preoccupation with the idea that identical electrode configurations should have identical field distributions for the same potential difference. This comes from the concept that configurations of the electric field constitute equivalent circuits of capacitances. But, of course, this is not correct. Some posit a grounded conductor plane between the two spheres in Fig. C1-1(b), thereby replacing the configuration with that of a sphere facing the ground. Then, they consider the field by doubling the gap length and increasing the voltage V/2 to V. This procedure is not wrong, but to keep the field distribution the same as in Fig. C1-1(b), the sphere radius also must be doubled. If this is done, however, the field distribution cannot be compared with that of Fig. C1-1(a). As a result, neither is this argument meaningful nor the comparison of the field successful using this approach.

Application 1: Sphere Gap An electrode configuration consisting of two spheres as given in Fig. C1-1 is called a sphere gap. This configuration is important in high-voltage engineering because it is used to measure high voltages based on the breakdown (sparkover) voltage of a gap. Concerning the so-called “standard sphere-gaps” utilized in practice for voltage measurement, such items as the evenness of sphere surfaces, the thickness of supporting rods, and the distance from neighboring objects are minutely regulated in order to suppress fluctuations in the breakdown voltage. In most textbooks of high-voltage engineering, the breakdown voltage is given in a table as a function of the sphere radius and gap length.

Application 2: Utilization Factor The ratio of the average field strength to Em is called the field utilization factor. The field utilization factor f is given by f ¼ V=ðdEm Þ

ðC1-2Þ

where V is the applied voltage and d is the gap length (electrode separation).

3.2 Mono-dielectric Fields

27

This factor lies between 0 and 1 and indicates the field uniformity of a given configuration. When d is the same, the higher factor f means a more uniform configuration and usually gives the higher discharge inception voltage. When discharge occurs at a certain electric field Ed in a medium, the discharge inception voltage Vd for a certain configuration is given by Vd ¼ fdEd :

ðC1-3Þ

Answer C2 The highest Em is in Fig. C2-1(c) and the lowest is in Fig. C2-1(a). That is, the order of Em is Fig. C2-1(a) < (b) < (c). It should also be noted that Em approaches V/d in all three configurations when d decreases to zero, whereas it approaches V/r when d increases to infinity. Therefore, Em differs only in the intermediate range of d/r.

Detailed Explanation One smart way of answering this question is to consider the field at the opposing point of the gap, namely, on the opposite end of a line of electric force. The electrode surface at this opposing point is convex in Fig. C2-1(a), planar in (b), and concave in (c). We can infer from these differences of surface shape that the order of the field strength there is Fig. C2-1(c) < (b) < (a). Because the integral of the electric field along a line of electric force is a constant and equal to V for all fields, we can expect that the order of Em will be reversed. The electric field depends only on d/r in all three configurations. Of these, only the field in Fig. C2-1(c) can be described by an explicit formula as Em ¼ ð1 þ d=rÞ  V=d:

ðC2-1Þ

The concrete values of the maximum fields are given in Table C2-1 for d = r and d = 2r, respectively.

Table C2-1 Maximum electric field Em

Figure C2-1(a) Figure C2-1(b) Figure C2-1(c)

d=r

d = 2r

1.52 V/d 1.77 V/d 2.00 V/d

2.34 V/d 2.68 V/d 3.00 V/d

28

3

Questions and Answers

This question has a basic character, but has various applications, as explained below.

Comment According to our experiences of teaching and lecturing, most students and electrical engineers answer that the order of Em is Fig. C2-1(c) < (b) < (a), which is exactly the opposite of the correct order. Their inference is often as follows: in Fig. C2-1(c), lines of electric force run uniformly from everywhere on the sphere’s surface, which seems to give the least concentrated distribution. Similarly, the field looks more uniform, or less concentrated, on the lowest point of the sphere in Fig. C2-1(b) than in (a). A critical comment on this answer was once published in the Journal of the IEE (Institute of Electrical Engineers), Japan. This comment and the counterargument are summarized in Sect. 6.2 because they are considered useful for understanding the electric field behavior in Fig. C2-1.

Application 1: Breakdown (Sparkover) Voltage of a Non-uniform Gap in Air The breakdown (or so-called sparkover) voltage of a non-uniform gap with a rod electrode is used during the design of insulation for equipment in atmospheric air, such as transmission lines. A rod with a square cross section is often utilized. It should be noted that the edges of a square rod produce a very high field, and usually corona discharges arise before total breakdown. This means that the electric field changes at breakdown because of the preceding discharge. Positive polarity breakdown is more critical and is considered more important in the design of insulation because the breakdown voltage is lower for positive polarity than for negative polarity for a non-uniform gap in air. Two typical non-uniform configurations are a rod-to-rod gap and a rod-to-plane (grounded conductor plane) gap, as respectively shown in Fig. C2-2(a) and (b). If the gap lengths are the same, the breakdown voltage of a rod-to-rod gap is usually higher in positive polarity than that of a rod-to-plane gap. The reason for this difference is that the electric field near the tip of the upper rod is more intensified in a rod-to-plane gap than in a rod-to-rod gap, as can be explained by comparing Fig. C2-1(a) and (b). Most high-voltage engineers consider a rod-to-plane gap to be the most non-uniform, resulting in the lowest breakdown voltage for a given gap length. However, the authors infer that a rod electrode might result in a still lower breakdown voltage than a rod-to-plane gap (for the same gap length) when located inside a container or an enclosure at zero potential, as shown in Fig. C2-2(c) and (d). Although it has not been experimentally demonstrated, the authors believe that the lowest breakdown voltage arises in Fig. C2-2(d), given identical gap lengths.

3.2 Mono-dielectric Fields

29

d

(a)

(b)

(c)

(d)

Fig. C2-2 Rod-gap configurations with the same gap length, d

Furthermore, it should be noted that even if a rod is not fully inside an enclosure, neighboring grounded objects work in a similar way to reduce the breakdown voltage by intensifying the field near the rod.

Application 2: Optimal Shape Design Optimal design in insulation of high-voltage equipment, for example, usually involves changing the shape of some parts to reduce the maximum field strength as much as possible. Numerous articles have been published on such optimal shape design. One example is to change the shape of the enclosure in a coaxial configuration while the shape of the inner stressed conductor is fixed as a hemi-spherically capped rod. As shown in Fig. C2-3(a), the shape of the enclosure is adjusted between two fixed points C and D so as to reduce the field strength on the inner conductor as much as possible. Figure C2-3(b) shows the optimized shape. It should be noted that Fig. C2-3(b) is somewhat more magnified than Fig. C2-3(a) to show the result more clearly. The optimized shape bulges conically at point D to increase the field there on the enclosure surface, thereby reducing the maximum field on the lowest point of the conductor’s surface (Ref. C2-2).

Application 3: Designing Discharge Gaps As described in Application 1 to Answer C1, a sphere gap discharges (leads to breakdown) at a specific voltage, the value of which is given in a table for different gap lengths and sphere radii. An electrical engineer used this table to design discharge gaps to protect a bank of high-voltage condensers from possible overvoltage. To reduce the vertical size, he used a sphere-to-plane gap instead of a

30

3

Questions and Answers

C, D: fixed Optimized between C and D

(Unit: cm) (a) Original shape

(b) Optimized shape [enlarged compared to part (a)] (Broken line: Original Solid line: Optimized)

Fig. C2-3 Optimal design for enclosure shape when the central conductor’s shape is fixed

sphere-to-sphere gap, as schematically shown in Fig. C2-4(b). Because the breakdown voltage of a sphere-to-plane gap is not given, he made the gap length shorter than the value for the corresponding sphere-to-sphere gap in the table. He believed that a sphere-to-plane gap would result in a more uniform field with a higher breakdown voltage than a sphere-to-sphere gap. Needless to say, this was a mistake. He should have made the gap length longer than the value in the table.

Answer C3 It is wrong. A simple rebuttal is the case where the distance d is sufficiently large in Fig. C3-1(a) or the two spheres are sufficiently separated in Fig. C3-1(c). The electric field strength on the upper sphere becomes that of an isolated sphere, namely V/R, in Fig. C3-1(a) and 2V/R in Fig. C3-1(c). The reason why it is wrong is that the potential Фc (=Фb + V) does not satisfy the boundary condition at infinity and, therefore, is not a justifiable potential.

3.2 Mono-dielectric Fields

31 R

Rectification

(a) High-voltage DC generation circuit (R ; Resistance)

(b) Protection gap (used in an upsidedown position)

Fig. C2-4 Schematic of a DC condenser bank (condensers in series) and a protection gap

Detailed Explanation The maximum field Em appears at the lowest point of the upper sphere. Only when the gap length is very small will Em be the same as V/d both in Fig. C3-1(a) and in Fig. C3-1(c). It is essential that the configuration to be superposed on Fig. C3-1(b) is not a constant potential V, but that shown in Fig. C3-2. In this figure, the potential is not constant everywhere, and there are lines of electric force extending from the spheres to infinity. If we express the potential in this figure as Фd, their relationship is not expressed by Eq. (C3-1), but by Uc ¼ Ub þ Ud

Fig. C3-2 Sphere-to-sphere gap with both spheres at the same voltage V

ðC3-2Þ

32

3

Questions and Answers

This underlines the fact that the electric field, i.e., the derivative of potential Фc, is not identical to the derivative of Фb.

Subject D: Electric Field at a Conductor Wedge Tip Question D1: Electric Field at a Sharp Conductor Wedge Tip We consider, for simplicity, a two-dimensional wedge-shaped conductor with a wedge angle h0, as shown in Fig. D1-1. Explain why the electric field is infinitely high at wedge tip P when it is considered to be perfectly sharp. Question D2: Zero Field at a Conductor Wedge Tip We also consider a two-dimensional case in this question. Figure D2-1 shows a conductor with a right-angled wedge tip. When a voltage is applied to this conductor, the electric field usually becomes infinitely high at wedge tip P. The situation of infinitely high field holds true not only for a right-angled wedge but also for any perfectly sharp wedge. Fig. D1-1 Two-dimensional conductor with a sharp wedge tip

Fig. D2-1 Conductor wedge tip P above a grounded plane (two-dimensional configuration)

Fig. D2-2 Inappropriate answer

3.2 Mono-dielectric Fields

33

Now, are there any configurations for which the electric field is zero at the tip of a perfectly sharp wedge? Solutions in which a conductor with a sharp tip is enclosed by another conductor, such as in Fig. D2-2, are excluded.

Question D3: Infinitely High Field at a Conductor Pit In contrast to Question D2, we here consider a conductor pit. If point P of the concave pit is perfectly sharp, as shown in Fig. D3-1, the electric field is usually zero at P. Now, are there any configurations where the electric field is infinitely high at P?

Answer D1 There are two possible approaches to this question. One is to consider the direction of the electric field on the conductor surface. The direction is always perpendicular to the surface. If an electric field has a finite value at P, it must have two directions there because that is where the two sides of the conductor meet. To avoid this irrational conclusion, the field must be either zero or infinitely high at P. Another approach is to consider an explicit formula for the field behavior near P. If we apply the variable separation method to Fig. D1-2 in two-dimensional polar coordinates (r, h), the potential near P is given as U ¼ Kr n sin nh;

n ¼ p=ð2p  h0 Þ:

ðD1-1Þ

The derivation of Eq. (D1-1) is given in Sect. 6.3. The electric field is proportional to rm, where m ¼ n  1 ¼ ðh0  pÞ=ð2p  h0 Þ. When wedge angle h0 is smaller than p, making a convex tip, m is negative and the field becomes infinitely high as r decreases to zero.

Fig. D3-1 Two-dimensional conductor pit with angle h0 > p

34 Fig. D1-2 Expression of Fig. D1-1 in two-dimensional polar coordinates

3

Φ (r,θ) P

Questions and Answers

E

θ =0

θ0 E

Answer D2 Three configurations could conceivably produce zero electric field at a sharp wedge tip. (1) Utilizing symmetric planes (Ref. D2-1) We consider two parallel conductor planes; one is at voltage 2 V and the other is at zero potential. A conductor having a sharp tip is located exactly in the middle of the parallel planes, as shown in Fig. D2-3. When voltage V is applied to the middle conductor, the middle plane (dotted line in the figure) and the conductor surface make an equipotential surface at voltage V. Because the field does not change when an equipotential surface is replaced with a conductor at the same voltage, P can be considered the tip of a sharp concave pit. As a result, the field is zero at P, as explained in Sect. 6.3 for the preceding Answer D1. (2) Applying the potential of another conductor This answer was suggested by Prof. Masao Kitano (Kyoto University). In addition to a conductor A with a sharp wedge and a grounded conductor plane, a third conductor B is introduced, as shown in Fig. D2-4. Conductor B is a plane in this figure. The point is that when a conductor with a sharp wedge is grounded, an infinitely high field at the edge is caused by an external field acting or entering there.

Fig. D2-3 Realization of zero electric field at point P

2V

V P

3.2 Mono-dielectric Fields

VB

35

VB

VB

VA

VA

VA

(a)

(b)

(c)

Fig. D2-4 Utilization of a third conductor at voltage VB

In Fig. D2-4(a), conductor A has voltage VA and conductor B is grounded. In Fig. D2-4(b), A is grounded and B has voltage VB (higher than zero). The field distribution of Fig. D2-4(c) is obtained by superposing Fig. D2-4(a) and (b). If VA is of positive polarity in Fig. D2-4(a), the field at the wedge tip P is positive infinity. That is, lines of electric force run out from the conductor A in Fig. D2-4(a). In Fig. D2-4(b), on the other hand, the field at P is negative infinity, i.e., lines flow into A. The latter situation of negative infinity occurs also in Fig. D2-4(c) when VB is much higher than VA. It follows that the direction of the electric field is reversed in Fig. D2-4(a) and (c), although the field is usually infinitely high in both cases at wedge tip P. Thus, we can make the field at P zero by increasing VB to a certain appropriate voltage in Fig. D2-4(c). Above-mentioned method (1) is difficult to apply if the shape of the conductor with the sharp tip is not symmetric, whereas in method (2), the shape of a conductor does not need to be symmetric. (3) Changing the orientation of the conductor tip In Fig. D2-5, a conductor with a sharp wedge tip is located under a plane electrode at positive voltage V. When the potential of the conductor is lower than V, for example, V/2, the field at P is negative infinity in Fig. D2-5(a), where wedge tip P is directed to the upper positive electrode. In contrast, the field at P is positive infinity in Fig. D2-5(b), where the conductor tip is turned to face the lower electrode. This suggests that if the conductor is rotated from the orientation in Fig. D2-5(a) to that in (b), the field at P becomes zero at some point. In this method (3) also, the shape of the conductor with the sharp wedge tip does not need to be symmetric.

Detailed Explanation: Numerical Field Calculation The field behavior was studied for the above method (2) by numerical field calculation, which was performed by the late Dr. Tadashi Kawamoto (CRIEPI). Figure D2-6(a) illustrates a sharp-tipped conductor located between two parallel-plane electrodes. The angle at the tip is p/4, or 45°. Two orientations were

36

3

(a) Tip points upward

Questions and Answers

(b) Tip points downward

Fig. D2-5 Conductor with a sharp wedge tip P between parallel-plane electrodes

VB

VA

(a)

(b)

Fig. D2-6 Results of numerical calculation of electric fields on the surface of a conductor with a sharp wedge tip

selected to produce a zero field at the tip. The calculations found that zero field occurs at the tip for VA = 0.868363VB for h = 37.5° and for VA = 0.768162VB for h = 67.5°. Figure D2-6(b) shows the calculated field strengths along both sides of the conductor surface LA and LB, and these decrease close to wedge tip P. Comment 1 It is to be noted that a strictly defined condition must be satisfied in all the above-mentioned methods (1), (2), and (3). In method (1), the conductor A with a sharp tip must be positioned in perfect symmetry. In contrast, in methods (2) and (3),

3.2 Mono-dielectric Fields

37

either the voltage of conductor B or its orientation must be exactly correct. This suggests that an experimental confirmation would be very difficult. An additional comment is that making the voltage VB equal to VA in Fig. D2-4(c) does not lead to zero field at P because there are lines of electric force running from P to the lower ground. If all three conductors are at a same voltage, the field is zero not only at P but everywhere. Such cases are outside the scope of this question, as mentioned at the end of Question D2.

Comment 2 The reason why the electric field can become zero at a sharp tip is briefly discussed here. As explained in the preceding Question D1, the principal term of the potential is Anrnsinnh near wedge tip P. The field behaves as rm where m = n –1 = (h0 − p)/ (2p − h0). When wedge angle h0 is smaller than p, making a convex tip, m is negative and the field becomes infinitely high at P in ordinary configurations. When the above-mentioned conditions are strictly satisfied in Figs. D2-4, D2-5, and D2-6, the term rm corresponding to the smallest positive n vanishes and a term with positive m is predominant.

Answer D3 It is impossible to realize an infinitely high field at a conductor pit.

Detailed Explanation In Question D2, three methods are explained to realize zero field at the sharp tip of a conductor. In method (2), another conductor produces an electric field (although infinitely high) in the opposite direction by applying different voltages. However, it is evident that any outside conductor cannot cause infinitely high fields at a conductor pit. Thus, method (2) cannot be applied to Question D3. This situation is evidently the same also for method (3). Method (1), which utilizes symmetry, is now discussed. We consider the angle at a conductor tip, as shown in Fig. D3-2. For a sharp tip, h0 is smaller than p, whereas for a sharp concave pit h0 is larger than p. In the former case of a sharp tip, the angle a (=p − h0/2) is larger than p/2. Therefore, we can realize an equipotential surface forming a concave pit with angle a (=p − h0/2) as shown in Fig. D3-2. In the latter case of a sharp pit where h0 is larger than p, we need to realize an equipotential surface with an angle a larger than p/2 in the space that causes an infinitely high field, as in Fig. D3-3. This is evidently impossible since the only remaining angle is (2p − h0), which is smaller than p.

38

3

Questions and Answers

Conductor

P

θ0 

Equipotential surface Fig. D3-2 A sharp conductor tip and an equipotential surface

Conductor P

θ0

α

Equipotential surface

Fig. D3-3 The case of a sharp conductor pit

Subject E: Electric Field for a Conducting Needle or a Foil on a Grounded Plane Question E1: Needle on a Grounded Plane Under a Uniform Field Figure E1-1 shows a slender rod (or needle) conductor standing on a grounded conductor plane under a uniform electric field E0. The field is expressed in axisymmetric coordinates (r, z). The needle has height h and radius a. It is self-evident that the potential of the rod surface is zero for any value of h and a. What is the electric field distribution when radius a becomes infinitesimally small, i.e., for an infinitesimally thin needle, assuming that no discharge occurs?

Question E2: Thin Foil Standing on a Grounded Plane Under a Uniform Field We now consider the two-dimensional case of Fig. E1-1 of Question E1: a flat plate conductor of thickness 2a standing on a grounded plane under a uniform electric field. For this question, (x, y) coordinates are used for the field. How is the electric field distribution for an infinitesimally thin foil or when thickness 2a becomes infinitesimally small?

3.2 Mono-dielectric Fields Fig. E1-1 A rod conductor on a grounded plane under a uniform field

39

z or y E0

h r or x

2a

Answer E1 The resulting field is ubiquitously uniform except on the needle surface. Figure E1-2 depicts the equipotential surfaces in (r, z) coordinates. At the tip of the needle where r = 0 and z = h, the potential suddenly changes from 0 to hE0 and the field from infinity to E0.

Detailed Explanation When the radius a of a needle is not zero, i.e., when it has finite thickness, the maximum field strength occurs at the needle tip. As a decreases, the maximum field strength increases and the area of the high field concurrently diminishes. At its Fig. E1-2 Equipotential surfaces when an infinitesimally thin rod conductor is on a grounded plane

z E0 Φ= 2.5hE0 2hE0 1.5hE0 hE0 0.5hE0 Φ =0

h r

40

3

Fig. E1-3 A spheroid in a uniform electric field

Questions and Answers

E0

h Equipotential surface

limiting state of a = 0, the field reaches infinity at the tip, but the impacted area is zero. Consequently, the presence of the needle does not affect the outside field. It is theoretically possible to study the field in this question using ellipsoidal coordinates. If a long spheroid is oriented in the direction of a uniform external field, E0, as shown in Fig. E1-3, an equipotential surface can be assumed to cross the middle of the spheroid horizontally. This configuration is identical to Fig. E1-1. The outline of the calculation is explained in Sect. 6.5, and the equations for the potential and electric field on the z-axis and the maximum field strength are also given there. It is easy to derive the field distribution of Fig. E1-2 from these equations.

Application An infinitesimally thin needle is a hypothetical object that cannot be realized in practice. Consequently, it seems impossible to experimentally verify the resulting field behavior. In gas-insulated equipment, such as gas-insulated substations (GIS), the presence of fine projections or small foreign particles can be a serious problem because they may weaken the insulation of the equipment. Question E1 represents a simplified, ultimate model for the field behavior of such fine projections or foreign particles in an insulation system. Answer E2 In contrast to Answer E1, the field behavior in this question is influenced by the foil, even though the foil is infinitesimally thin. Figure E2-1 depicts the equipotential surfaces in (x, y) coordinates. Detailed Explanation The field in this question can also be obtained by applying ellipsoidal coordinates to the two-dimensional configuration shown in Fig. E2-2. The outline of the

3.2 Mono-dielectric Fields

41

Fig. E2-1 Equipotential surfaces when an infinitesimally thin foil conductor is standing on a grounded plane

y E0

h x

calculation is described in Sect. 6.6. When the thickness 2a is very small, i.e., approaching zero, the potential and the absolute value of electric field on the y axis are, respectively, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðE2-1Þ U ¼ E0 y2  h2 ; E ¼ E0 y= y2  h2 These equations can be easily derived from Eq. (6.6-3) in Sect. 6.6. The potential and electric field E (absolute value) are respectively E0y and E0 at a remote location where y is sufficiently large. If we rewrite the electric field using distance L from the top of the plate in the y direction, i.e., L = y − h, then, when L is small compared to h, we obtain E ¼ E0

pffiffiffiffiffiffiffiffi pffiffiffi h=2= L;

ðE2-2Þ

This confirms the assertion in Answer D1 that the field behaves as rm and m is −1/2 for an infinitely thin conductor foil of h0 = 0, where r corresponds to L in Eq. (E2-2).

Fig. E2-2 An ellipsoidal plate under a uniform electric field

y E0

h

a

x

42

3

Questions and Answers

Subject F: Electrode Configurations and Electric Fields Question F1: Electric Field Decreasing Radially on a Plane Figure F1-1(a) shows an electrode configuration in axisymmetric coordinates (r, z) in which a stressed doughnut-shaped ring electrode lies inside a grounded conducting cylindrical enclosure. For simplicity, the configuration is considered to be symmetrical in respect to the plane z = 0, and the connecting lead is omitted. Lines of electric force on the z = 0 plane run radially from the central ring to the cylindrical enclosure surface, as depicted in Fig. F1-1(b). It could be concluded from this that on the z = 0 plane, which is axisymmetric, the electric field decreases with r as a function of 1/r. Is this conclusion correct? If not, explain the reason.

Question F2: Uniform Field-Forming Electrodes 1: Square Electrodes A uniform field means a field for which the direction and magnitude are both constant in space. The concept of a uniform field appears theoretically in electromagnetism, in particular in electrostatics. Uniform field-forming electrodes are important practically for discharge experiments and gas lasers, among others. Perfectly uniform fields do not exist in practice because the realization of such fields requires infinitely large electrodes. In practice, a uniform field is realized only in a limited region of space. In this question, we consider a square electrode, shown in Fig. F2-1, as a design for creating a uniform field in a limited central space. The electrode is placed at height H parallel to a grounded plane. The side length of the square electrode is L. Such a set-up is used, for example, when calibrating a meter that measures electric fields in the open air under AC transmission lines.

z

r r

(a) Configuration

(b) Electric field on z = 0 plane

Fig. F1-1 A doughnut-shaped ring electrode inside a grounded cylindrical conductor

3.2 Mono-dielectric Fields

43

z

z Φ =V

L

L

y x

Φ =V H/2

H Φ =0

0

Φ =0

Fig. F2-1 Configuration with a square electrode

The electric field distribution of Fig. F2-1 depends on the ratio L/H. The electric field on the z-axis (x = y = 0) approaches V/H (the average electric field) when ratio L/H increases. Estimate, without resorting to numerical field calculation, the value of L/H to realize an electric field higher than 99% of V/H at z = 0 (on the ground plane) and at z = H/2 (the center of the gap) as shown in Fig. E2-1(b).

Question F3: Uniform Field-Forming Electrodes 2: Why Is the Borda Profile Not Utilized in Practice? As mentioned above, uniform field-forming electrodes are necessary for such purposes as discharge experiments and gas lasers. A parallel-plane square-electrode as explained in Question F2 is not appropriate for these applications because the edges and corners of the electrodes produce a much higher field than in the central part, resulting in the inception of discharge at these locations. There have been various kinds of electrodes proposed in the past, almost all of which are axisymmetric with rounded edges. They are called uniform field-forming electrodes. In two-dimensional conditions, conformal mapping or conformal representation can be used to realize a constant electric field strength from the central part to the rounded edge of the electrode surface. One thus theoretically obtained profile is called the Borda electrode, or Borda profile, after its proposer. However, the Borda profile is not used in practical discharge experiments or lasers. Consider the reason.

Answer F1 It is not correct. Although the lines of electric force seem to decrease with 1/r on the z = 0 plane, its three-dimensional density must take into account the lines above and below the z = 0 plane, which do not behave as 1/r.

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3

Questions and Answers

z

r

(a) Configuration with sharp protrusion

(b) Schematic of lines of electric force on the z = 0 plane.

Fig. F1-2 Configuration with a sharp-edged ring protrusion on the grounded enclosure surface

Detailed Explanation A self-evident contradictory case is shown in Fig. F1-2(a) where an axisymmetric sharp-edged ring protrusion exists on the grounded enclosure surface. Figure F1-2(b) is a rough sketch of the lines of electric force and demonstrates that the field is rather higher near the enclosure than on the ring surface. This behavior is not at all consistent with a 1/r decrease in field strength. Another contradictory example is a configuration of concentric spheres. The field behavior is also axisymmetric on the z = 0 plane, but the field does not decrease with 1/r, but rather with 1/r2. In conclusion, a field distribution related to 1/r is valid only in perfectly coaxial configurations not dependent on z.

Application A doughnut-shaped ring conductor is used to shield high-voltage components, for example, a terminal in open-air equipment. This configuration reduces the original high field there, thereby suppressing the occurrence of corona discharge. The larger the radius of the ring section, the lower the electric field on the surface. Such a design is finalized using numerical field calculations.

Answer F2 As mentioned in Question F2, the electric field of Fig. F2-1 depends only on ratio L/H, where L is the side length of the square electrode and H is the height of the electrode. Values of L/H larger than about 3.2 and 1.5, respectively, are necessary

3.2 Mono-dielectric Fields

45

to realize an electric field higher than 0.99 V/H at z = 0 (on the grounded plane) and at z = H/2 (at the center of the gap). Although a precise estimation is not the intention of this question, it should be noted that realizing a uniform field requires relatively large square electrodes.

Detailed Explanation Figure F2-1 is neither two-dimensional nor axisymmetric but reflects a threedimensional field. Therefore, a numerical field calculation is necessary to analyze the electric field correctly. However, the following approximate method is available for estimating the field behavior. Figure F2-2 shows a configuration in which a semi-infinite plane conductor extending from x = 0 to infinity is at height H parallel to a grounded conductor plane. This idealized two-dimensional field can be solved in the complex plane (x, jy) by using the Schwarz-Christoffel transformation, which is a conformal mapping method. The outline of the calculation is described in Sect. 6.7. The resulting field E on the right-hand surface of the grounded plane is, for a positive value of x,   x 1 1 1f ¼ þ ln H p f f

ðF2-1Þ

where f = E/E0 and E0 (=V/H) is the average electric field strength. Evidently, E becomes E0 for very large x. The edge effect g refers to how much E deviates from E0 at x. g ¼ ðE  E0 Þ=E0 ¼ f  1:

ðF2-2Þ

The central point in the original three-dimensional configuration of Fig. F2-1 is surrounded by four sides of the square. Therefore, the edge effect is about four times that of the two-dimensional value from Eqs. (F2-1) and (F2-2). Figure F2-3 shows the approximately obtained field at the center (x = y = z = 0) of the grounded plane and the results of a numerical three-dimensional calculation. Their agreement is considerably good (Ref. F2-1).

jy

Φ =V H

x=0

x

Fig. F2-2 A semi-infinite plane conductor above a grounded plane in complex coordinates

46

3

Questions and Answers

L Numerical results Approximate results

0

4

2

6

L/H Fig. F2-3 Electric field at the origin (x = y = z = 0) of Fig. F2-1

Application Various AC field strength meters are now used to measure electric fields under AC transmission lines. Because the ratio of the induced signal to the original electric field is complex for most meters, they are calibrated in a well-defined quasi-uniform field produced by a parallel-plate configuration such as that shown in Fig. F2-1. Two kinds of calibration configurations have been used: a vertically symmetric configuration (configuration A) and a non-symmetric configuration (configuration B). Configuration A: Fig. F2-4(a). This set-up consists of two parallel, equal-sized, square conductor plates with a separation of 2H. The two plates are stressed to a polarity-reversed AC voltage V of the same amplitude by a center-tapped transformer. A field meter is calibrated at the central position P, i.e., at x = y = z = 0. The dimensions L = 1.5 m and H = 0.375 m are used in practice (Ref. F2-2). Configuration B: Fig. F2-4(b). This set-up consists of an upper parallel square conductor plate at height H above a sufficiently large grounded conductor plate on the floor plane. Only the upper plate is stressed to AC voltage V. A field meter is calibrated at the center of the mid-plane P, i.e., at x = y = 0 and z = H/2. The dimensions L = 1 m and H = 1 m have been used. It has been made clear that these dimensions give an electric field significantly lower than V/H and are therefore unsuitable for calibration (Ref. F2-1). It should be noted that the field distributions are identical in A and B if L and H are the same, but the field is not the same at the calibration point P in the two configurations.

3.2 Mono-dielectric Fields

47

z

z Φ =V L

y

Φ =V

H

P

x 0

Φ = –V

Φ = –V

(a) Symmetric configuration A

z

z

Φ =V L

y

H

Φ =V P

x Φ =0

Φ =0

(b) Non-symmetric configuration B Fig. F2-4 Configurations for calibrating AC electric field meters

Answer F3 The conformal mapping method gives a two-dimensional profile of the Borda electrode. It is a true two-dimensional shape that is infinitely large; therefore, it cannot be realized in practice. If an axisymmetric electrode with the same cross section is substituted for the two-dimensional shape, the rounded edges make the electric field there higher than in the original two-dimensional profile. As a result, the axisymmetric shape does not create constant electric fields on the surface. Further details are described in the following: Section 6.8: Relationship Between the Surface Shape of an Electrode and the Electric Field, Section 6.9: Comparison of Electric Fields Between Two-Dimensional and Axisymmetric Configurations, Section 6.10: Axisymmetric Uniform Field-Forming Electrodes in Practice.

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3

Questions and Answers

Comment: Borda profile As already mentioned, this profile was derived by J. Ch. Borda by applying the conformal mapping method, not in electrical engineering but in fluid mechanics. He conceived of the profile of a jetting hole to realize a constant fluid velocity. There are two kinds of Borda-profiles: p/2-Borda and p-Borda, as shown in Fig. F3-1. The names give the range of angles for which the electric field remains constant. Each range is shown in the figures with arrow marks. These two profiles are expressed in explicit formulae with a medium parameter u as: p/2-Borda: x ¼ 2fsin u  ðln tanðu=2 þ p=4ÞÞg;

y ¼ 2ð1  cos uÞ

ðF3-1Þ

p-Borda: x ¼ sin u  u;

y ¼ 1  cos u þ 2 ln cosðu=2Þ:

ðF3-2Þ

φ

y

y

In these formulae, u indicates the angle between the direction of the electric field and the central axis, as shown in the upper part of Fig. F3-1(a). The electric field increases significantly when a Borda profile is substituted with a cross-sectionally identical shape in axisymmetric coordinates. Table F3-1 summarizes the (maximum) electric field on the corner (edge) for W/d = 10 in axisymmetric Borda profiles, where W is the electrode width (diameter) and d is the gap length. The field is normalized with V/d where V is the applied voltage.

(a) π /2-Borda

(b) π -Borda

Fig. F3-1 Profiles of p/2-Borda and p-Borda electrodes

Table F3-1 Maximum electric field on the corner of an axisymmetric Borda profile

Profile

Normalized field

p/2-Borda 1.15 p-Borda 1.46 W/d (electrode diameter divided by gap length) = 10

3.3 Electric Charge, Space Charge, and Drifting Charge

3.3

49

Electric Charge, Space Charge, and Drifting Charge

The electrostatic fields that this book deals with are those generated by electric charge. Section 3.2 principally handles stressed conductors, but not charge itself. This section now poses questions related to charges existing either in space or on a boundary surface and questions related to electric current produced by moving charge. An important point to note is the difference between the electric field generated by charge and the field produced by stressed conductors.

Subject G: Charges and Electric Fields Question G1: Electric Field on a Single Sheet of Charge We consider a single sheet uniformly charged with charge density r, as shown in Fig. G1-1. Applying Gauss’s theorem, the electric field E on the sheet is E ¼ r=ð2e0 Þ

ðG1-1Þ

as explained in Sect. 6.11. This formula holds even if the distance to the surface decreases to zero. In contrast, the electric field of a point charge Q at a distance r from the charge behaves as Q/r2. This means that the field increases to an infinitely high value for very small values of r. Why does the electric field not increase infinitely when the distance to the surface of a single sheet of charge decreases to zero?

Fig. G1-1 Single sheet of charge with charge density r

ε0 E

S

ε0

50

3

Questions and Answers

Question G2: Electric Field on a Conductor Surface For stressed conductors, the field does not change when all the conductors are removed leaving only the surface charge. This charge corresponds to the single sheet of charge described in Question G1. On a conductor surface, the electric field on the surface of a single sheet of charge with charge density r might be considered to be r/(2e0). As explained in Sect. 6.11, however, the electric field is r/e0 on a conductor surface. Explain the reason for this difference.

Question G3: Laplace’s Equation and Poisson’s Equation When space charge exists, we must apply Poisson’s equation instead of Laplace’s equation. In gas-filled spaces and even in solid dielectrics, there always exists some space charge. In the atmosphere, for example, positive and negative ion pairs are generated and disappear incessantly to give some space charge. Estimate the density of space charge or its current which is large enough to distort the original uniform electric field in a parallel-plane electrode set-up. How would these differ in a non-uniform field?

Answer G1 If a sheet is divided into infinitesimally small areas, the charge in each area becomes correspondingly small because the charge density is constant. Further, the electric field does not increase as the distance to the surface decreases. In the case of a point charge, the size of which is considered to be zero, the field becomes infinitely large as we approach the charge because the quantity of charge is finite. However, we must also consider that, viewed microscopically, a single sheet of charge is composed of discrete point-like charges such as electrons and ions. Based on our current experience, ordinary macro-level electromagnetism emerges from the integrated and averaged actions of individual particles. In the present analysis, the microscopically small characteristics of elementary particles need not be considered. To study such minute structures and the actions of individual particles, another branch of physics, such as quantum theory, is applied.

Answer G2 The electric field at a point on an electrically charged sheet is r/(2e0) in two perpendicular directions. On a conductor surface, the field of all the other charges in the configuration is r/(2e0) in the direction from inside to outside (if the charge is positive) at the point under consideration. These charges cancel the field to zero inside and double it to r/e0 outside.

3.3 Electric Charge, Space Charge, and Drifting Charge

51

This situation is explained as follows. Because of the presence of free electrons, the electric field inside a conductor must be zero everywhere. In other words, free electrons flow so as to cancel any electric field inside. As a result, charges are distributed on the conductor surfaces to make the inside field zero. It should be noted that this charge distribution is automatically and very rapidly established corresponding to the conductor potentials of the configuration.

Answer G3 First, it is worth noting that when both positive and negative charges exist, the net amount of charge plays a decisive role on an electric field. That is to say, if the densities of positive and negative charges are the same, the field distribution is given not by Poisson’s equation but by Laplace’s equation, The point of this question is not to derive an actual value but to consider how to estimate the threshold or boundary value for space charge to have an effect. Two methods are considered. One simple method [method (1) below] is to calculate the space charge density that changes the original applied field. The other [method (2) below] is to compare the amount of space charge with the charge on the electrode surface. Analysis of an ion flow field in parallel-plane electrodes is explained in Sect. 6.12, while estimation of the threshold space charge for a coaxial configuration in a non-uniform field is described in Sect. 6.13.

Detailed Explanation Here, we postulate a steady ion flow current in a simple one-dimensional configuration of parallel-plane electrodes. The origin of the ions is not considered here. Method (1): Calculation of the ion flow current or density that is capable of distorting the original field In an ion flow field, the fundamental relationship between current density j and electric field E is j ¼ lqE

ðG3-1Þ

where l is the ion mobility and q is the charge density. For a steady state, the current density is constant everywhere in the gap because, otherwise, charge would accumulate somewhere in the gap. In contrast, q and E vary from place to place. When current flows uniformly, as shown in Fig. G3-1, the electric field can be obtained by solving Poisson’s equation in one-dimension. The derivation is given in Sect. 6.12.

52

3

Questions and Answers

Ion flow changes or distorts the electric field in the gap. As explained in Sect. 6.12, the largest change in the electric field occurs on the electrode surface. The condition in which the largest change of an original field is 10% gives the threshold of current density jq for field distortion as jq ¼ 0:2e0 lE02 =d ¼ 0:2e0 lV 2 =d 3

ðG3-2Þ

where d is the gap length, V is the applied voltage, and E0 is the original electric field in the gap, i.e., equal to V/d. The threshold qq of charge density is  qq ¼ jq ðlE0 Þ ¼ 0:2e0 E0 =d ¼ 0:2e0 V=d 2

ðG3-3Þ

Method (2): Comparison of space charge with the charge on the electrode surface The total amount of space charge on a unit area is qd, while the charge on the electrode surface is approximately e0E0 where E0 is the original electric field without space charge. These two values are compared to obtain the threshold of space charge density or current density that distorts the field. If 10% is taken as the threshold ratio for field distortion, then qd ¼ 0:1e0 E0

ðG3-4Þ

which gives an approximate value of the threshold current as jq ¼ 0:1e0 lE02 =d:

ðG3-5Þ

This value is half of that given by Eq. (G3-2), but has the same dependency on E0 and d. It is worth noting that in Ref. G3-1, the above ratio is taken not as 10% but as 100%.

Application: Space Charge in Atmospheric Air Ions in atmospheric air are generated and disappear incessantly. The net ion density is several tens of pC/m3, i.e., in the order of 10−10 C/m3 or 10−16 C/cm3 in positive polarity. This value is sufficiently small compared to the threshold of charge density described below. As a result, unless discharge is present, we can safely apply Laplace’s equation for electric field calculations in air. Next, we consider cases in which more ions than those normally present in air are generated by discharge or other sources. Ion mobility l in air is 2  10−4 m2/Vs. From Eqs. (G3-2) and (G3-3), the threshold for field distortion are

3.3 Electric Charge, Space Charge, and Drifting Charge

53

Φ=0

Fig. G3-1 One-dimensional ion flow between parallel-plane electrodes

Φ=V

j d

jq ¼ 4  109 A=cm2 ; qq ¼ 2  1012 C=cm3 for E0 ¼ 1kV=cm and d ¼ 10 cm; and jq ¼ 4  1012 A=cm2 ;

qq ¼ 2  1014 C=cm3

for E0 ¼ 100 V=cm and d ¼ 100 cm: It is noteworthy that small ion current densities in the range of pA/cm2 to nA/cm2 or a charge density of pC/cm3 or less may affect the field distribution.

Subject H: Electric Current Caused by Drifting Charge Question H1: Ion Current in a Uniform Field Figure H1-1 shows parallel-plane electrodes with electrode separation d and voltage V, forming a uniform field V/d. We assume that a positive ion and a negative ion are generated at a distance d/4 from the grounded lower electrode. The positive ion drifts to ground, and the negative ion moves to the upper electrode, both at a constant velocity proportional to the field strength V/d. Figure H1-2 depicts the possible waveform of two impulse currents occurring when the ions arrive at the grounded or the upper electrode, where T is the time required for an ion to cross the whole electrode separation distance d. Is this current waveform correct? If not, explain the reason and show the correct waveform.

Question H2: Ion Current in a Non-uniform Field In contrast to Question H1 for a uniform field, consider the current caused by ion drift in a non-uniform field.

54

3

Questions and Answers

Fig. H1-1 Drift of ions in a uniform electric field

V

d d/ 4

q

x

–q

0

Fig. H1-2 Tentative current waveform (T is the time for an ion to travel distance d)

I

0

T/4

3T/4

t

Question H3: Current Based on the Energy Balance Figure H3-1(a) shows a point charge Q moving at velocity v between parallel plane electrodes separated by d at applied voltage V. We apply the energy balance principle to derive the current in an outside circuit, i.e., we equate the instantaneous work done by the moving charge to the corresponding input energy (Ref. H3-1). The instantaneous work done is QEDx and the input energy is VIDt, where Dx is the distance Q covers during a short interval Dt. This results in the following current I ¼ QEDx=ðVDtÞ ¼ vQE=V

ðH3-1Þ

where v is the drift velocity of Q. When space charge q is present in addition to Q, as shown in Fig. H3-1(b), the field E in Eq. (H3-1) changes depending on the magnitude and position of q. As a result, the outside current differs between Fig. H3-1(a) and (b), even if the velocity v is the same. Is this appropriate?

3.3 Electric Charge, Space Charge, and Drifting Charge

55

V

V Q

Q

v

(a) Without a space charge

q

v

(b) With space charge q

Fig. H3-1 Point charge Q drifting between parallel-plane electrodes

Answer H1 The current waveform shown in Fig. H1-2 is wrong. Ion charge existing between the two electrodes induces charge on the electrode surfaces during the drift motion as the action of lines of electric force. The amount of induced charge changes because of the movement of the ions, and this induced charge is observed as a continuous current in an outside circuit.

Detailed Explanation This is fundamental knowledge for measuring discharge current and is well known to researchers and engineers working in the field of gas discharge. As far as the authors are aware, the first source that mentioned this current waveform is a 1955 book by von Engel (Ref. H1-1). He explained the case of a single ion drifting between parallel-plane electrodes. The charges Q1 and Q2, which are induced on the two electrodes by a positive ion Q located at distance x, can be obtained using Green’s reciprocity theorem, as explained in Sect. 6.14: Q1 ¼ Qx=d

and

Q2 ¼ Qðd  xÞ=d

ðH1-1Þ

When the drift velocity v of Q is proportional to the electric field with mobility l, v = lE = lV/d. Thus, current I1 is I1 ¼ dQ1 =dt ¼ Qv=d ¼ lQV =d 2 ¼ Q=T

ðH1-2Þ

The current I2 (=dQ2/dt) is the same as I1 but in the opposite direction, as expected. If the polarity of a drifting ion is reversed (i.e., a negative ion with charge −Q), the polarity of the induced charge is also reversed. But because the direction of ion drift is also reversed, this produces the same current, so long as the mobility of a negative ion is the same as that of a positive one.

56

3

Questions and Answers

In conclusion, because T is equal to d/v, the current I (absolute value) is as follows. From t = 0 (time of ion pair generation) to t = T/4, I = 2lQV/d2,=2Q/T, and from t = T/4 to t = 3T/4, I = lQV/d2 = Q/T.

Application: Current Caused by the Continuous Flow of Ion Pairs The ionization of air by cosmic rays and radioactive substances frequently occurs, producing a positive ion and an electron. The electron almost immediately forms a negative ion by attaching to a neutral molecule, thus creating an ion pair. We consider ion current in a parallel-plate electrode configuration of area S and separation d in air. When N pairs of ions are produced every second in a unit volume, dSN positive and negative ions are generated per second. As a result, dSN positive ions arrive at the negative electrode and dSN negative ions arrive at the positive electrode every second. It may be suggested that if each positive or negative ion has the same amount of charge, Q or −Q, the current is 2QdSN, i.e., equal to the total charge reaching the electrodes every second. However, the current is not equal to the charge arriving at the electrodes every second but depends on the drift time of the charge. This is explained as follows: The current caused by an ion pair is derived in Answer H1. Integrating the current leads to Q/2 from t = 0 to t = T/4 and also Q/2 from t = T/4 to t = 3T/4. The sum is not 2Q, but Q in total for one ion pair, although two ions arrive at the electrodes from opposite directions. Evidently, an ion that is generated at a position close to an electrode and arrives there almost immediately barely contributes to the current. Another explanation is as follows. A plane at distance x from the anode (positive electrode) is considered. When N ion pairs are generated in a unit volume every second, SxN positive ions are generated between 0 and x and cross plane x per second. This generates current QSxN (Q is equal to the electron charge for a single-valent ion). In contrast, S (d – x)N negative ions cross plane x in the opposite direction. The resulting total current is QSdN, which corresponds to half of all the ions generated per second.

Answer H2 The basic principle of a current induced by drifting ions is the same in a non-uniform field as in a uniform one. Ion charge induces charge on the electrode surface, the changing of which in time generates current in an outside circuit. However, a changing electric field results in a varying velocity for an ion even if its mobility is constant. Consequently, the current is large when an ion moves near an electrode with a higher electric field. For example, in such an extremely non-uniform field as that in a needle–plane configuration, the current rapidly increases when an ion approaches or leaves the needle tip.

3.3 Electric Charge, Space Charge, and Drifting Charge

57

Detailed Explanation As an example of a non-uniform field, concentric spheres are considered for which the field distribution can be given as an explicit formula. Section 6.15 explains the current waveform observed in an outside circuit for a concentric-sphere configuration. When the radius of the outer sphere is sufficiently large compared with that of inner sphere radius R, the current waveform as a function of the position of the ion from the inner sphere surface (x) is  4 I ¼ lR2 VQ ðR þ xÞ

ðH2-1Þ

where l is the ion mobility, Q is the ion charge, and V is the applied voltage. As a function of time t, when Q begins to drift from the surface of the inner sphere toward the outer sphere at time t = 0,  4=3 I ¼ lR2 VQ= 3lRVt þ R3 :

ðH2-2Þ

If the current and time are normalized using I0 and t0, respectively, the current waveform is given as I ¼ I0 =ðt=t0 þ 1Þ4=3

ðH2-3Þ

where I0 = lVQ/R2 and t0 = R2/(3 lV). The current waveform is shown in Fig. H2-1, in which the current and time are both given as normalized values.

Fig. H2-1 Ion current in a concentric-sphere configuration

58

3

Questions and Answers

The current is evidently large near x = 0 or t = 0 when Q is close to the surface of the inner sphere.

Answer H3 The charge induced by Q on each electrode surface can be obtained by applying Green’s reciprocity theorem, which is explained in Sect. 6.14. It is also described in the Detailed Explanation to Answer H1. The induced charges are −Qx/d and –Q(d − x)/d, respectively, when Q is at a distance x from one electrode surface. Differentiating the charge with time t gives the current. Because distance x changes as vt, the current is I ¼ vQ=d:

ðH3-2Þ

When no other space charge exists and the effect of the induced charge (image charge) is negligibly small, E = V/d in parallel-plane electrodes. In this case, Eq. (H3-1) coincides with Eq. (H3-2). However, Eq. (H3-1) is not correct when a space charge q exists because the presence of q modifies the electric field E so that it is no longer equal to V/d. It has been pointed out that the above-mentioned energy balance equation is not correct because it does not take into account the change in potential energy involved in the motion of Q (Ref. H3-2).

Comment When the charge velocity is given by a constant mobility l, Eq. (H3-2) gives I ¼ lEQ=d:

ðH3-3Þ

The presence of the additional space charge may change the electric field E in Eq. (H3-3), thereby influencing the charge velocity and current in an outside circuit.

3.4

Composite Dielectrics

Under an electric field, dipole charges in a dielectric move from their respective positions, a process known as polarization. This effect is expressed by vector P (the polarization vector), which is the moment of the electric dipoles per unit volume in a polarized dielectric. Composite dielectrics arise when two or more dielectrics exist. When the dielectric constant (relative permittivity) e is constant everywhere inside each dielectric, we can apply Laplace’s equation to the whole space by considering the polarization charge on their boundaries. The governing equations and the boundary

3.4 Composite Dielectrics

59

conditions involved are explained in Chap. 5. As also mentioned there, e is regarded as constant in each medium throughout this book.

Subject I: Polarization Charge and Accumulated Charge on a Dielectric Surface Question I1: What Is Polarization Charge? Figure I1-1(a) and (b) both show configurations consisting of two dielectrics with dielectric constants e1 and e2. We consider the polarization charge on the interface of the two dielectrics. How should the dielectric constants of the two media be treated? Are they still e1 and e2? Question I2: The k Function in Probe Measurement Figure I2-1 represents a probe sensor that measures the quantity of accumulated charge on a solid dielectric surface. It should be noted that the figure is only a schematic: the shape of a practical probe is usually a square rod with a guard around the sensor. When accumulated charge DQ exists on a minute surface area DS, DQ induces charge DQP on a grounded probe sensor at potential zero. Because DQP is proportional to DQ, DQP ¼ kDQ:

ðI2-1Þ

The coefficient k is also the potential of DS without charge when a unit voltage is applied on the sensor. Confirm this relationship. V

V

ε

ε

ε2 (a) Fig. I1-1 Configurations of composite dielectrics

(b)

ε

60

3

Fig. I2-1 Schematic of a probe for measuring accumulated charge on a solid dielectric surface

Questions and Answers

Probe (Sensor)

Q

S Solid

Question I3: Solid Supporting Dielectric 1 Mono-dielectric fields consisting of only one dielectric rarely exist in practice. Some form of solid dielectric (insulating material) is usually indispensable for supporting stressed conductors. When the shape of a solid surface coincides at each point with the lines of electric force that would exist without the solid, the field does not change from the original field without the solid. That is to say, both the electric field and lines of electric force keep their original distribution even if such a solid dielectric is introduced. One example is schematically shown in Fig. I3-1, where the lower solid dielectric (ed) is specially formed so that its surface coincides everywhere with the original lines of electric force existing without the solid. Explain why the field does not change when the surface of a solid coincides with the original lines of electric force.

Fig. I3-1 Configuration of a spherical electrode and a specially shaped solid dielectric

Φ

εg εd

V

3.4 Composite Dielectrics

61

Question I4: Solid Supporting Dielectric 2 Figure I4-1 shows a floating conductor A (a simplified model of an electric field meter) without any solid support. It acquires a voltage as a result of electrostatic induction by a uniform field E0, a power frequency AC field in this example. When a solid dielectric support is inserted under A, as shown in Fig. I4-2, the field becomes that of a composite dielectric. The surface of the solid is formed to coincide with the original lines of electric force in Fig. I4-1. Does the electric field in Fig. I4-2 remain unchanged or does it change from that in Fig. I4-1?

Fig. I4-1 A conductor located above a grounded plane

Fig. I4-2 A conductor supported by a solid insulator

E0

E0

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3

Questions and Answers

Answer I1 When we consider a polarization charge of density r on the boundary, the original dielectrics should both be regarded as equivalent to a vacuum, for which the dielectric constant is e0.

Detailed Explanation As explained in Sect. 5.3, the well-known boundary condition on the interface of dielectrics e1 and e2 is e1 E1n ¼ e2 E2n ;

ðI1-1Þ

where En is the normal component of the electric fields in each dielectric on their boundary or interface. However, if we consider a polarization charge of density r, as in Fig. I1-3, the boundary condition can be written as e0 E1n  e0 E2n ¼ r:

ðI1-2Þ

This means that the dielectric constant is treated as e0 in each dielectric. It is assumed here that no true charge is present at the boundary. If the dielectric constants e1 and e2 are both constant and homogeneous inside each dielectric medium, then divP = 0, where P is the polarization vector, which is the moment of the electric dipoles per unit volume in a polarized dielectric. The polarization charge does not appear inside the dielectrics but is present only on their interface. Unlike true charge, the polarization charge does not move under an electric field, but its magnitude does change depending on the field.

V

V

ε

σ

ε

ε (a)

σ

ε

(b)

Fig. I1-2 Incorrect interpretation of the situation when polarization charge with a charge density r is introduced

3.4 Composite Dielectrics

63 V

ε0 ε0 (a)

V

σ

ε0

σ

ε0

(b)

Fig. I1-3 Correct interpretation of composite dielectrics when a polarization charge is introduced

Comment A considerable number of students are confused by these facts. Their misunderstanding is probably because the concept of polarization charge is introduced and explained for a figure of two dielectrics with e1 and e2. We sometimes see figures such as Fig. I1-2(a) and (b). However, Fig. I1-3(a) and (b) are, in fact, the correct ones.

Application: Field Calculation Including a Floating Conductor (Ref. I1-1) When a configuration includes a conductor the potential of which is undefined (constant on its surface, but not given), it is called a floating potential or a floating conductor problem. An example is the field strength meter discussed in Answer I4 below. In the field calculation, the potential of a floating conductor is treated as an unknown. To compensate for this additional unknown, a supplementary condition is usually added that the net charge on the conductor is zero. In the charge simulation method, which is a method of numerical calculation for electric fields, fictitious charges inside a conductor are used to equivalently express the field outside the conductor. When a floating conductor is in contact with two dielectrics, for example, a gas and a solid, in this method, the condition that the total sum of the fictitious charges is zero produces an error. This is because the fictitious charges include the action of the polarization charge. The correct supplementary condition is that the total sum of the true charge (excluding the polarization charge) is zero. The same consideration is necessary also in the surface charge method, another numerical method. The surface charge that creates the electric fields includes the polarization charge, which must be excluded in a supplementary condition for a floating conductor.

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Answer I2 The simplest way to confirm this is to apply Green’s reciprocity theorem, as explained in Sect. 6.14. We denote three locations and two states as follows: Location 1: probe sensor; location 2: DS; and location 3: grounded conductor plane. They are, respectively, designated P, S, and G. State A: The sensor is grounded and DQ exists on DS. State B: A unit voltage is applied to the sensor and no charge exists on DS. Summarizing these conditions gives the following combinations of potentials and charges on P, S, and G: State A: (0, QPA), (USA, DQ), (0, QGA) State B: (1, QPB), (USB, 0), (0, QGB) Applying Eq. (6.14-1) in Sect. 6.14, QPA þ USB DQ ¼ 0:

ðI2-2Þ

Because QPA = DQP, k is equal to USB, the potential of DS in State B, i.e., the potential of DS without charge when a unit voltage is applied on the sensor. Another confirmation is described in Sect. 6.16.

Comment The concept of the k function was proposed by Prof. A. Pedersen and others (The Technical University of Denmark) to make clear the relationship between accumulated charge on a dielectric surface and induced charge on a probe sensor (Refs. I2-1 and I2-2).

Answer I3 When a solid dielectric is introduced, the normal component of the electric flux density, which must be continuous at the boundary, may change the original field. However, the normal component is zero on the surface of the solid because the surface at each point is formed along the original lines of electric force. The other boundary condition on the tangential component of the boundary as well as that on the potential of the conductor surfaces are evidently satisfied. Moreover, Laplace’s equation is also satisfied, as was the case for the original field without the solid. Consequently, the original lines remain unchanged when the solid dielectric is introduced.

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65

Answer I4 The field changes. In contrast to the case (Fig. I3-1) in Question I3, the potential of conductor A is not given and changes when the solid is inserted under it. Accordingly, the field distribution also changes. This is because the introduction of the solid reduces the impedance between the conductor and the grounded plane, resulting in a decrease of the conductor potential and an increase of the induced charge. Comment: Floating Conductor In Figs. I4-1 and I4-2, the potential of conductor A is constant but not given, making it a so-called floating potential or floating conductor problem. Some factors to consider in field calculations including a floating conductor are explained in the Application to Answer I1. Application: Field Strength Meter A field strength meter is utilized near the ground plane to measure the electric field under an AC transmission line. The field is considered to be almost uniform near the ground. Figure I4-3 is a schematic of such a field strength meter (called a “probe”) supported by a solid dielectric column. The dimensions given in the figure are not those of a practical probe. A field strength meter is also discussed in Question F2.

E0

z

R

Example sizes (in mm) and constant εs D = 80, 2R = 100, 2Rs = 50 h = 0~960, εs = 4 (The left figure is not to this scale.)

Probe

D

Supporting solid insulator

h

εs

Rs

r

Fig. I4-3 An electric field strength meter supported by a solid insulator under a uniform field (es is relative dielectric constant of the solid insulator)

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Subject J: Triple-Junction Effect A triple-junction or triple-joint is defined as a point where the interface of two dielectrics meets the surface of a conductor (electrode). When one of the dielectrics is a gas (or a vacuum), the triple-junction is the surface point of a solid dielectric in contact with a conductor surface. The triple-junction effect is the (often peculiar) electric field behavior near a junction. When a solid dielectric has a cross-sectionally straight surface, the triple-junction effect is called the Takagi effect. Question J1: Triple Junction 1: Finite Contact Angle (Takagi Effect) Figure J1-1 shows a cross-sectionally linear interface of two dielectrics e1 and e2 with a finite contact angle h0 on a grounded conductor plane. In this two-dimensional configuration, the electric field should be infinitely high or zero at contact point P. Can you explain why? Question J2: Sharp Wedge-Shaped Tip of a Solid Dielectric Figure J2-1 shows a wedge-shaped interface of two dielectrics (e1 and e2) with a finite angle. If a solid dielectric e1 forms the wedge-shaped surface, then, when e1 > e2, the field is usually infinitely high at tip P under an external electric field. Is there any configuration in which the field becomes zero at the tip?

Fig. J1-1 Linear interface of two dielectrics e1 and e2 lying on a grounded plane in two-dimensions

Fig. J2-1 Wedge-shaped interface of two dielectrics in two-dimensions

ε2 ε1

ε2 ε1

3.4 Composite Dielectrics

67

Question J3: Sharp Dielectric Pit We consider the case e2 > e1 in Fig. J2-1 of Question J2, which corresponds to a sharp pit in a solid dielectric. When a conductor has a sharp pit with a finite angle, it is impossible to realize an infinitely high electric field at tip P, as explained in Answer D3. Consider whether the field can be infinitely high at tip P of a solid dielectric pit.

Question J4: Triple-Junction Effect 2: Zero Contact Angle Smooth contact with a zero contact angle means that a conductor surface and a dielectric interface have a common tangent at the point of contact. It also means that the conductor surface or the interface (or both) must be curved at the contact point. One example is shown in Fig. J4-1 where a stressed spherical conductor lies on the flat surface of a solid dielectric with thickness d. In this figure, ed and eg are the dielectric constants of the solid and the gas, respectively. Consider the electric field near the contact point P in this configuration. An important consideration is whether the field near P becomes higher or lower compared to the case where the solid dielectric does not exist.

Answer J1 We consider the direction of the electric field near tip P of the boundary. The direction of the electric field is always perpendicular to the conductor surface in both dielectrics e1 and e2. However, lines of electric force starting from the plane in e1 (to the right side of P) are refracted from h1 to h2 to satisfy the following relationship: tan h1 e1 ¼ tan h2 e2

Fig. J4-1 Spherical conductor lying on the surface of a solid dielectric

ðJ1-1Þ

ε d

P

ε

68

3

Fig. J1-2 Refraction of a line of electric force at the interface of two dielectrics

Questions and Answers

ε2 ε1 θ2

θ0

θ1

P

as shown in Fig. J1-2. The derivation of this equation is explained in Sect. 6.17. From this relationship, we can determine that, if an electric field has a finite value at P, it must have two directions, depending on whether it originated in dielectric e1 or e2. This is unreasonable, and it therefore follows that the electric field must be either infinitely high or zero at P.

Detailed Explanation The above explanation is the same in principle as that of Answer D1 for a wedge-shaped conductor with a sharp tip. The field behavior in Fig. J1-1 can be analyzed by applying the variable separation method with the configuration described in polar coordinates (r, h), as shown in Fig. J1-3. Section 6.18 explains in detail the derivation of the governing equation as well as its solution relative to the ratio e2/e1 and h0. The following description is only a summary of that content. When the electric potential is expressed as rn as a function of distance r from P, the following transcendental equation can be derived:

Fig. J1-3 The configuration of Fig. J1-1 in polar coordinates

Φ (r,θ ) Φ2

(ε2)

Φ1

θ0 P

(ε1)

θ =0

3.4 Composite Dielectrics

69

e2 tan nh0 þ e1 tan nðp  h0 Þ ¼ 0

ðJ1-2Þ

Similarly, as for a wedge-shaped conductor, the smallest positive solution (nmin) of n is predominant near P. The magnitude of the electric field is proportional to rm, where m = nmin − 1. When e2 < e1, then nmin > 1 and m > 0, meaning that the electric field becomes zero as P is approached. In contrast, when e2 > e1, nmin > 1 and m < 0, which translates to an infinitely high field at P. The field behavior for the contact of three dielectrics with straight interfaces is described in Sect. 6.19.

Comment: The Takagi Effect Professor Jun-ichi Takagi (Waseda University, Japan) first analyzed this problem to find an infinitely high field near a contact point. He applied the extended conformal transformation method and also tried to obtain experimental confirmation based on the birefringence of solid dielectrics (Ref. J1-1). One of the authors (T. T.) and others analyzed the phenomenon using analytical and numerical field calculation methods (Refs. J1-2 and 3). They also named the occurrence of zero or infinitely high electric fields the Takagi effect after his pioneering work published in 1939. This field behavior was once referred to as “Einbettungseffekt” (the German for embedding effect) (Ref. J1-4). This name was based on an experimental set-up that consisted of a round rod electrode embedded obliquely in an insulating column. However, this experiment showed only an intensified field near the inside contact point and a weakened field near the outside point; it did not demonstrate an infinitely high or zero field.

Application: Significance of the Takagi Effect Solid dielectrics are inevitably used in high-voltage equipment to support stressed conductors. However, the Takagi effect is not a critical issue in outdoor air insulation. This is because pollution and humidity in the atmosphere lead to conduction on solid surfaces, thereby mitigating the high field near the contact point. However, the increased field strength at contact points can become a significant problem for the development of insulation systems with oil, compressed SF6, or vacuum in almost perfectly dry conditions. In these systems, much higher electric fields are applied, and the performance of the insulation is more sensitive to the electric field. Furthermore, more precise computation of field behavior using numerical methods has underlined the importance of the Takagi effect. In addition to high-voltage engineering, the peculiar field behavior of the Takagi effect is also important in such areas as waveguides and semiconductors.

70

3

Questions and Answers

To reduce the field strength near contact points, the adoption of appropriate contact configurations for solid supports has been proposed (Ref. J1-5). Answer J2 It should be noted that the configuration shown in Fig. J2-1 does not constitute a triple-junction because it consists of only two dielectrics without a conductor. However, its field behavior is analyzed similarly to that of Question J1. Section 6.20 explains the analysis of the configuration in Fig. J2-1 using the variable separation method. Hereafter, we rewrite e1 as ed and e2 as eg in Fig. J2-1. This question is also quite similar to Question D2, which involves a wedge-shaped conductor. However, here we consider two dielectrics in two-dimensional configurations. Two methods are applicable for realizing zero electric field at the sharp tip P of a solid dielectric. (1) Utilizing a symmetric plane We consider two parallel plane conductors with one plane at voltage V and the other at voltage −V. A solid dielectric ed with a sharp tip is located exactly in the center of the parallel planes, as shown in Fig. J2-2. Based on symmetry, the potential of the middle plane (dotted line in the figure) is zero. Because the field does not change when an equipotential surface is replaced with a conductor at the same potential, the field in Fig. J2-3 is identical to that in Fig. J2-2. The field is zero at the contact point P as a result of the Takagi effect. It should be noted that the field strength ought to be infinitely high at the other contact point, Q, if the dielectric forms a sharp tip there.

Fig. J2-2 Configuration of vertical symmetry for a solid dielectric with a sharp tip between two parallel plane conductors

Q

εd

Fig. J2-3 Modified configuration of Fig. J2-2

εd

3.4 Composite Dielectrics

71

εg

εg εd

(a) Tip points upward

εd

(b) Tip points downward

Fig. J2-4 Solid dielectric with a sharp wedge-shaped tip P between parallel-plane electrodes

(2) Changing the direction of the tip of a solid dielectric When a solid dielectric does not have a symmetric shape, above-mentioned method (1) is difficult to apply. An infinitely high field at the tip results from an external field acting there. In Fig. J2-4(a), where tip P is directed toward the upper positive electrode, the field at P is negative infinity. In contrast, the field is positive infinity in Fig. J2-4(b), where the solid dielectric is turned upside down. This suggests that if the solid dielectric is rotated from its orientation in Fig. J2-4(a) to that in (b), the field at P would become zero somewhere en route. Comment: Cone-Shaped Interface of a Solid Dielectric Questions J1 and J2 both consider two-dimensional configurations. Figure J2-5 shows an axisymmetric configuration with a cone-shaped solid dielectric (ed > eg). When ed is sufficiently large compared with eg, the field behavior becomes that of a

Fig. J2-5 Axisymmetric configuration with a solid cone-shaped interface

z

Φ ε θ0 Φ P

ε

72

3

Questions and Answers

sharp needle conductor existing in space. The analysis of the field in Fig. J2-5 is described in Sect. 6.21. Answer J3 We now consider the two-dimensional configuration shown in Fig. J2-1. The field at tip P is infinitely high when e2 > e1. This result is completely different from the field behavior near the tip of a sharp conductor pit, as explained in Answer D3. At the tip of a sharp conductor pit, the field is zero; in a solid dielectric pit the field is not usually zero but is in fact infinitely high.

Detailed Explanation Rewriting ha/2 as h0 in Eq. (6.20-4) gives the following two equations: e2 tan nh0 þ e1 tan nðp  h0 Þ ¼ 0

ðJ3-1Þ

e1 tan nh0 þ e2 tan nðp  h0 Þ ¼ 0:

ðJ3-2Þ

and

The solution of Eq. (J3-1) is explained in Sect. 6.18. The analysis shows that Eq. (J3-1) has a solution for n between 0 and 1 when e2 > e1, whereas n is larger than 1 when e2 < e1 in Fig. J2-1. In contrast, for the case of a diamond-shaped dielectric not in contact with a conductor plane, as shown in Fig. J3-1, it is demonstrated in Sect. 6.20 that the governing equation Eq. (6.20-4) has a solution for n between 0 and 1 when e2/e1 is replaced by e1/e2. This means that an infinitely high electric field occurs at tip irrespective of the ratio of e1 and e2 in usual configurations. Only in special cases, however, does the solution with n between 0 and 1 disappear, and the smallest solution of n then becomes larger than 1. Figure J3-2 shows an example of such a special case: a configuration symmetrical horizontally as well as vertically. It is evident from symmetry that the middle plane passing through P constitutes an equipotential surface, whereas a vertical plane passing through Q coincides with the direction of the electric field. In this configuration, when e2 > e1, the electric field is zero at tip Q of the dielectric pit, but is infinitely high at tip P of the dielectric pit.

Fig. J3-1 Diamond-shaped interface of two dielectrics

ε2 ε1

3.4 Composite Dielectrics

73

Fig. J3-2 Diamond-shaped interface of two dielectrics with both vertical and horizontal symmetry

ε2 ε1

Application: Electric Field When the Dielectric Constant of a Solid is Sufficiently High Textbooks on electromagnetism explain that when the dielectric constant of a solid dielectric is sufficiently large, the field outside assumes the distribution that would be formed by a conductor rather than a dielectric. But this rule does not hold for the case of Fig. J2-1. Another such case is a straight interface of e1 and e2 on a conductor plane, as shown in Fig. J1-1. If e2 > e1, the electric field is infinitely high at the contact point P, irrespective of the value of e2. If dielectric e2 is replaced with a conductor, the field is zero at P. This difference arises because a conductor forms an equipotential surface, whereas a dielectric remains nonconductive even for a sufficiently high dielectric constant. Answer J4 We denote the dielectric constants of the solid and the gas dielectrics as ed and eg, respectively. The dielectric constant of a gas is substantially equivalent to that of the vacuum. The case without a solid dielectric corresponds to ed being equal to eg. When ed increases from eg, the field strength near P increases in the gas, as expected, but it increases also in the solid dielectric in the configuration shown in Fig. J4-1.

Detailed Explanation When an external electric field is applied to two dielectrics ed and eg (ed > eg) in series, the electric field E is usually higher in eg and lower in ed than the value for a single dielectric medium. This difference of E between eg and ed is ascribed to the continuity of the normal component of dielectric flux density eEn, as given in Eq. (J4-1), eg Egn ¼ ed Edn :

ðJ4-1Þ

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3

Questions and Answers

Figure J4-2 shows a one-dimensional example of a solid dielectric and gas located in series in the direction of an electric field. The electric field Eg in the gas and Ed in the solid can easily be derived as   Eg ¼ ed V= ed dg þ eg dd

ðJ4-2Þ

  Ed ¼ eg V= ed dg þ eg dd

ðJ4-3Þ

V is the applied voltage, and dg and dd are the gap distance of the gas and the thickness of the solid dielectric, respectively. These formulae clearly show that with increasing ed the field grows in the gas and decreases in the solid. When ed is sufficiently large, Eg = V/dg and Ed = 0, which corresponds to the field when the solid dielectric is replaced with a conductor. In the case of zero-angle contact in Fig. J4-1, however, the electric field near the contact point increases not only in dielectric eg but also in ed. The field behavior for Fig. J4-1 with infinitely large thickness d has been analyzed analytically using the successive image charge method and numerically using the charge simulation method (Refs. J4-1 and 2). The former analysis, which applied the image charge method, is explained in Sect. 6.22. The electric field at the contact point P, Egc in the gas (eg side) and Edc in the solid (ed side), are respectively given by Egc ¼ es ðes þ 1ÞEc =2

ðJ4-4Þ

Edc ¼ ðes þ 1ÞEc =2

where es = ed/eg and Ec is the field strength at P when ed = eg, i.e., when the solid dielectric is not present. It should be noted that because no gas gap exists at P, Egc is the limiting value as P is approached. The above expressions clearly show that the field at P increases not only in the gas but also in the solid dielectric. The value of Edc in the solid is a linear function of es. As explained in Sect. 6.23, Eq. (J4-4) also holds for the two-dimensional case of a cylindrical conductor lying on a thick solid dielectric plane.

V

εg

dg

εd

dd

Fig. J4-2 Configuration of a solid dielectric and gas in series

3.4 Composite Dielectrics

75

When the thickness d of the solid dielectric is finite, the analytical method is difficult to apply. One of the authors (T. T.) and others analyzed the field behavior in such cases by applying numerical method (Refs. J4-1 and 2). Approximate formulae for the electric field at the contact point are   Egc ¼ es eks þ 1 Ec =2   Edc ¼ eks þ 1 Ec =2

ðJ4-5Þ

where k is a constant between 0 and 1 that depends on the configuration of the dielectrics but not on their dielectric constants. Typical values of contact-point electric fields are given in Sect. 6.24.

Comment 1: Analysis Using the Variable Separation Method As explained in Questions J1 and J2, when the interface of two dielectrics and the surface of a conductor are both straight lines in cross section, the variable separation method can be applied to analyze the electric field. In configurations with zero contact angle accompanied by a rounded interface or surface, the variable separation method is difficult to apply.

Comment 2: Application of an Equivalent Circuit The field behavior in composite dielectrics is often explained using an equivalent circuit consisting of partial capacitances. Figure J4-3 shows such an equivalent circuit of capacitances near the zero-angle contact point P. However, this concept is not applicable to the field behavior at P, because the resulting field strength would decrease in the solid dielectric (ed side) with increasing ed.

Fig. J4-3 Equivalent circuit of electrostatic capacitances near contact point P

εg

Cg

εd

P Cd

76

3

Questions and Answers

εs = 8

εs

θ (degrees) Fig. J4-4 Normalized electric field on the surface of a doughnut-shaped ring conductor supported by a solid dielectric column (es = ed/eg)

Application 1: Solid Dielectric Supports The field behavior near contact points is important in various insulation systems because the use of solid dielectric supports is inevitable. Figure J4-4 shows the normalized electric field on the surface of a doughnut-shaped ring conductor that is supported by a solid dielectric column with relative dielectric constant es. The horizontal axis is the angle on the cross-section of the conductor, as shown in the figure. Without a solid dielectric, which corresponds to es = 1, the maximum field appears at h  130°, but the electric field near the contact point increases rapidly with increasing es and surpasses the field near 130° when es is larger than approximately 6.

Application 2: Electrorheological Fluid Field enhancement or concentration close to contact points can create a strong adhering force in multiple dielectric particles; this is referred to as the pearl-chain-forming force. One application is in electrorheological fluids in the

3.4 Composite Dielectrics

77

form of micron-sized dielectric particles suspended in a host insulating liquid. Electrorheological fluids exhibit significant change in rheological properties when subjected to an electric field because the particles rapidly form chains in the direction of the applied electric field. Electrorheological fluids have the advantage of very short response times and may be used in various devices such as dampers, clutches, and actuators.

Subject K: Electric Fields Including Conduction Several types of conduction exist in solids, but Subject K deals solely with ohmic conduction. This is the most fundamental type of conduction in which the current is proportional to the applied voltage. Throughout this book, conductivity or its reciprocal, resistivity, is regarded as constant and uniform throughout a solid. When no conduction is involved in a composite dielectric configuration, the electric field for AC stress is much the same as that for DC stress, provided the AC frequency is not too high. If a solid possesses conductivity, the field varies depending on the AC frequency. A detailed explanation of the effect of conduction is given in Chap. 5. When the dielectric constant and conductivity are constant and uniform ubiquitously in a solid, Laplace’s equation holds, and the effect can be expressed by the following complex dielectric constant: e_ ¼ e þ r=ðjxÞ

ðK-1Þ

where e is the dielectric constant, r is the volume conductivity, and x is the angular frequency (equal to 2pf, where f is the AC frequency). As a result, the electric potential and field are given as complex numbers.

Question K1: Ohmic Conduction and Ion Flow Explain the difference between the ohmic currents in cases of solid resistance and the current due to ion flow in air under DC stress.

Question K2: What Is Surface Conduction? There are two kinds of conduction in a solid: volume and surface conduction. In cases of volume conduction, AC current tends not to flow uniformly but mainly flow in a region closer to the surface. This phenomenon, called “the skin effect,” is more marked for higher AC frequencies and higher conductivities. Under DC or low frequency AC stress, however, current flows uniformly inside a solid with constant volume conductivity.

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If surface conduction occurs in an infinitely thin current path, volume conductivity must be infinitely high there. Explain what surface conduction or surface conductivity is.

Question K3: Proposal for Field Calculations Including Surface Conduction The following procedure was once proposed for calculating electric fields when surface conduction is present. (1) The electric field is calculated neglecting conduction, or with the solid having infinitely high resistivity, for a given configuration. (2) Electrostatic capacitances are computed by considering a minute surface area ds using the following formula: Z C ¼ e0 En ds=U ðK3-1Þ S

where e0 is the dielectric constant in a vacuum or gas, En is the normal component of the electric field on ds, and U is the potential at the corresponding location. Figure K3-1 schematically shows the partial capacitances C12, C24, and so forth, to be computed. (3) The electric field is computed for an equivalent circuit, as shown in Fig. K3-1(b), by adding surface resistance. Consider if this procedure is appropriate or not.

C24 C34

(a) Computation of partial capacitances

(b) Equivalent circuit consisting of surface resistances R and partial capacitances C

Fig. K3-1 Proposed calculation method including surface resistance

3.4 Composite Dielectrics

79

Fig. K4-1 Spherical conductor located above a hemispherical solid dielectric with conductivity on a grounded plane

d/ 5 Φ =V orV cos

d

εg

d

P

εd

ρ

d

Question K4: Comparison of Electric Fields for DC and AC Applications Figure K4-1 shows a composite dielectric configuration where a spherical conductor is located above a hemispherical solid dielectric on a grounded conductor plane. Consider the electric field at the highest point P of the lower hemispherical solid dielectric. When the hemispherical solid dielectric in Fig. K4-1 has volume conductivity, is the field at P in the gas (eg) higher for (a) DC voltages or (b) AC voltages?

Question K5: Electric Fields with and Without Conduction In the configuration shown in Fig. K4-1, an AC voltage is applied on the upper spherical electrode. Is the field at P in the gas (eg) higher (a) for a perfect dielectric (no conduction) or (b) for a solid with some volume or surface conduction?

Answer K1 The relationship between the current and the applied electric field is totally different for ohmic conduction and ion flow. In ohmic conduction, the current is proportional to the applied voltage, or, equivalently, the current density is proportional to the electric field. For DC voltages, the electric field is given by Laplace’s equation.

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For ion flow in air, the electric field is given by Poisson’s equation, and the current has a nonlinear relation with the applied field. It is also necessary to consider the effect of air currents. Furthermore, in the case of bipolar fields (where both positive and negative ions coexist), disappearance of ions as a result of recombination must be considered.

Detailed Explanation Ohmic conduction is the most fundamental type of conduction in a solid. The resistance of a solid material sometimes changes depending on the applied electric field, but an ordinary resistor usually has a constant value R. In a solid in which Ohm’s law is operative, the current I is equal to the applied voltage difference V divided by R, or, equivalently, the current density j is equal to the conductivity j multiplied by the applied electric field E. That is I ¼ V=R

or

j ¼ jE:

ðK1-1Þ

For ion flow in air, the current density is j ¼ qv;

ðK1-2Þ

where q is the charge density and v is the drift velocity. Using mobility l, v is equal to lE. Thus, j ¼ lqE:

ðK1-3Þ

Because q also depends on the field E, the relationship between j and E becomes nonlinear. The governing equations for a monopolar ion field are as follows. Poisson’s equation: div E ¼ q=e0

ðK1-4Þ

div J ¼ divðlqEÞ ¼ 0

ðK1-5Þ

Current continuity equation:

In these equations, E and J are both vector quantities. When mobility l is constant, the relationship between q and E is as follows. grad q  E þ q2 =e0 ¼ 0:

ðK1-6Þ

In bipolar fields (where both positive and negative ions are present), simultaneous equations for both ions are necessary, and these must include the effect of recombination. The effect of air currents must also be considered.

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Application: Calculation of Ion Flow Fields Analysis of ion flow fields is important for studying the environmental problems associated with DC transmission lines and the properties of electric precipitators. In contrast to the usual Laplacian fields, not only the electric potential (or electric field) but also the amount of charge q are unknown quantities in ion flow fields. A number of numerical calculation methods for ion flow fields have been published. These calculations are accompanied by two main difficulties. First, it is not easy to set up a proper boundary condition for q. One frequently applied boundary condition is that the electric field remains as the corona inception field on the surface of ion generating electrodes. The other difficulty is that numerical computation is liable to become unstable when executing repetitive solutions of simultaneous equations.

Answer K2 Conduction always occurs in a finite thickness, in reality. However, surface conduction is the conduction in which, for convenience, the thickness is regarded as zero.

Detailed Explanation As is well known, the resistance R is expressed as R ¼ qL=S

ðK2-1Þ

where q is the volume resistivity and L and S are, respectively, the length and sectional area of the current path. For the resistance of a surface layer, as shown in Fig. K2-1, for example, the sectional area of the current path S is the thickness D multiplied by the length of the circumference, W (=2pR0), and so

Fig. K2-1 Surface resistance on a cylinder

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3

R ¼ qL=ðD  WÞ

Questions and Answers

ðK2-2Þ

In practice, neither q nor D is zero. However, when q and D are both small, we rewrite Eq. (K2-1) or Eq. (K2-2) with qs = q/D as R ¼ qs L=W:

ðK2-3Þ

Conductivity is the reciprocal of resistivity. The condition of q being small corresponds to high conductivity j. The introduction of qs or the surface conductivity js makes it easier to describe a conductive layer. Such a layer results from a surface coating of conducting material, and, in practice, it can result from pollution or humidity being adsorbed on a solid surface. The advantage of considering surface conduction is that we need not use q or D, values which often cannot be clearly defined.

Comment: Infinitely High Conductivity This question is very easy to answer. However, it should be noted that in Eq. (K2-2), surface conductivity js = jD, where j is the volume conductivity. Some students are puzzled by the fact that the volume conductivity must be infinitely high to realize finite surface conductivity when thickness D is zero.

Answer K3 The electric field has only a normal component on a conductor surface, but this property does not hold on a solid dielectric surface. The proposed method neglects the tangential field component on dielectric surfaces. As an example, we take the configuration shown in Fig. K3-2(a), in which a cylindrical post-type solid dielectric is inserted between parallel-plane electrodes. Because the side of the solid coincides everywhere with lines of electric force, there is no normal component of the electric field there. As a result, there is no electrostatic capacitance on the solid surface computed from Eq. (K3-1). In the calculation for Fig. K3-2(b), which includes surface resistance, a normal component of the electric field appears on the solid surface when the surface resistance is not constant along its side. Furthermore, as a result of non-constant surface resistance, the potential difference arises, for example, between points A and B in the figure. These effects are not taken into account in the proposed method.

3.4 Composite Dielectrics

(a) Cylindrical solid dielectric between parallel-plane electrodes

83

(b) Equivalent circuit

Fig. K3-2 Applied example of the proposed method for field calculations when surface conduction is present

Answer K4 We do not compare effective electric field strength, but rather the peak values for both DC and AC. Therefore, the applied AC voltage is V cosh, as shown in Fig. K4-1, and the applied DC voltage is V. The field strength at P in gas (eg) is higher for DC than for AC application. This is because under DC stress, the presence of conduction decreases the impedance of the solid to zero, even if it has low conductivity, thereby making the divided voltage lower in the hemispherical solid and higher in the gas.

Detailed Explanation Figure K4-2 shows an equivalent circuit for Fig. K4-1. The solid hemisphere is expressed by a parallel impedance Z consisting of a capacitive impedance ZC and a resistive impedance ZR. For low frequency AC, ZR is dominant. Under DC stress, impedance ZR becomes zero, and as a result, the potential at P becomes zero. Therefore, the voltage applied to the gaseous space becomes higher than that under AC stress. For high frequency AC, ZC plays a decisive role. The transition from the DC state to the AC state depends on the ratio of ZR/ZC. More detailed field behaviors are shown in Fig. K4-3, which depicts the numerically calculated electric field (E) at the lowest point of the upper sphere as well as at P on the gas (eg) side and on the solid (ed) side. The vertical axis E is normalized by V/d. The horizontal axis is q (volume resistivity) or qsd (surface resistivity multiplied by d) of the solid.

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(Spherical conductor)

(Air)

Hemispherical dielectric

Fig. K4-2 Equivalent circuit for Fig. K4-1

Tip of the upper sphere electrode

Surface resistivity Volume resistivity Tip of the hemispherical dielectric (εg side)

Tip of the hemispherical dielectric (εd side)

ρ or ρsd (Ω cm) Fig. K4-3 Numerically calculated electric fields for the configuration shown in Fig. K4-1 (f = 50 Hz). The dashed lines show the electric field for surface conduction, i.e., for qsd on the horizontal axis

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As shown in Fig. K4-1, the length d is the radius of the hemisphere, and also the radius of the sphere and the distance between the sphere and the hemisphere. For the calculation results shown in Fig. K4-3, the applied voltage is Vcos(2pft), where the frequency f is 50 Hz, and the relative dielectric constant of the solid is 4. The distance d is a scaling factor, and does not affect the normalized result in Fig. K4-3. The results for low q correspond to the field for DC, whereas those for high q correspond to the field for high-frequency AC. The results in Fig. K4-3 clearly show that the transition from the DC state to the AC state occurs in the resistivity range of 108–1011 X cm.

Answer K5 The electric field strength at P in the gas (eg) is higher for the case with a perfect dielectric, i.e., no conduction. This is evident from the equivalent circuit for Fig. K4-1 shown in Fig. K4-2, as explained in Answer K4. The higher the conductivity is, the lower ZR is, which makes the divided voltage lower in the hemispherical solid and higher in the gas. This is also apparent in the field behaviors shown in Fig. K4-3 in relation to q or qsd on the horizontal axis. Concerning the differences resulting from the effects of volume conduction and surface conduction, Fig. K4-3 shows that qsd (surface resistivity multiplied by d) on the horizontal axis has much the same effect as q (volume resistivity). As shown in Fig. K4-1 and also described in the Detailed Explanation to Answer K4, d refers to three different lengths, but in the term qsd, d corresponds to the radius of the hemisphere. These points are more clearly understood in configurations allowing for an analytical solution of the electric field. One of these configurations is a hemispherical solid dielectric lying on a grounded plane under a uniform field. The solution for this configuration is given in Sect. 6.25.

Comment As explained in Comment 2 to Answer J4, an equivalent circuit such as that in Fig. J4-3 is not applicable for analyzing the field near a contact point. However, the use of an equivalent circuit is applicable in cases without contact, as in Fig. K4-1.

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Other Topics

This section collects questions associated with electric force and induction on a human body.

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Subject L: Electric Force Electric fields exert force on charges. The magnitude of the electric or electrostatic force is given by QE, where Q is the charge and E is the electric field applying force on Q. For a conductor, the force per unit area is r2/(2e0) of surface, where r is the local charge density. This is because the field created by all the other charges existing in the configuration is r/(2e0), as explained in Answer G2. As r is equal to e0E, the resulting force is e0E2/2 per unit area of conductor surface.

Question L1: Force on a Charge in an Electric Field We consider a point charge Q for simplicity. Is it true that the electric force exerted on Q is QEo, where Eo is the electric field without Q, which is to say, the field before Q is introduced?

Question L2: Significant Force Caused by Balanced Forces? Electric forces such as the floating force exerted on a conductor or a conducting particle lying on a grounded plane are explained in Application 2 of Answer L1. Here we consider a spherical conductor located between parallel-plane electrodes, as shown in Fig. L2-1. Figure L2-1(a) shows a sphere with no charge located between an upper electrode at voltage V and a grounded lower electrode. When the sphere is located in perfect symmetry on the middle plane, its potential Va = V/2, and the same amount of negative and positive charge, −Q and Q, are induced respectively on the upper and lower hemispherical surfaces. The upper charge −Q exerts a force F1 upward, whereas the lower charge Q exerts a force of the same magnitude downward, resulting in zero net force on the sphere. In Fig. L2-1(b), the two parallel-plane electrodes are both grounded, and the sphere is charged with 2Q. Considering symmetry, the forces exerted upward and downward are of the same magnitude and cancel each other, resulting in zero net force on the sphere. The potential of this sphere is designated Vb. In Fig. L2-1(c), the sphere is at the same location as in Fig. L2-1(a) between the electrode at V and the grounded electrode. When the potential of the sphere Vc = Va (=V/2) + Vb, the electric field is the superposed fields of configurations (a) and (b). In this configuration (c), a non-zero force F3 seems to appear. For example, if the value of Q is selected so that Vb is equal to V/2, Vc becomes V. As a result, lines of electric force run only from the sphere to the lower electrode, but not to the upper one. Is it not strange that the superposition of two configurations with zero net force creates a non-zero force?

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-Q Q

Q Q

2Q

F

(a)

(b)

(c)

Fig. L2-1 Spherical conductor between parallel-plane electrodes

Question L3: Electric Field and Force on a Charged Spherical Dielectric Shell We consider the electric field for a spherical shell made of solid dielectric uniformly charged with density r on the surface. Compare its electric field with that of a spherical conductor of the same size when the shell (or spherical conductor) are either isolated in space or at height H above a grounded conductor plane, as shown in Fig. L3-1(a). The effect of the grounded plane is expressed as an image charge, as shown in Fig. L3-1(b). Also, consider the force exerted on the shell for both configurations shown in Fig. L3-1.

σ

σ R H

(a) Above a grounded plane

σ

(b) Expression with an image charge Fig. L3-1 Uniformly charged spherical dielectric shell

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λ Q

Q

Q

Q

(a) Four point charges

λ

λ

λ

(b) Four line charges

Fig. L4-1 Two configurations for holding a point charge q

Question L4: Holding a Charge with Electrostatic Force 1: Surrounding Point or Line Charges Earnshaw’s theorem states that a charged particle cannot be held statically in a stable equilibrium by electrostatic forces alone. Here, the definition of stable equilibrium is noteworthy. A state is stable when a slight (hypothetical) displacement is given to a particle and a restoring force acts on the particle so as to return it to its initial position. When a slight displacement is magnified, the state is not stable even if all the forces acting on the particle are balanced in its initial position. Consider whether a charged particle can be stably held in the following configurations: (a) Four point charges, one on each corner of a square (b) Four infinitely long parallel line charges, one on each corner of a square in cross section. All charges are of positive polarity. These configurations are shown schematically in Fig. L4-1(a) and (b).

Question L5: Holding a Charge with Electrostatic Force 2: A Cage with Uniform Surface Charge Consider whether a charged particle can be stably held in the following cages enclosed with uniform surface charge. These configurations are shown schematically in Fig. L5-1(a) and (b): (a) A cubic cage enclosed with uniform surface charge (b) A spherical cage with uniform surface charge and a small opening. These surface charges and the introduced charges are all of positive polarity.

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q

(a) Uniformly charged cubic cage Opening

q

Three-dimensional figure

Cross section

(b) Uniformly charged spherical cage with a small opening Fig. L5-1 Two configurations for holding a charged particle (point charge q)

Answer L1 It is not always true. Three kinds of force can be considered in this question, namely, the original field E0, the field that the charge Q itself exerts, and the field due to charges that Q induces. These are schematically shown in Fig. L1-1. Introducing charge Q changes the original field from E0 to a different value, but this field distortion usually need not be considered as a force acting on Q. If we substitute a small sphere for a point charge, it is evident that the forces exerted by its own charge are everywhere of the same magnitude and in a normally outward direction, thereby cancelling out as a whole. In contrast, the field created by charges that an introduced charge Q induces on nearby conductors cannot be neglected. This is because Q induces charges on conductor surfaces that change the original field and exert force on Q itself.

90

(a) The original uniform field

3

(b) Field resulting from Q itself

Questions and Answers

(c) Field caused by the charge induced by Q on conductor surfaces

Fig. L1-1 Electric fields related to a charge in an electric field

Detailed Explanation: Example of an Induced Field A simple example is when only an infinitely wide grounded conductor plane is present. The original field E0 with no charge is everywhere zero. An introduced point charge Q induces charge on the grounded surface, which produces a field and exerts force on Q. An exception is the case where outside conductors are all sufficiently remote from Q. For example, a uniform field formed by sufficiently large extremely remote electrodes. The introduction of Q induces charge on the remote electrodes but does not significantly change the original field. Also, when the original electric field E0 is formed not by stressed conductors but by isolated charges, such as point charges, the electric force is given as QE0. In conclusion, a required condition for the electric force to be given by QE0 is that an introduced charge does not affect the original field.

Application 1: Effect of Induced Charge in Gas Discharge When charge Q exists at a distance L from a grounded conductor plane, the electric field resulting from the induced charge is equivalently given by an image charge at a distance 2L (the mirror image point). This electric field E caused by the image charge is   E ¼ Q= 16pe0 L2

ðL1-1Þ

at the location of charge Q. Discharge inception takes place in a gas when a sufficiently high electric field accelerates electrons, which successively ionize gas molecules and produce a certain number of additional electrons (called an electron avalanche). This number of electrons required for discharge inception is considered to be roughly 108. Because

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the charge of an electron is 1.6  10−19 C, it follows from Eq. (L1-1) that when L is 1 cm, the electric field induced by 108 electrons is about 400 V/m or 4 V/cm. This value is much lower than the field causing discharge in air, which is about 30 kV/cm. Consequently, the field produced by the image charge can usually be neglected.

Application 2: Floating Force When a conductor or conducting particle lies on a grounded conductor plane, the particle may float up under a sufficiently high electric field. Such a phenomenon is important because it can result in detrimental effects by foreign conducting particles in gas insulated equipment, for example. The original electric field is usually regarded as a uniform field E0, which induces charge Q on a particle. The floating force is not equal to E0Q, but remains always lower than this value because there is a counteracting force from the image charge of opposite polarity that Q induces on the grounded plane. However, the force approaches E0Q when the particle floats above the grounded plane. The force on the particle is computed by integrating e0E2/2 over the total surface of a conducting particle, where E is the electric field on the conductor surface. Section 6.26 summarizes induced charge Q and floating force F for three conductor shapes.

Answer L2 The electric force on a conductor cannot be superposed, even when the electric field can be superposed.

Detailed Explanation We discuss this question using the expression for the electric force acting on a spherical conductor. When we express the electric fields in Fig. L2-1(a), (b), and (c) as Ea, Eb, and Ec, we get Ea þ Eb ¼ Ec

ðL2-1Þ

This corresponds to the principle of superposition. As explained in the introduction to Subject L, the force exerted on the sphere can be obtained by integrating the square of the normal component of the electric field on the sphere’s surface. It should be noted that electric fields have only a normal component on a conductor surface and can be treated as scalars. Thus, the force for Fig. L2-1(c) is

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I 2F ¼ e0

I Ec2 ds

¼ e0

I 2

ðEa þ Eb Þ ds ¼ e0

Questions and Answers

I Ea2 ds þ e0

I Eb2 ds þ 2e0

Ea Eb ds ðL2-2Þ

from the explanation given in the beginning of Subject L. These integrals must be performed considering the direction of the force on the surface. In the right of the last equals sign in Eq. (L2-2), the first and second integrals are both zero because the integral is identical and in opposite directions on the upper and lower hemispheres due to symmetry. However, the third term is not zero, because the product EaEb has a different sign on the upper and lower hemisphere, and, as a result, the integral yields a non-zero force. This force is in the downward direction.

Comment: Charge and Potential of a Spherical Conductor When the charge on the sphere is 2Q in Fig. L2-1(b), its potential is V/2. As a result, the potential of the sphere becomes V in Fig. L2-1(c), with charge 2Q (refer to Question A3). It should be noted that the charge need not be 2Q on the sphere in Fig. L2-1(b). However, the direction of the resulting force in Fig. L2-1(c) depends on the charge in Fig. L2-1(b).

Answer L3 (1) Electric Field When a spherical shell made of solid dielectric is isolated in space, its electric field distribution is identical to that for an isolated spherical conductor with the same charge density. This holds true outside the shell or the spherical conductor. However, there is no electric field inside the shell or spherical conductor. The electric fields are different when a spherical charged dielectric shell or a spherical conductor is located at height H above a grounded plane. The presence of the grounded plane is taken into account by replacing it with an image charge of opposite polarity at the mirror image position, as shown in Fig. L3-1(b). In the case of a conductor, its surface charge moves to keep the potential constant throughout the surface. This is easily understood as an extreme case when H is very small. For a conductor, decreasing H moves the surface charge to the lower part and increases the electric field there to a higher value. In the end, the field becomes infinitely high when H is zero. In contrast, for a spherical shell of solid dielectric, the surface charge is fixed, and the field does not increase markedly as H approaches to zero. The electric field is given equivalently by the effect of two point charges, that is, charge Q (=4pR2r) at the sphere center and another charge −Q at the mirror image position separated

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by 2(H + R) from the sphere center. This means that the electric field does not increase markedly when H approaches zero. (2) Electric Force When a dielectric shell is isolated in space, the outward acting (expanding) stress affects on all parts of its surface. Because the shell is uniformly charged, the force exerted is everywhere symmetric, and only radial components remain. The radial components cancel each other out because of the symmetry on the total sphere surface. This means that the net force is zero. Both the charge distribution and the resulting force are the same for a dielectric shell and an isolated, charged spherical conductor. When a dielectric shell exists above a grounded plane, as in Fig. L3-1(a), the force exerted by the charge on its own shell surface is cancelled out, just as it is in an isolated shell. Thus, we have only to consider the force exerted by the image charge. The force exerted on the shell is F¼

Q2 1  : 16pe0 ðH þ RÞ2

ðL3-1Þ

The derivation is given in Sect. 6.28. This force is equivalent to the attracting force between two point charges separated by 2(H + R).

Answer L4 One method of evaluating stability is to examine the electric potential in a given configuration. A positive charge moves to a position at a lower potential, because the electric field is the space derivative of the potential and is directed toward a lower potential. Evidently, a negative charge moves to a position at a higher potential. Configuration (a) Four point charges of equal magnitude are placed at (1, 1), (–1, 1), (–1, –1), and (1, –1) in the (x, y) coordinates on the z = 0 plane. A charged particle is placed at the center (0, 0). At the initial position, the forces on the introduced particle are balanced, irrespective of its polarity. Figure L4-2 shows the electric potential in the (x, y) plane (on the z = 0 plane) for Configuration (a). The potential is lowest in the center (0, 0), which indicates that a positive charge is stable at the initial location on this plane. However, the electric field on the z-axis is in the upward direction for positive z and in the downward direction for negative z. Therefore, a positive charge moves upward or downward after a slight displacement along the z-axis.

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V/ (Q/4πε0)

Fig. L4-2 Cross-sectional diagram of the electric potential on the z = 0 plane for the configuration shown in Fig. L4-1(a)

In contrast, a negatively charge particle moves away from the center on the z = 0 plane toward one of four point charges. It is evident that the potential of each of the four positive point charges is higher than that of the center. Configuration (b) Each line charge is infinitely long and has the same charge density. A charged particle is placed at the center of the four line charges. In Configuration (b) also, at the initial position, the forces on an introduced particle are balanced irrespective of its polarity. It is evident that a negative particle will move toward one of the line charges, the potential of which is higher than that at the center. Therefore, only a positive particle is considered below. In Configuration (a) with four point charges, a positive particle moves upward or downward on the z-axis. This instability does not occur in Configuration (b) consisting of four line charges. Because a positive particle moves toward a position with a lower potential, we have only to search for a position with a potential lower than that of the initial position. The potential of a line charge is expressed as C0 − ln(L) excepting the proportion constant, where C0 is a constant and L is distance from the line charge. The potential of four line charges at the initial position, namely, the center (0, 0) is U/C

X n

pffiffiffi ln Ln ¼ C  4 ln 2 ¼ C  1:39

ðL4-1Þ

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where C is a constant, n is the number of line charges, and Ln is the distance between the center and a line charge. The potential at the midpoint between two line charges is U/C

X

pffiffiffi ln Ln ¼ C  2 lnð1Þ  2 ln 5 ¼ C  1:61:

ðL4-2Þ

n

Therefore, the potential at the midpoint of two line charges is lower than that at the center of four line charges. This means that a positive particle escapes through the midpoint of two line charges. Evaluating the potential distribution is a simple and convenient method for determining whether a point charge is stable. However, the force on an introduced charge (particle) can be also evaluated directly with the electric fields acting on the particle. Two examples of such computations are given in Sect. 6.29.

Answer L5 Configuration (a) An electric field exists inside the cubic domain in this configuration. An introduced charged particle moves due to this electric field and either collides with or passes through a sheet wall of surface charge. Configuration (b) If a sphere with uniform surface charge has no opening, no electric field exists inside it. Under this condition, no force is exerted on an introduced charge, and the situation is beyond the scope of Earnshaw’s theorem. However, when a small opening exists on the sphere’s surface, an electric field appears inside the sphere. An introduced positive charge escapes out through this opening.

Detailed Explanation Configuration (a) It might appear strange that a positive charge moves toward the region of positive surface charge. This movement is caused by the electric field inside the cubic cage. Although the surface charge is of positive polarity, the direction of the electric field at the surface may not always be outward, but may be inward toward one of the surfaces. To demonstrate a simple example of such an electric field, we consider two parallel electric sheets A and B, as shown in Fig. L5-2. When they are infinitely wide with charge densities r and 2r, the electric fields are, respectively, r/(2e0) and

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Questions and Answers

3E

E

E A

E

E

E

σ

σ

B

2σ

B

E

A

σ

3E

Fig. L5-2 Two parallel electric sheets with surface charge density r and 2r

r/e0, extending from each sheet surface to infinity. Superposing these fields yields the electric field r/(2e0) in the inflowing direction on the lower surface of sheet A. Figure L5-3 shows the electric field distributions on two cross sections inside the cubic cage. These field distributions were calculated by Dr. Akiyoshi Tatematsu (CRIEPI, Japan) using the surface charge method. These figures indicate that an electric field exists in the outward direction on diagonal lines inside the cage; an introduced charged particle may move as a result of this electric field. Configuration (b) For the spherical cage without an opening, the electric field is zero inside and r/e0 on the outside surface. As explained in Answer G2, the outward field is r/(2e0), which is caused by its own charge on the corresponding surface element, plus r/(2e0) caused by the other charges. If a small opening is made in the sphere surface, a non-zero field appears at the opening, the value of which is r/(2e0), caused by the other charges. This field is in the outward direction. A positive charge will escape through this opening, although the charges on nearby surfaces are all of positive polarity.

Comment: Problem Setting in the Journal of the Institute of Electrical Engineers of Japan In the Journal of IEE (Institute of Electrical Engineers), Japan, there is a column entitled “Coffee Break,” to which members of the IEE of Japan may submit short essays or similar items. One of the authors (T.S.) submitted a quiz consisting of Question L4, and later, with the other authors, he submitted Question L5. These submissions were accepted.

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A)

B)

Fig. L5-3 Cross-sectional diagram of the electric field for the uniformly charged cubic cage shown in Fig. L5-1(a)

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Questions and Answers

The first question concerning four point charges at the corners of a square [Configuration (a) in Question L4] was published in the column and the IEE of Japan offered a prize of approximately $30US to student members who could solve the question correctly (Refs. L5-1 and 2). Several answers were submitted from student members to the IEE of Japan and most of them were correct. The second question, concerning the cubic cage [Configuration (a) in Question L5], was also published, and readers of the Journal of the IEE of Japan were asked to submit their answers (Refs. L5-3 and 4). This time, only one answer was submitted, and it was from a vice president of a Japanese electric power company. He sent in the correct answer to the first question. However, his answer to the second question was that there was no electric field inside the cubic cage.

Subject M: Induction on the Human Body Electrostatic induction is the voltage or current caused by other energized conductors or charges. Induction occurs because an equal amount of charge must exist at both ends of lines of electric force. If the source (voltage or charge) is invariant in time, the induced voltage is also constant in time. However, if the source changes in time, a current is induced that is usually in phase with the changes in the source. Nonetheless, in some cases, for example, when a medium has conductivity, the induced current many not be in phase with the source.

Question M1: Shaking Hands Under Atmospheric Electricity Electric fields called atmospheric electricity exist in the open air and are directed toward the surface of the Earth. Their magnitude changes considerably depending on atmospheric conditions, but the magnitude is between 80 and 200 V/m in fair weather. When it is 100 V/m, the voltage or potential amounts to 100 V at the height of 1 m and 200 V at 2 m in air. Despite this potential difference, however, no current flows when a very tall person shakes hands with a very short person when both are standing outside on the ground. Explain the reason why no current flows.

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Question M2: Induction on a Falling Body Explain the induction caused by atmospheric electricity when a human body falls in the atmosphere and finally reaches the ground. It is postulated that the body falls in the standing position. For the sake of simplicity, the use of a parachute is not considered. The effect of surrounding objects is also neglected.

Question M3: Induction from a Power-Frequency High-Voltage Source Explain what kinds of electric fields and currents are induced on the human body near a power-frequency high-voltage source, for example, when standing below a high-voltage AC transmission line.

Question M4: Induction from a High-Voltage DC Source Explain what kind of current is induced on the human body near a high-voltage DC source, for example, when standing below a high-voltage DC transmission line.

Question M5: Boundary Conditions for Calculating Induced Current The electrical characteristics [e.g., the complex dielectric constant given in Eq. (K-1)] of a human body are explained in Question M3. A human body can be treated as an almost perfect conductor in respect to DC or low-frequency AC fields existing outside the body. Furthermore, the electrical characteristic (roughly, the reciprocal of the impedance) of a human body is much higher (in the order of 107 times) than that of the atmosphere (e0) without conductivity. This suggests that such outside fields have only a normal component on the surface of the human body. Using subscript A to indicate the atmospheric air outside a human body where the electrical characteristic is represented by dielectric constant e0, and using subscript P for a human body whose electrical property is represented by volume conductivity j, then we have the boundary conditions for calculating induced current inside the human body as follows: Normal component: jAn = jPn, that is, jxe0EAn = jEPn Tangential component: jAt = jPt = 0, that is, EAt = EPt = 0. Are these boundary conditions correct?

100

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Answer M1 Atmospheric electricity can be regarded as a DC field because it does not change rapidly. The human body is an almost perfect conductor in a DC field. As a result, the whole body is at zero potential when a person stands on the ground, and no potential difference arises between two people, irrespective of their heights. Comment It should be noted that the presence of a human body distorts the original atmospheric electric field, just as a conducting rod standing on grounded conductor plane does in Fig. E1-1. This distortion is shown schematically in Fig. M1-1, where a human body is simulated by a conducting hemispheroid for simplicity. It is reported that a hemispheroid of B/H = 0.15 (B is the base radius and H is the height) may give a good estimate of the AC induced current for a standing human with the same height (Ref. M1-1). The electric field on the top of a human head is highly dependent on the local geometry of the head, but it rises to roughly 15 times that of the original field (atmospheric electricity). In the case of a pig, which has a shorter and fatter body than a human, the highest increase in the electric field appears on its back and amounts to about seven times the original value (Refs. M1-1 and 2).

Fig. M1-1 Distortion of atmospheric electricity by a human body (Dotted lines indicate equipotential surfaces.)

z E0

H

0 B

r

3.5 Other Topics

101

Answer M2 As explained in Answer M1, the human body is regarded as a conductor in atmospheric electricity. If the body has no overall charge, its potential is the same throughout the whole body. Consequently, the problem is the change of field distribution with time when a conductor is moving inside a uniform field. This is summarized as follows: Potential: The potential of the body is roughly equal to the space potential at the height of the center of the body. As the body approaches the ground, its potential decreases and finally reaches zero when it arrives at the ground. Charge: As schematically shown in Fig. M2-1(a), atmospheric electricity E0 induces negative charge −Q1 on the upper part of the body and positive charge Q1 on the lower part, respectively. As the body approaches the ground, the absolute value of each induced charge increases somewhat to −Q2 and Q2, as in Fig. M2-1(b). When the body arrives at the ground, Q2 vanishes by discharging there. Current in the earth: As explained in Question H1, charge moving in the air induces current in the earth through the changing lines of electric force. But the magnitude of the current is usually very small, because −Q1 and Q1 on the body move at the same speed. The discharge of Q2 on arrival causes an instantaneous current flow. However, this current is also small inside the earth because the discharge occurs only at a specific location on the surface between Q2 and the charge that has been induced before arrival. If the body has well-insulated shoes, the discharging of Q2 does not occur instantly on arrival at the ground, but is usually delayed by a short time.

Eo

Eo

-Q1 Q1

-Q2

Q2

(a) High above the ground

(b) Just before arrival on the ground

Fig. M2-1 Charge on a human body induced by atmospheric electricity

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Questions and Answers

Comment For a body height of 1.8 m, the amount of charge released when a body touches the ground is as small as 5nC for atmospheric electricity at E0 = 100 V/m.

Answer M3 AC sources such as transmission lines induce charge on the surface of a human body as displacement current acting perpendicularly on the body. Unlike a DC source, the induced charge changes with the same frequency as the AC source. This brings about a current that flows depending on the distribution of impedance (overwhelmingly conductivity) inside the body.

Detailed Explanation

α dispersion

Relative permittivity

β dispersion

Conductivity

Conductivity κ (S/m)

Relative dielectric constant εs

First, we must determine the electrical characteristics of a human body for a power-frequency AC field. Figure M3-1 shows these electrical characteristics, i.e., the relative dielectric constant es and the volume conductivity j of human muscle tissue, in relation to the AC frequency (Ref. M3-1). The human body is often approximated as a substance of homogeneous and constant characteristics, as given in Fig. M3-1. It should be noted that the numerical values are given in logarithmic scales in this figure.

γ dispersion

Frequency (Hz) Fig. M3-1 Relative dielectric constant and volume conductivity of human muscle tissue

3.5 Other Topics

103

As explained in Sect. 5.4 and also given in Eq. (K-1), for a quasi-static AC field, the electrical characteristics are typically represented by complex dielectric constant ec as e_ c ¼ es e0 þ j=ðjxÞ ¼ es e0 þ 1=ðjxqÞ

ðM3-1Þ

where x = 2pf, f is the frequency of the AC source, j is the volume conductivity, and q is the volume resistivity (the reciprocal of j). From Fig. M3-1, es and j are roughly 106 and 0.1 S/m, respectively, in the range 50–60 Hz. Comparing the two terms in the complex dielectric constant, es e0  9  106 ;

j=x  3  104 F=m

ðM3-2Þ

for 50 or 60 Hz. The latter is roughly 30 times higher than the former, indicating that conduction is the predominant contribution to the impedance of a human body in the power frequency range. Furthermore, the electrical characteristic of the human body (j/x) is roughly 107 higher than that of the atmosphere (e0). Consequently, the human body can be regarded as conducting for an AC source of 50–60 Hz, similar to the case of the DC field discussed in Question M1.

Application 1: Calculation of Induced Current Inside a Human Body Various numerical computations have been performed to analyze the induced current inside a human body. Figure M3-2 shows one such example, which was performed in 2000 by Prof. Boonchai Techaumnat (Chulalongkorn University) by applying the boundary element method. He used 576 elements with 4778 nodes.

Current density (μA/m2)

Fig. M3-2 Numerically calculated distribution of induced current inside a human model

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Questions and Answers

At present, more detailed computations with divisions finer by several orders of magnitude are possible. The division of about 6  107 elements (with 0.5 mm square elements) is reported already in a 2014 paper (Ref. M3-2).

Application 2: Induction by an AC Transmission Line Regulations in Japan dictate that the electric field should be below 3 kV/m near the ground under AC transmission lines. In such a field, the induced current density is about 2 mA/m2 near a person’s neck. The current density is four to five times higher in the ankle because the whole induced current flows to ground through this narrow body part. There is also electromagnetic induction due to load current flowing in a transmission line. The induced current is an eddy current, so-called as it flows whirling around the magnetic field. The magnetic flux density caused by an AC transmission line is usually no more than 20 lT near the ground in Japan, which induces about 0.4 mA/m2 for 50 Hz inside the human body. This value is considerably lower than the current electrostatically induced by an AC voltage, as mentioned above.

Answer M4 Three kinds of induced currents are theoretically likely. One is caused by a DC electric field, which induces charge on the surface of a human body, although no current flows when the body is not in motion. A slight current may flow when a person moves, for example, if the person jumps. However, its magnitude is usually small compared with the current induced by a power-frequency AC field that changes from its positive peak value to its negative peak value 50 or 60 times per second. The second is the current that magnetic fields induce in a moving conductor. However, this current is also usually small. A concrete example is mentioned in the Application below. The third is the current of ions flowing into a body, which are generated at the high electric field region on high-voltage conductors. Ion flow is discussed in Question K1.

Application: Current Induced by the Magnetic Field of a DC Transmission Line The load current of a DC transmission line creates a magnetic field in the neighborhood. Even if this field is invariant in time, an electric field is induced in a conducting substance moving in the magnetic field. In the human body, this holds true for blood flow. The induced electric field E is

3.5 Other Topics

105

E ¼uB

ðM4-1Þ

where u is the speed of blood flow and B is the magnetic flux density. This electric field is induced in the direction perpendicular to both the blood flow and the magnetic field. The blood flow speed u is about 0.2 m/s on average for arteries, but can be as high as about 0.6 m/s in a major artery. As mentioned in Question M3, the magnetic flux density caused by an AC transmission line is usually no more than 20 lT near the ground. This magnitude is also much the same for a DC transmission line. However, the Earth’s terrestrial magnetism exists everywhere; its value is about 50 lT in Japan. If we consider this value and a blood flow speed of 0.6 m/s in Eq. (M4-1), the induced electric field is roughly 30 lV/m.

Answer M5 These boundary conditions are not correct. The tangential component is not zero inside the human body.

Detailed Explanation For two dielectric media A and B with conductivity, the boundary condition for the normal component including conduction is explained in Sect. 5.5 as ðeA þ jA =jxÞE_ An ¼ ðeB þ jB =jxÞE_ Bn

ðM5-1Þ

Because the conductivity is zero in atmospheric air (subscript A) and the conductivity is predominant in the human body (subscript P) for 50 or 60 Hz, Eq. (M5-1) becomes jxe0 EAn ¼ jEPn :

ðM5-2Þ

This means that the normal component of displacement current in the air is equal to that of the conduction current inside the body. Concerning the tangential component, the continuity of potential or the tangential component is applied: EAt ¼ EPt :

ðM5-3Þ

As explained in Question M3, j for a human body is about 0.1 S/m and j/(xe0) is roughly 3–4  107 in Eq. (M5-2). This means that the normal component EPn inside a human body is overwhelmingly small compared to EAn in the air. However, the tangential component EPt inside the body should not be regarded as much smaller than the normal component EPn, but rather of the same order as EPn. Thus,

106

3

Questions and Answers

Fig. M5-1 Incorrect boundary conditions for calculating induced current inside the human body

JAn JPn

jωε0EAn

κEPn

Fig. M5-2 Correct boundary conditions for calculating induced current inside the human body

the current or electric field on the boundary is not like that in Fig. M5-1 but, rather, like that in Fig. M5-2. This point is more clearly explained in Sect. 6.25 with a configuration that has an analytical solution for the electric field distribution.

Chapter 4

Three Simple Questions Not Related to Electric Fields

As a thought-provoking pastime for readers, the authors here offer three simple questions in technology or physics. No answers to these questions are given in this book. Question 1: Active Filter An active filter is also sometimes called active noise control. It is a method of eliminating unwanted or harmful noise by emitting sound with the same amplitude but the opposite phase, thereby cancelling out the original noise. However, the noise or sound has the energy of an acoustic wave. Where has this sound energy gone when the noise is cancelled out? Question 2: Wind Turbine We frequently see large wind turbines for wind power generation. These turbines usually have two or three blades, much fewer than the vanes used in water turbines for hydraulic power generation. Why is the number of blades in wind turbines so small? Don’t turbines fail to catch the energy of the wind that passes between the blades? Question 3: Superconducting Coil Current keeps flowing at the same value as its initial state in a superconducting coil. This is the principle of SMES (Superconducting Magnetic Energy Storage) systems in which no ohmic losses occur because of the coil’s zero resistance. However, it is well known that electrons in circular motion lose energy by radiating electromagnetic waves. Why can the current keep flowing in a superconducting coil?

© Springer Nature Singapore Pte Ltd. 2020 T. Takuma and T. Shindo, Problems and Puzzles in Electric Fields, https://doi.org/10.1007/978-981-15-3297-9_4

107

Chapter 5

Fundamentals of Electrostatic and Quasi-electrostatic Fields

5.1

Governing Equations

As described in Sect. 2.1, an electric field is created by either an electric charge or a magnetic field changing with time. The former corresponds to the electrostatic and quasi-electrostatic fields covered in this book. The latter, which is formed by a changing magnetic field, is outside the scope of the book. The basic equation for electric field E is derived from one of Maxwell’s equations: rot E ¼ 

@B @t

ð5:1-1Þ

In quasi-electrostatic cases, with the magnetic field B being either zero or constant on the right-hand side of Eq. (5.1-1), the electric field E has a scalar potential Ф, i.e., E ¼ grad U

ð5:1-2Þ

This scalar function Ф is called the electric potential. It has a constant value on the surface of a conductor, equal to the conductor voltage. Combining Eq. (5.1-2) with another of Maxwell’s equations relating electric flux density D and charge density q, div D ¼ q

ð5:1-3Þ

and the constituent equation for the materials, D ¼ eE

© Springer Nature Singapore Pte Ltd. 2020 T. Takuma and T. Shindo, Problems and Puzzles in Electric Fields, https://doi.org/10.1007/978-981-15-3297-9_5

ð5:1-4Þ

109

110

5 Fundamentals of Electrostatic and Quasi-electrostatic Fields

leads to the following equation for Ф, divðe grad UÞ ¼ q

ð5:1-5Þ

In this equation, e is the dielectric constant (permittivity) of the material or medium and q is the space density of the charge. In a vacuum and approximately in a gas, e is equal to the electric constant e0, which has a value of 8.854  10−12 F/m. Throughout this book, e is treated as a constant inside each medium, although in practice it changes somewhat depending on the applied electric field and temperature, among others. Under the approximation of constant e and in the absence of space charge, the governing equation for Ф inside each medium becomes the following: divðgrad UÞ ¼ DU ¼ 0:

ð5:1-6Þ

This equation is called Laplace’s equation after Pierre Simon Laplace who first derived the equation in polar coordinates in the field of celestial mechanics. The main task in most electric field calculations is to obtain the electric potential Ф which satisfies Eq. (5.1-6) under the corresponding boundary conditions.

5.2

Uniqueness Theorem in Electrostatic Fields

When a function satisfies both Laplace’s equation and the all boundary conditions, this function is the unique correct solution for the electric field; this is called the uniqueness theorem. When two functions Ф and X both satisfy Laplace’s equation DU ¼ 0 and DX ¼ 0:

ð5:2-1Þ

Evidently, the difference uð¼U  XÞ also satisfies Laplace’s equation, because Du ¼ DðU  XÞ ¼ 0:

ð5:2-2Þ

The well-known Green’s Theorem converts the volume integral in a domain v to the integral on the closed surface S of the corresponding domain, as I u S

@u dS ¼ @n

Z n

o uDu þ ðgraduÞ2 dv:

ð5:2-3Þ

v

When the two functions Ф and X have the same value on surface S, the left side of the equation is zero, while the first term of the right-hand integral is also zero, thus

5.2 Uniqueness Theorem in Electrostatic Fields

Z

Z " 2  2  2 # @u @u @u ðgraduÞ dv ¼ þ þ dv ¼ 0: @x @y @z 2

v

111

ð5:2-4Þ

v

This equation shows that the function uð¼U  XÞ must be a constant everywhere. However, it is zero on the boundary, which proves that Ф is equal to X everywhere. It is easy to understand that this uniqueness theorem can be applied to the solution of Poisson’s equation.

5.3

Boundary Conditions for Composite Dielectric Fields

Composite dielectrics mean configurations where two or more dielectrics exist. More details are explained in the beginning of Sect. 3.4. The boundary condition on a conductor surface is a constant electric potential, Ф = Ф0, the so-called Dirichlet condition. At the interface of two dielectrics eA and eB, however, two conditions hold. One is the continuity of Ф, or equivalently, the continuity of the tangential field strength Et, UA ¼ UB

ð5:3-1Þ

EAt ¼ EBt

ð5:3-2Þ

or

where subscripts A and B indicate the corresponding media or materials. The other condition is the continuity of the normal component Dn of the electric flux density, DAn ¼ DBn :

ð5:3-3Þ

From Eq. (5.1-4), this can be rewritten as eA EAn ¼ eB EBn :

ð5:3-4Þ

These equations, together with the constant potential condition on conductor surfaces, complete the boundary conditions for composite dielectric configurations. However, it should be noted that these boundary conditions at dielectric interfaces are only macroscopic expressions and should not be generalized to minute surface structures such as those involving molecules or atoms.

112

5.4

5 Fundamentals of Electrostatic and Quasi-electrostatic Fields

Basic Equations for Capacitive–Resistive (Mixed) Fields

Any real material exhibits a certain degree of conductivity, producing what are usually called capacitive–resistive or mixed fields. Apart from special conduction environments involving such phenomena as electrical discharges or ion flow, the most basic conductivity is ohmic conduction. Two fundamental kinds of conduction are possible in ohmic conduction: volume and surface conduction. These are characterized by volume and surface conductivities, j and js, respectively. See Question K2 for an insight into the nature of surface conduction. Although both volume and surface conductivity may change locally depending on the conditions, they are considered to be constant in each medium throughout this book. Before explaining capacitive–resistive (mixed) fields, we will briefly mention the approach using complex variables (phasor notation) for steady AC fields. As explained in Sect. 2.1, when all the materials involved are either perfect conductors or perfect dielectrics (i.e., without conductivity) and only one kind of voltage waveform (source) is present, the field distribution is identical to the distribution of a DC field at an instantaneous voltage. This holds true as long as the applied voltage does not change very fast compared with the traveling time of the field. That is to say, in slowly changing field conditions, the instantaneous potential and field strength are always similar in any part of the region, i.e., they are merely proportional to the applied voltage at that instant. When several voltage sources exist, the field must be computed not as a DC field but by superposing the instantaneous effects of all the sources. If all the waveforms of the sources are sinusoidal with the same angular frequency x, we can conveniently express the instantaneous field in the system directly with complex field values. We can calculate mixed fields by taking into consideration the true charge induced by volume or surface conduction. In a region with dielectric constant e and volume conductivity j, the electric field E is formulated from Eqs. (5.1-3) and (5.1-4) as follows: divðeEÞ ¼ q:

ð5:4-1Þ

On the other hand, div J ¼ divðjEÞ ¼ 

@q @t

ð5:4-2Þ

where J is the current density and q is the charge density in the space. These equations lead to   @E div jE þ e ¼ 0: @t

ð5:4-3Þ

5.4 Basic Equations for Capacitive–Resistive (Mixed) Fields

113

In the quasi-steady state for AC fields of angular frequency x (=2pf, where f is the corresponding frequency), the phasor notation (complex number expression) can be used to represent the electric field distributions. Thus, Eq. (5.4-3) becomes   div ðj þ jxeÞE_ ¼ 0

ð5:4-4Þ

pffiffiffiffiffiffiffi where j ¼ 1 (also called the imaginary unit) and the over-dot indicates a complex expression (phasor notation) of the variable for AC fields. The field, including the effect of volume conduction, can be expressed simply by substituting for dielectric constant e a complex expression, such as e_ ¼ e þ

j : jx

ð5:4-5Þ

In cases without conduction, j = 0 (infinitely high resistivity) and the complex dielectric constant simply reduces to e. However, Eq. (5.4-5) indicates that the effect of volume conduction is predominant for a DC field or AC fields where j  jxe. If dielectric constant e and volume conductivity j do not change in a medium or a part of the region, the governing equation, Eq. (5.4-4), once again becomes Laplace’s equation as given in Eq. (5.1-6) for the complex electric field. Under this condition, an induced charge does not exist inside each medium but only at its boundaries (i.e., on conductor surfaces and at dielectric interfaces).

5.5

Boundary Conditions for Mixed Fields

The boundary conditions given in Eqs. (5.3-1)–(5.3-4) must be somewhat modified at the dielectric interfaces in mixed fields. The continuity of electric potential Ф, Eq. (5.3-1), or the continuity of tangential field strength Et, Eq. (5.3-2), still applies. However, the continuity of the normal component of the electric flux density, Eq. (5.3-3) or (5.3-4), changes as follows [Ref. 5.5-1]. For systems with two dielectrics with volume conductivity, i.e., dielectric A (dielectric constant eA, volume conductivity jA) and dielectric B (eB, jB), as shown in Fig. 5.5-1, the continuity condition of the normal component of the electric flux density at the interface is expressed as

Fig. 5.5-1 Interface of dielectrics with volume conduction

JBn B (εB , σB)

A (εA, σA) JAn

114

5 Fundamentals of Electrostatic and Quasi-electrostatic Fields

DAn  DBn ¼ qs :

ð5:5-1Þ

In this equation, subscript n means a component normal to the interface, and qs is the surface charge density, which is given by Z ð5:5-2Þ qs ¼ ðJBn  JAn Þdt: Since current density J is equal to jE, Z qs ¼ ðjB EBn  jA EAn Þdt:

ð5:5-3Þ

We can rewrite the integral with 1/jx for steady AC fields with an angular frequency x in the phasor notation, thus q_ s ¼

 1  _ jB EBn  jA E_ An : jx

ð5:5-4Þ

This equation, when combined with Eq. (5.5-1), gives the final boundary condition as     jA _ jB _ eA þ EAn ¼ eB þ EBn : jx jx

ð5:5-5Þ

Equation (5.5-5) is the same as that for the interfaces of usual dielectrics without conductivity with e replaced with the complex dielectric constant given in Eq. (5.4-5). When surface conduction exists, a more general conservation of charge is expressed (Refs. 5.5-2 and 3) as e_ B E_ Bn  e_ A E_ An ¼

1 rs  J_ s jx

ð5:5-6Þ

The effect of surface conduction along a dielectric interface is included in the right-hand side of Eq. (5.5-6), where J_ s ¼ js E_ t is the surface current density at the interface and rs  J_ s is the tangential divergence of Js, which is treated as a vector. The normal is taken in the direction from medium B to medium A.

5.6 Summary of Equations in Mixed Fields

5.6

115

Summary of Equations in Mixed Fields

When characteristic variables e and j are constant in each medium, Eq. (5.4-5) is classified into the following three categories according to the value of the defining parameter xe/j. (a) xe/j  1: high frequency or capacitive fields   div eE_ ¼ 0

ð5:6-1Þ

This equation corresponds to usual electric fields where an AC voltage is applied. (b) xe/j  1: DC (low frequency) or resistive fields   div jE_ ¼ 0

ð5:6-2Þ

This equation is the same as Eq. (5.6-1) where e is replaced with j, and the corresponding field is often called a steady current field. (c) General cases of mixed fields The field equation, Eq. (5.4-4), is equal to Eq. (5.6-1) where e is replaced with the complex dielectric constant given in Eq. (5.4-5). If both e and j are constant everywhere, Eq. (5.4-4) again leads to div E_ ¼ 0, i.e., Laplace’s equation.

Chapter 6

Supplementary Explanations

6.1

Image Charge Method (in Answer A3)

The image charge method is used to express the equivalent field distribution by replacing a conductor with a hypothetical charge, called an image charge. As explained in most textbooks on electromagnetism, a simple example is the case where a point charge Q exists at height H above a plane conductor (ground), as shown in Fig. 6.1-1(a). Figure 6.1-1(b) depicts the corresponding image charge, a point charge situated at depth H below the original plane. When using the image charge technique, the conductor is removed from consideration, and then the original point charge and its image charge together precisely express the field above the plane. In the case of Fig. 6.1-1, the magnitude of the image charge is equal to the total amount of charge induced on the original plane surface. This method is also applicable to the case of a point charge with a spherical conductor. Answer A3 explains the case in which a point charge exists either outside a spherical conductor or inside a hollow spherical conductor (conducting shell).

© Springer Nature Singapore Pte Ltd. 2020 T. Takuma and T. Shindo, Problems and Puzzles in Electric Fields, https://doi.org/10.1007/978-981-15-3297-9_6

117

118

6 Supplementary Explanations

Q H

H Ground

Ground

H Image charge (a) A point charge above the ground

(b) Arrangement of the image charge

Fig. 6.1-1 Point charge above the ground

6.2

Critical Comment on Answer C2

This critical comment was provided by an anonymous writer during a round-table talk in the Journal of IEE (Institute of Electrical Engineers), Japan (Ref. 6.2-1). It addresses the Detailed Explanation in Answer C2 that the order of Em is easily inferred or estimated based on the field at the opposite point (lowest point Q in Fig. 6.2-1). This is based on the relationship Zd E dl ¼ V

ð6:2-1Þ

0

That is, the integral of the electric field along any line of electric force over the whole gap length is the same and is equal to potential difference V. In the above-mentioned article, the following two arguments, (1) and (2), are stated. (1) In Fig. 6.2-1(a), as a point on the surface moves away from Q, the corresponding gap becomes larger. That is to say, the surface is more remote from the opposite electrode than, for example, in Fig. 6.2-1(b). This should result in the electric field being lower at Q in Fig. 6.2-1(a) than in Fig. 6.2-1(b). (2) Because charges of the same polarity repel each other, the field should not be concentrated at point Q. As a result, it can neither be asserted that the electric field at lower point Q becomes high nor that the field at Q is higher in Fig. 6.2-1 (a) than in Fig. 6.2-1(b).

6.2 Critical Comment on Answer C2

V

V P

119

P Q d

(a)

V Q

(b)

P

Q

(c)

Fig. 6.2-1 The three electrode configurations considered in Question C2

Counterarguments on (1) and (2) were published in an issue of the same journal 4 months later by one of the authors (T. T.) (Ref. 6.2-2). These are summarized as follows. Concerning the above argument (1), the point at issue is that the (smallest) gap length is the same and equal to d in all three configurations. The shape near Q plays a more decisive role on the electric field there than the gap length near Q. The larger gap distance near point Q means that the shape (consequently, potential also) is more protruding at Q. Evidently, this should cause a higher electric field at Q, which is contrary to argument (1). Concerning argument (2), it is true that charges of the same polarity repel each other. However, the point at issue is that charge distributes itself on a conductor surface irrespectively of its polarity. At the top edge of a sharp conductor, for example, the field can be very high, theoretically infinitely high, although the charge is of the same polarity there. In a word, the value of the electric field has nothing to do with the polarity of the charge existing there. As supplementary information, the distribution of the electric field is shown along a line of electric force from Q to P for d = R in Fig. 6.2-2(a) and for d = 2R in Fig. 6.2-2(b). The vertical axis is normalized with the average field V/d, while the horizontal axis (distance) is given as l/d. It is evident from these figures that the order of the maximum electric field (Em) at P is reversed from that of the field at Q.

6 Supplementary Explanations

E/(V/d)

120

R d

l

R

R l

d d (b)

(a)

l/d

Q

l (c)

P

(a) d/R=1

R R d

l

d

l E/(V/d)

(a)

Q

R

d

l (c)

(b)

l/d (b) d/R=2

Fig. 6.2-2 Electric field distribution across the gap from Q to P

P

6.3 Variable Separation Method Applied to Question D1

6.3

121

Variable Separation Method Applied to Question D1

This method expresses the electric potential in question as a multiplication of functions, each of which depends on a single variable. For two-dimensional polar coordinates (r, h) as in Fig. 6.3-1, Laplace’s equation is   1@ @U 1 @2U r þ 2 2 ¼ 0: r @r @r r @r

ð6:3-1Þ

Applying the variable separation method gives the solutions An r n sin nh and Bn r n cos nh, where An and Bn are constants. The potential is defined in the range from h0 to 2p, or from 0 to (2p − h0). From this, the boundary conditions are U = 0 at h =0 and at h = 2p − h0. Because the solution Bn r n cos nh does not satisfy the boundary condition at h = 0, the potential is expressed as U ¼ Kr n sin nh;

ð6:3-2Þ

where K is a constant depending on the configuration. The exponent n is determined from the boundary condition at h = 2p − h0 as n ¼ p=ð2p  h0 Þ:

ð6:3-3Þ

This is the smallest positive value, and it plays a predominant role when r is small, i.e., close to tip P. The electric field is proportional to rm, where m ¼ n  1 ¼ ðh0  pÞ=ð2p  h0 Þ. The smaller the value of h0, i.e., the sharper the tip, the smaller the value of m. This causes a more rapid increase of the field as P is approached. The value of m is –1/3 for a right-angled tip (h0 = p/2), whereas m is –1/2 for an infinitely thin conducting foil or sheet. When h0 = p, m = 0 and no singularity of the field occurs. When tip angle h0 is larger than p, it forms a sharp concave pit, and m is larger than zero. Consequently, the field decreases to zero as r decreases to zero. The electric field behavior near a sharp conical tip is summarized in Sect. 6.4. Fig. 6.3-1 Expression of Fig. C1-1 in two-dimensional polar coordinates

122

6.4

6 Supplementary Explanations

Field Behavior Near a Sharp Conical Tip (in Sect. 6.3)

The field behavior near a sharp conical tip is explained in Ref. 6.4-1. When the angle at a conical tip is 2h0 and this angle is small, the electric field near P is roughly proportional to rm. This behavior is fundamentally the same as that near a two-dimensional tip explained in Sect. 6.3. The value of m is expressed as m ¼ 1=f2 lnð2=h0 Þg  1:

ð6:4-1Þ

m ¼ 1=ð2 ln h0 Þ  1:

ð6:4-2Þ

When angle h0 is small,

Furthermore, when the angle h0 is very small, m is close to −1. This means that the field increases almost in inverse proportion to the distance from P when P is approached.

6.5

Ellipsoidal Coordinates Applied to Question E1

Details of the ellipsoidal coordinate system are omitted here (Ref. 6.5-1). For a spheroidal conductor (or an ellipsoid of revolution) in the coordinates (r, z) shown in Fig. 6.5-1, equipotential surfaces are given by the surfaces of n = constant in the following equation

h2

Fig. 6.5-1 Spheroid in ellipsoidal coordinates

z2 r2 þ 2 ¼ 1: þn a þn

ð6:5-1Þ

z

E0

h

0

a ξ=0

r

6.5 Ellipsoidal Coordinates Applied to Question E1

123

In this equation, height h and radius a mean the cross-sectional dimensions of the spheroid, as shown in Fig. 6.5-1. The conductor surface corresponds to n = 0. Under a uniform field E0 in the downward (–z) direction, the potential U is as follows: ( U ¼ E0 z 1 

tanh1

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k ¼ 1  a2 =h2

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi) ðh2  a2 Þ=ðn þ h2 Þ  ðh2  a2 Þ=ðn þ h2 Þ ; tanh1 k  k

ð6:5-2Þ

where h > a. The absolute value of electric field on the z-axis (r =0)is ( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi) 2 2 E0 z h2  a2 1 h  a E ¼ E0 þ  tanh z tanh1 k  k z2  ðh2  a2 Þ

ð6:5-3Þ

where the downward direction of a field is positive. The electric field Em at the needle tip, where r = 0 and z = a, is Em ¼

k3 E  01 : ð1  k2 Þ tanh k  k

ð6:5-4Þ

When the radius a approaches 0, j becomes 1, as can be seen from Eq. (6.5-2), and Em becomes infinitely high. However, it can be confirmed that when a approaches 0, the potential U and electric field (absolute value)E become E0z and E0, respectively, for any z larger than h. This means that the electric field distribution is a uniform field E0 everywhere outside the needle.

6.6

Ellipsoidal Coordinates Applied to Question E2

Figure E2-2 shows the ellipsoidal coordinates for a two-dimensional ellipsoidal configuration with the cross-sectional shape (x, y) uniform in the z-direction. In the following equation x2 y2 þ ¼ 1; n þ a2 n þ h2

ð6:6-1Þ

n = constant expresses equipotential surfaces, and n = 0 corresponds to the surface of the ellipsoidal conductor, where a and h are the minor and major axes of the ellipse, respectively, as shown in Fig. E2-2. When a uniform field E0 exists in the downward (–y) direction, the potential is given as follows:

124

6 Supplementary Explanations

E0 y h U¼ ha

sffiffiffiffiffiffiffiffiffiffiffiffiffi ! n þ a2 a : n þ h2

ð6:6-2Þ

When a approaches zero, U ¼ E0 y 

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi n=ðn þ h2 Þ

ð6:6-3Þ

This equation indicates that the potential does not become a simple formula as it did in Question E1 involving a thin needle.

6.7

Electric Field for a Semi-infinite Plane Electrode Above a Grounded Plane (in Answer F2)

Applying the Schwarz-Christoffel transformation (Ref. 6.7-1) gives the following relationship between the z-plane (=x + jy) as shown in Fig. F2-2 and a uniform W-plane (=U + jU) as follows: z¼

np o H H W þ exp ðW þ jV Þ þ H p p V V

ð6:7-1Þ

The upper semi-infinite plane corresponds to U = V, y = h, whereas the grounded plane corresponds to U = 0, y = 0. The coordinate x on the grounded plane is expressed with a parametric variable U as x¼

  H H pU U  exp  þH ; p p V V

1  U  1:

ð6:7-2Þ

The electric field on the grounded plane is E¼

V H



  pU 1 þ exp  : V

ð6:7-3Þ

Eliminating U from these equations and rewriting a normalized electric field E/ (V/H) as f gives the following equation.   x 1 1 1f ¼ þ ln : H p f f

ð6:7-4Þ

6.8 Relationship Between the Surface Shape of an Electrode …

6.8

125

Relationship Between the Surface Shape of an Electrode and the Electric Field (in Answer F3)

The following relationship exists between the average curvature H of an electrode surface and the electric field E @E ¼ 2H  E @n

or

@ ln E ¼ 2H @n

ð6:8-1Þ

where n is the unit vector in the normal direction for the corresponding location on the surface. The average curvature H is given as  H¼

1 1 þ q1 q2

 2

ð6:8-2Þ

where q1 and q2 are the principal (maximum and minimum) curvature radii for a corresponding point on the surface (Refs. 6.8-1 and 6.8-2). In the case of a spherical electrode with radius R, H is equal to 1/R, whereas H is 1/(2R) for a cylindrical electrode with radius R. This suggests that when the cross-sectional profile is the same for a two-dimensional electrode and an axisymmetric electrode, the curvature is larger in the axisymmetric case than in the two-dimensional case. As a result, the maximum electric field is usually higher for the axisymmetric electrode than for the corresponding two-dimensional electrode. For a two-dimensional configuration in (x, y) coordinates that is uniform in the zdirection, Eq. (6.8-1) can be easily derived as shown in Fig. 6.8-1. Applying

Fig. 6.8-1 Relationship between the electric field and the electrode surface shape in two-dimensional (x, y) coordinates

y

R E E-

x

126

6 Supplementary Explanations

Gauss’s theorem to four sides of a quadrilateral space, only the normal components of the electric field contribute to the integral of the closed domain. Therefore, ERDh ¼ ðE þ DE ÞðR þ DRÞDh:

ð6:8-3Þ

which readily results in Eq. (6.8-1).

6.9

Comparison of Electric Fields Between Two-Dimensional and Axisymmetric Configurations (in Answer F3)

Section 6.8 explains the relationship between the curvature of an electrode surface and the electric field E. The larger the curvature (equivalently, the smaller the curvature radius), the higher the electric field is in general. Therefore, the maximum electric field is usually higher in an axisymmetric electrode than in a two-dimensional electrode. However, it should be noted that this characteristic is not always true because the maximum field is also influenced by any opposing electrode. Only if the opposing electrode and the gap length are the same is the maximum electric field higher in an axisymmetric electrode than in a two-dimensional electrode for the same cross-sectional profile. The electric field is examined for a simple electrode configuration consisting of a semicircular edge (radius R) connected to a flat part with radius L in axisymmetric configuration (length 2L in two-dimensional cases) in cross section, as shown in Fig. 6.9-1. The maximum field always appears on the curved edge. 1.2

W/2

E/E0

y(z)

1.1

R

W/D=10 R/d=5/3

RS

1.0

V d

L

2D

x(r)

0.9

0.8 0

2

x/d, r/d

4

6

Fig. 6.9-1 Electric field on the surface of an electrode consisting of a flat part connected to a semicircular edge

6.9 Comparison of Electric Fields Between Two-Dimensional … Table 6.9-1 Maximum electric field (normalized) for the configuration shown in Fig. 6.9-1

R/d

2D

127 AS

1 1.24 1.27 5/3 1.15 1.17 2 1.12 1.14 2D two-dimensional; AS axisymmetric. W/d (electrode diameter divided by gap length) = 10

The numerically calculated field strength on the electrode surface is shown on the right side of this figure for R/d = 5/3 and W/d = 10, where W is the electrode diameter, equal to 2(L + R), and d is the gap length. The electric field E in the figure is normalized by E0 (=V/d). Table 6.9-1 summarizes the maximum electric field on the semicircular edge of the 2D (two-dimensional) and AS (axisymmetric) configurations. The field is normalized with V/d. Although the radius R of the semicircle is rather large, it still causes an increase of the field larger than 10% on the edge, and this increase is more conspicuous in an axisymmetric profile than in a corresponding two-dimensional one.

6.10

Axisymmetric Uniform Field-Forming Electrodes in Practice (in Answer F3)

A practical uniform field-forming electrode always consists of a central plane smoothly connected to a rounded edge. In Sect. 6.9, it is explained that the simple profile shown in Fig. 6.9-1 produces a considerably higher electric field on the edge than in the central part. About ten electrode profiles, each of which is named after its proposer, have been proposed and utilized in high-voltage engineering and laser technology. As shown in Fig. 6.10-1, two configurations are basically possible: a vertically non-symmetric configuration with an electrode facing a grounded conductor plane and a symmetrical configuration with two identical electrodes facing each other. Table 6.10-1 gives the maximum electric field in three typical uniform field-forming profiles for W/d = 10 where W is the maximum width of an electrode and d is its gap length, as shown in Fig. 6.10-1. The field is normalized with V/d, that is, the field in the central part. It is evident that all of these profiles produce a slightly higher field on its edge surface than in the center. At present, the profiles for axisymmetric uniform field electrodes can be determined by numerical field calculations. Such an approach for a so-called axisymmetric Borda-profile has been published (Ref. 6.10-1). Furthermore it is worth noting that even if the field strength is constant or the same on the electrode surface, the gap length is different along each line of electric force in the center and on the edge. This may produce different discharge inception

128

6 Supplementary Explanations

y (z) y (z)

W/ 2 V

W/ 2

2d

V

V

d

x (r) 2D(x,y) or RS(r,z) (a) Non-symmetric

(b) Symmetric

Fig. 6.10-1 Axisymmetric uniform field-forming electrodes

Table 6.10-1 Maximum electric field on the edge of three axisymmetric uniform field-forming electrodes

Profile Rogowski (p/2) Chang Ernst (with two parameters) W/d (electrode diameter divided by gap

Electric field 1.005 1.005 1.085 length) = 10

or breakdown voltages for each point of the surface. Numerical optimization of an electrode profile which takes into account the discharge inception voltage has been also reported (Refs. 6.10-2 and 6.10-3). Note: Because the original Rogowski and Chang profiles extend to infinity, W for these electrodes is taken as twice the radius at which the electric field decreases to 99% of the value at the center of the electrode in each two-dimensional profile.

6.11

Application of Gauss’s Theorem (in Answer G1)

This theorem states that the total sum of electric flux density D integrated on a closed surface of a domain is equal to the total charge Q existing inside the domain. This means that

6.11

Application of Gauss’s Theorem (in Answer G1)

129

I D ds ¼ Q:

ð6:11-1Þ

Figure G1-1 shows a small closed surface with side lengths DS and DL that encircles a small portion of a single sheet of charge with charge density r in a medium with dielectric constant e0. Because electric field E exists only perpendicular to the surface, 2e0 EDS ¼ r DS

ð6:11-2Þ

E ¼ r=ð2e0 Þ

ð6:11-3Þ

This results in

For a conductor surface with charge density r, the same procedure gives e0 E DS ¼ r DS

ð6:11-4Þ

because the electric field is zero inside the conductor. Therefore, the electric field E on the conductor surface is E ¼ r=e0

6.12

ð6:11-5Þ

Electric Field in a Uniform Ion Flow Field (in Answer G3)

Applying Poisson’s equation to uniform ion flow in a one-dimensional configuration, as shown in Fig. G3-1, electric field E is given as dE q j ¼ ¼ ; dx e0 e0 lE

ð6:12-1Þ

dE2 2j ¼ e0 l dx

ð6:12-2Þ

that is

where l is the ion mobility and q is the charge density. As mentioned in the Detailed Explanation to Answer G3, current density j is constant everywhere in the gap. Solving Eq. (6.12-2) under the boundary condition that the potential is 0 at x = 0 and V at x = d gives

130

6 Supplementary Explanations

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi2  E x 1þ 1  n ¼ n þ E0 d 2

ð6:12-3Þ

2jd ; E0 ¼ V=d. e0 lE02 Evidently, when ion flow is present, the expression for the electric field is exceptionally complicated even for the simplest case of uniform one-dimensional flow. Figure 6.12-1 shows the field distribution in the gap for various values of n. In Eq. (6.12-3), E increases monotonously with x. The deviation of E from E0 is largest on the electrode surfaces and smallest in the middle of the gap. The value of n is roughly 0.4 when the largest deviation of the field from E0 is 10%. This gives the threshold value of current density for field distortion as where n ¼

jq ¼ 0:2e0 lE02 =d ¼ 0:2e0 l V 2 =d 3 :

ð6:12-4Þ

When n is small, Eq. (6.12-3) is approximated as   E x 1  : ¼ 1þn E0 2d 4

ð6:12-5Þ

This approximate equation also gives n = 0.4 for E = 1.1E0 at x = d and E = 0.9E0 at x = 0.

Fig. 6.12-1 Normalized electric field involving unipolar ion flow between parallel-plane electrodes

= 0.4

110

E/E 0 (%)

0.2 0.1 0.01 0

100

90 0

0.5 x/d

1

6.13

6.13

Estimation of Threshold Space Charge in a Non-uniform Field …

131

Estimation of Threshold Space Charge in a Non-uniform Field Condition (in Answer G3)

In highly non-uniform conditions, corona discharge occurs in high-field regions that generate space charge. As a simple non-uniform field, we take a coaxial configuration consisting of two cylinders with radius R1 and R2 (R1 < R2). Poisson’s equation in cylindrical coordinates (r, z) is 1 @ ðrE Þ q ¼ r @r e0

ð6:13-1Þ

This equation can be solved under the condition that the current j per unit length (2prj = 2prqlE) is constant in the gap. However, the resulting equation is extremely complicated. In method (2), mentioned in the Detailed Explanation to Answer G3, the amount of space charge is compared with the charge on the electrode surface. The threshold value qq of charge density for field distortion is given from the condition that the space charge is 10% of the electrode charge. In the coaxial configuration mentioned above, this condition gives   qq p R22  R21 ¼ 0:2pe0 V= lnðR2 =R1 Þ

ð6:13-2Þ

where V is the applied voltage. Thus, when R1  R2,   qq ¼ 0:2e0 V= R22 lnðR2 =R1 Þ :

ð6:13-3Þ

The corresponding current density is n o jq ¼ 0:2e0 l V 2 = R1 ðR2 lnðR2 =R1 ÞÞ2

ð6:13-4Þ

on the surface of the inner electrode. Exemplary values are qq = 4  10−13 C/cm3 and jq = 2  10−8 A/cm2 for R1 = 1 mm, R2 = 10 cm and V = 10 kV. Ion mobility l is taken as 2  10−4 m2/ Vs in atmospheric air. Compared with the corresponding values in parallel-plane electrodes (explained in Answer G3) with the same applied voltage, 10 kV, and the same gap length, 10 cm, qq is of much the same order, but jq is several times larger in non-uniform coaxial electrodes. This is caused by the very narrow inner conductor and the resulting high field there.

132

6.14

6 Supplementary Explanations

Green’s Reciprocity Theorem (in Answers H1, H3 and I2)

This theorem expresses the relationship between the electric potential P and charge Q for two different states in configurations with n conductors. When we denote the two different states at identical positions with subscripts A and B, the following relationship holds, n X

PAi QBi ¼

n X

i¼1

PBi QAi :

ð6:14-1Þ

i¼1

As a simple example, we consider Fig. 6.14-1 in which point charge Q is present between two parallel-plane conductors. We denote three positions and two states as follows: Positions 1 and 3: two plane conductors Position 2: position of Q State A: the two plane conductors are grounded, and Q is present State B: conductor 1 is at potential V, conductor 3 is grounded and Q is absent. These conditions of electric potential and charge are summarized in Table 6.14-1. Applying Eq. (6.14-1), we can derive QA1 V þ Q ðd  xÞ V=d ¼ 0:

ð6:14-2Þ

As a result, QA1 ¼ ðd  xÞ Q=d;

Fig. 6.14-1 Point charge Q between two parallel-plane conductors

QA3 ¼ x Q=d:

ð6:14-3Þ

6.14

Green’s Reciprocity Theorem (in Answers H1, H3 and I2)

Table 6.14-1 Electric potential and charge for two states in Fig. 6.14-1

State State State State

A A B B

133

Conductor

1

2

3

Potential Charge Potential Charge

0 QA1 V QB1

PA2 Q (d − x)V/d 0

0 QA3 0 QB3

In composite dielectric configurations, both polarization charge and true charge may exist, but only the true charge should be considered in Eq. (6.14-1) (Ref. 6.14-1).

6.15

Ion Current in Concentric-Sphere Configuration (in Answer H2)

It is explained in Answer A3 that the charge on the surface of a sphere induced by an ion with charge Q outside the sphere is equal to its image charge. When Q is at a distance L from the center of the inner sphere, its image charge Qi is Qi ¼ RQ=L

ð6:15-1Þ

if the outer sphere is sufficiently large. From this, the induced current is   I ¼ dQi =dt ¼  RQ=L2 ðdL=dtÞ:

ð6:15-2Þ

In this equation, dL/dt is the drift velocity of Q, that is, lE, where ion mobility l is assumed to be independent of the electric field. The electric field at distance L near the inner sphere is E ¼ RV=L2 :

ð6:15-3Þ

The distance from the inner sphere surface is taken as x (=L − R). I ¼ l R2 VQ=ðR þ xÞ4 :

ð6:15-4Þ

In contrast to the current in parallel-plane electrodes, which is constant (i.e., independent of x), the current in a non-uniform configuration depends highly on x. When R is small, the current is particularly high when the ion is near the surface of the inner electrode. The current waveform as a function of time t is obtained from the relationship between x and t by integrating v = dx/dt = lE  4=3 I ¼ l R2 VQ= 3lRVt þ R3

ð6:15-5Þ

134

6 Supplementary Explanations

Fig. 6.16-1 Electrostatic capacitances between a probe and DS CP

CPS

CS

6.16

Electrostatic Capacitances in Fig. I2-1

Because Fig. I2-1 constitutes a static field, its potentials and charges can be expressed by electrostatic capacitances. Figure 6.16-1 shows such capacitances connecting a probe sensor, a small surface area DS, and the ground. If the probe is grounded, the charge DP which DQ on DS induces on the sensor is DP ¼ 

CPS DQ CPS þ CS

ð6:16-1Þ

These capacitances are easy to understand from Fig. 6.16-1. The potential ФS of DS when voltage V is applied to the probe sensor is US ¼

CPS V: CPS þ CS

ð6:16-2Þ

The coefficient on the right-hand side of Eq. (6.16-1) corresponds to the (negative) k function as shown in Eq. (I2-1), which also conforms with the coefficient of Eq. (6.16-2).

6.17

Refraction Law for a Line of Electric Force (in Answer J1)

As explained in Sect. 5.3, two boundary conditions hold on the interface of two dielectrics e1 and e2. One is the continuity of electric potential Ф, or, equivalently, the continuity of tangential field strength Et, E1t ¼ E2t

ð6:17-1Þ

6.17

Refraction Law for a Line of Electric Force (in Answer J1)

135

E 2t

Fig. 6.17-1 Refraction of a line of electric force on the interface of two dielectrics

E 2n

E 1n

E 1t

where subscripts 1 and 2 indicate the corresponding media or dielectrics e1 and e2. The other condition is the continuity of the normal component Dn of the electric flux density (Fig. 6.17-1), D1n ¼ D2n :

ð6:17-2Þ

e1 E1n ¼ e2 E2n :

ð6:17-3Þ

This can be rewritten as

From these formulae, E1t E1n

6.18



E2t tan h1 e1 ¼ ¼ E2n tan h2 e2

ð6:17-4Þ

Analysis of the Field in Fig. J1-1 by the Variable Separation Method

(1) Equation for exponent n We consider the configuration shown in Fig. J1-3 in two-dimensional polar coordinates (r, h), where the origin of the coordinate axes is contact point P. This analysis is similar to that explained in Question D1, but here we consider electric potentials U1 and U2, respectively, in two dielectrics e1 and e2.

136

6 Supplementary Explanations

By applying the variable separation method to Fig. J1-3, the series solution that satisfies Laplace’s equation is U1 ; U2 ¼

1 X

r n ðan cos nh þ bn sin nhÞ:

ð6:18-1Þ

n¼1

The boundary conditions are as follows: (a) U1 ¼ 0 at h = 0, (b) U2 ¼ 0 at h = p, and (c) U1 ¼ U2 ; e1 ð@U1 =@hÞ ¼ e2 ð@U2 =@hÞ at h = h0. From boundary conditions (a) and (b), U1 ¼ U2 ¼

1 X n[0 1 X

an r n sin nh ð6:18-2Þ bn r sin nðp  hÞ n

n[0

From boundary condition (c), the following transcendental equation is derived: e2 tan nh0 þ e1 tan nðp  h0 Þ ¼ 0:

ð6:18-3Þ

Similarly as in the case of a wedge-shaped conductor, the smallest positive solution of n, nmin, is predominant near P. (2) Solution of exponent n We introduce the following function to examine the solution of n in Eq. (6.18-3). f ðnÞ ¼ ðe2 =e1 Þ tanðnh0 Þ þ tan nðp  h0 Þ:

ð6:18-4Þ

Figure 6.18-1 shows the first and second terms of this function in relation to n when 0 < h0 < p/2. The first term (e2/e1)tan(nh0) changes from positive infinity to negative infinity at n1 = p/(2h0) where n1 > 1. In contrast, the second term, tan n (p − h0), changes from negative infinity to positive infinity at n2 = p/[2(p − h0)], where n2 lies between 1/2 and 1. From these facts, it is evident that the function f (n) changes from positive infinity to negative infinity at n2 and increases monotonously in the range from n2 to n1. However, when n = 1, f (n) = (e2/e1 − 1)tanh0, which is positive if e2 > e1. In conclusion, Eq. (6.18-4) has a solution for n between 1/2 and 1 when e2 > e1. When e2 < e1, on the other hand, f (1) is negative, which means that the smallest solution is larger than 1.

6.19

Contact of Three Dielectrics with Straight Interfaces …

137

n1 = π/(2θ 0)

0

1

n2 1

n1 2

2n 2 2

2n 1 4

3

4

n

n

n2 = π/(2(π−θ0)) Fig. 6.18-1 Characteristics of (e2/e1)tan nh0, tan n(p − h0)

6.19

Contact of Three Dielectrics with Straight Interfaces (in Answer J1)

The field behavior for the contact of three dielectrics with straight interfaces is explained in detail in Refs. 6.19-1 and 2. Figure 6.19-1 shows a two-dimensional configuration in which three dielectrics with relative permittivity e1, e2, and e3 meet at central point P. The configuration of Fig. 6.19-1 includes cases of a medium (dielectric), such as a vacuum with dielectric constant e0 and a conductor with infinitely high dielectric constant. Consequently, Fig. J1-1 in Question J1 is also covered by Fig. 6.19-1. In short, the electric potential and field are expressed as functions of rn and rn−1, respectively, near P, where r is the distance from P. Exponent n is the smallest positive solution to the following equation:

138

6 Supplementary Explanations

Fig. 6.19-1 Contact of three dielectrics with straight interfaces

1

1 2 2

3

2    e1 þ e22 e3 sin A sinðA  BÞ cosðB  CÞ þ e21 þ e23 e2 sin A cosðA  BÞ sinðB  CÞ    e22 þ e23 e1 cos A sinðA  BÞ sinðB  CÞ  2e1 e2 e3 f1  cos A cosðA  BÞ cosðB  CÞg ¼ 0: ð6:19-1Þ In this equation, A = nh1, B = nh2, and C = 2pn. Angles h1 and h2 are given in Fig. 6.19-1. An alternative equation is given in Refs. 6.19-1 and 2, i.e., a linear combination of cosine functions, which is seemingly quite different from Eq. (6.19-1); however, the two equations are equivalent.

6.20

Analysis of the Configuration in Fig. J2-1 Using the Variable Separation Method

This analysis is for a wedge-shaped interface between two dielectrics e1 and e2 with opening angle ha, as shown in Fig. 6.20-1. Unlike the former analysis in Sect. 6.18, no conductor plane exists in this configuration. We consider electric potentials U1 and U2, respectively, in dielectrics e1 and e2 in two-dimensional polar coordinates (r, h). By applying the variable separation method to Fig. 6.20-1, an infinite series solution that satisfies Laplace’s equation is given in the same form as the aforementioned Eq. (6.18-1) as U1 ; U2 ¼

1 X

r n ½an cosðnhÞ þ bn sinðnhÞ

ð6:20-1Þ

n¼1

To ensure a finite value of the potentials near tip P, the exponent n must not be negative, and the smallest positive value of n is predominant near P. If we use such a value of n, the above potentials can be approximated as

6.20

Analysis of the Configuration in Fig. J2-1 Using the Variable …

Fig. 6.20-1 A wedge-shaped interface between two dielectrics in two-dimensional polar coordinates

139

Φ2

(ε2)

Φ1

(ε1)

U1 ¼ r n ½an cosðnhÞ þ bn sinðnhÞ U2 ¼ r n ½cn cosðnhÞ þ dn sinðnhÞ:

ð6:20-2Þ

The boundary conditions are as follows: U1 ¼ U2 ; e1 ð@U1 =@hÞ ¼ e2 ð@U2 =@hÞ

at h ¼ 0 and h ¼ ha :

Applying these conditions leads to simultaneous linear equations for constants an, bn, cn, and dn as unknowns. For these unknowns to have nonzero solutions, the determinant of the j4  4j matrix composed of the corresponding coefficients for an, bn, cn, and dn must be zero. This procedure gives the following equation for n: 

 e21 þ e22 sin A sinðA  BÞ  2e1 e2 f1  cosðA  BÞg ¼ 0 A ¼ nha ; B ¼ 2pn

ð6:20-3Þ

However, if we take the plane that bisects angle ha, as shown in Fig. 6.20-2, the above formula assumes a much simpler form: tan nðha =2  pÞ e1 e2 ¼ ; or ¼ : tanðn ha =2Þ e2 e1

ð6:20-4Þ

This formula is the same in principle as Eq. (J3-1) in Question J3 or Eq. (6.18-1) in Sect. 6.18.

Fig. 6.20-2 Equivalent configuration to Fig. 6.20-1

ε1

140

6.21

6 Supplementary Explanations

Analysis of the Field for a Cone-Shaped Dielectric Interface (in Answer J2)

We consider the axisymmetric coordinates (r, h), as shown in Fig. J2-5, where the z-axis is the axis of rotation, and U1 and U2 are the electric potentials in dielectrics e1 and e2, respectively. By applying the variable separation method to the configuration shown in Fig. J2-5, an infinite series solution that satisfies Laplace’s equation is given as 1 X

U1 ; U2 ¼

r n ðan cos nh þ bn sin nhÞ:

ð6:21-1Þ

n¼1

The boundary conditions are U1 ¼ U2 ; e1 ð@U1 =@hÞ ¼ e2 ð@U2 =@hÞ at h ¼ h0 : Because U1 and U2 respectively have no singularity at h = 0 and p (180°), they can be given as U1 ¼ U2 ¼

X X

an r n Pn ðcos h0 Þ bn r n Pn fcosðp  h0 Þg ¼

X

bn r n Pn ð cos h0 Þ

ð6:21-2Þ

where Pn is the Legendre function of order n. Eliminating the coefficients from the boundary conditions at h = h0 and expressing the derivative of Pn with a recurrence equation, the following equation is derived for n: e1 Pn ð cos h0 Þfcos h0 Pn ðcos h0 Þ  Pn1 ðcos h0 Þg ¼ e2 Pn ðcos h0 Þfcos h0 Pn ð cos h0 Þ þ Pn1 ðcos h0 Þg

ð6:21-3Þ

From this equation, the smallest positive solution of n, nmin, is obtained. The electric field near the cone tip behaves as rm, where m = nmin − 1. More details are explained in Ref. 6.21-1, which also gives values of m in relation to angle h0. The results are summarized for h0 < p/2 as follows. (a) The condition e2 < e1 means a (solid) dielectric projection, resulting in nmin < 1 and m < 0. As a result, the electric field is infinitely high at P. (b) The condition of e2 > e1 means a (sharp) dielectric pit where nmin > 1 and m > 0. As a result, the electric field is zero at P. The value of m is different for two-dimensional and axisymmetric cases, even if e2/e1 and angle h0 are the same in both cases. In the extreme condition of e1  e2, m = –1/2 in the two-dimensional case, which corresponds to a thin foil conductor. In the axisymmetric case, in contrast, m = –1, which corresponds to a thin needle conductor.

6.22

6.22

Analysis of the Axisymmetric Configuration in Question J4 …

141

Analysis of the Axisymmetric Configuration in Question J4 Using the Successive Image Charge Method

We consider the configuration of Fig. 6.22-1 for a solid dielectric of infinitely large thickness. The ratio ed/eg is the only major parameter in the computation. The configuration is axisymmetric with the origin of the coordinate axes z = 0 serving as the contact point. The electric field can be obtained by placing image charges alternately inside the spherical conductor and inside the solid dielectric, as follows. First, the boundary condition for an isolated spherical conductor is satisfied by placing a point charge of magnitude q ¼ 4peg RV at z = R. Second, to satisfy the boundary condition on the surface of the dielectric, a point charge of quantity −Pq is placed at z = −R, where P = (ed − eg)/(ed + eg). This procedure of placing image charges successively is continued to satisfy the two boundary conditions. Table 6.22-1 lists a series of image charges and their corresponding locations. The resulting electric field Egc in the gas at contact point P is ( V E gc ¼ R

1  X n¼1

nP

n1



þ

1 X n¼1

) ðnPÞ

¼

  V es es þ 1  R 2

ð6:22-2Þ

where es = ed/eg. As described in Answer J4, since no gas gap exists at P, Egc is the limiting value as P is approached.

Fig. 6.22-1 Spherical electrode on a solid dielectric plane (with infinitely large thickness d in Fig. J4-1)

z Φ =V R g

Z = 0 (P) d

142

6 Supplementary Explanations

Table 6.22-1 Disposition of image charges for Fig. 6.22-1 (spherical conductor on a solid dielectric plane) Inside the sphere

Position (z) Charge Inside the dielectric Position (z) Charge q ¼ 4peg RV, P = (ed − eg)/(ed + eg)

R q –R –Pq

R/2 Pq/2 –R/2 –P2q/2

R/3 P2q/2 –R/3 –P3q/3

   

R/n Pn−1q/n –R/n –Pnq/n

Evidently, the electric field Edc in ed at the contact point is Edc ¼

6.23

V es þ 1  : R 2

ð6:22-3Þ

Analysis of the Two-Dimensional Case in Question J4 Using the Successive Image Charge Method

If Fig. J4-1 is considered to be a two-dimensional configuration in (x, y) coordinates in which a cylindrical conductor lies on a solid dielectric plate of thickness D, a series of line charges can be applied instead of point charges. The potential V of the cylindrical conductor is given by the following line charge of density k when D is sufficiently large, k ¼ 2peg V= lnfð2D þ RÞ=Rg:

ð6:23-1Þ

Table 6.23-1 lists a series of these image charges and their corresponding locations. These image charges successively satisfy the boundary conditions on the surfaces of the cylindrical conductor and the solid dielectric. As a result, the electric field Egc in the gas at the contact point P is Egc ¼

  1

es es þ 1 1 X n n k :  Pn1 k  þ Pn k  ¼ 2 2peg n¼1 R R 2peg R

ð6:23-2Þ

Table 6.23-1 Disposition of image charges for a cylindrical conductor on a solid dielectric plane Inside the cylinder

Position (z) R Charge k Inside the dielectric Position (z) –R Charge –Pk k ¼ 2peg V= lnfð2D þ RÞ=Rg, P = (ed − eg)/(ed + eg)

R/2 Pk –R/2 –P 2k

R/3 P2k –R/3 –P3k

   

R/n Pn−1k –R/n –Pnk

6.23

Analysis of the Two-Dimensional Case in Question J4 …

143

In this equation, P = (ed − eg)/(ed + eg), just as in Sect. 6.22. When es = 1, i.e., the configuration without the solid dielectric, the electric field Ec at P is V/R for a spherical conductor and k/(2pegR) for a cylindrical conductor. Thus, the electric field at P is given as Egc ¼ es ðes þ 1ÞEc =2

ð6:23-3Þ

Edc ¼ ðes þ 1ÞEc =2

for both a spherical conductor and a cylindrical conductor lying on a solid dielectric, where es = ed/eg.

6.24

Contact-Point Electric Fields (in Answer J4)

Figure 6.24-1(a) and (b) show some typical values for electric fields in the ed (solid dielectric) side at the contact point as a function of es (=ed/eg). The electric field on the vertical axis is normalized with Ec, the field for es = 1, i.e., the value without a solid dielectric. Some plots of Eq. (J4-5) with the corresponding values of k are also included in the figures. For identical cross-sectional profiles, the value of k is larger in axisymmetric configurations than in two-dimensional cases, which indicates a more rapid increase

3

k=0.77

2

E0

1E

k =0

0

0

2

4

εs

R0

4

R/R 0 1/2

6

(a) Axisymmetric configurations

8

10

Edc /Ec

Edc /Ec

k=1

R0

4

k=1

3

k=0.60

2 1E 0

D

D=5R 0

E0 2

4

εs

6

8

k=0

10

(b) Two-dimensional configurations

Fig. 6.24-1 Normalized electric field at the contact point in the solid dielectric (ed) side (es = ed/eg)

144

6 Supplementary Explanations

in the electric field as the contact point is approached in axisymmetric configurations. The hatched area in Fig. 6.24-1(a) indicates the range of the electric field at the contact point (Ref. 6.24-1).

6.25

Hemispherical Solid Dielectric on a Grounded Plane Under a Uniform Field (in Answers K5 and M5)

(1) Analytical solutions of the electric potential Figure 6.25-1 shows a hemispherical solid dielectric with conductivity lying on a grounded conductor plane under a uniform field. The electric field for this configuration can be expressed as an analytical formula if the conductivity is constant everywhere inside the solid. The conductivity is zero in the air outside the solid. If the subscripts 1 and 2 indicate the region outside and inside the solid, respectively, the potentials are given in polar coordinates (r, h), as follows: outside the solid (e1 side)   3 1 e2  e1 2 R E0 cos h U1 ¼ rE0 cos h þ 1þ T 1 þ T2 r2 2e1 þ e2 3 3e1 T R E0 cos h þj   2 r2 2e1 þ e2 1 þ T inside the solid (e2 side) U2 ¼

Fig. 6.25-1 Hemispherical solid dielectric with conductivity on a grounded plane under a uniform field

3e1 T  ðT þ jÞrE0 cos h 2e1 þ e2 1 þ T 2

ð6:25-1Þ

Hemispherical Solid Dielectric on a Grounded Plane …

6.25

145

where q is the volume resistivity of the solid. Parameter T is equal to xqð2e1 þ e2 Þ, where x = 2pf (f is the AC frequency), and R is the radius of the hemisphere. The electric fields can be obtained by differentiating these potentials. It should be noted that the electric field is uniform in the z direction everywhere inside the solid, even if it has conductivity. If surface conduction exists instead of volume conduction in Fig. 6.25-1, the potential can be obtained merely by making T ¼ xqs R ð2e1 þ e2 Þ=2 in Eq. (6.25-1), where qs is the surface resistivity. In the configuration shown in Fig. 6.25-1, qsR/2 results in an identical field distribution as that for volume resistivity q. (2) Electric field on the boundary for high conductivity When the conductivity is very high, i.e., q is so small that conduction is predominant in the solid, the electric fields on the boundary are given as follows: Normal component outside the solid ðe1 -sideÞ: E1n ¼ 3E0 cos h inside the solid ðe2 -sideÞ: E2n ¼ j  3e1 x qE0 cos h Tangential component: E1t ¼ E2t ¼ j  3e1 x qE0 sin h:

ð6:25-2Þ

In air, e1 is equal to e0. When e0xq is much smaller than 1, as in Question M5, E1n is much larger than E1t, E2n, or E2t. However, E2t is not zero, but is comparable in size with E1t or E2n. In fact, the electric field inside the solid in Fig. 6.25-1 is uniform and constant, as can be seen in Eqs. (6.25-1) and (6.25-2). This means that the difference between the normal component En and the tangential component Et of the electric field or current density is only that of the sine and cosine functions inside the solid dielectric.

6.26

Examples of Induced Charge and Floating Force (in Answer L1)

Given below are the induced charge Q and floating force F for the three conductor shapes shown in Fig. 6.26-1. (a) Sphere [Fig. 6.26-1(a)] Q ¼ 6:58pe0 R2 E0 ;

F ¼ 5:48pe0 R2 E02

ð6:26-1Þ

This was first derived by N.-J. Felici by using the coordinate transformation method (Ref. 6.26-1).

146

6 Supplementary Explanations

E0

E0

(a) Sphere

(b) Hemisphere

(radius R )

E0

(c) Thin plate

(radius R )

(area S )

Fig. 6.26-1 Three conductor shapes (particles) on a grounded conductor plane under a uniform field

(b) Hemisphere [Fig. 6.26-1(b)] Q ¼ 3pe0 R2 E0 ; F¼ 2:25p e0 R2 E02

ð6:26-2Þ

The calculation process is summarized in Sect. 6.27. (c) Thin sheet (foil) [Fig. 6.26-1(c)] Q ¼ e0 SE0 ;

F ¼ e0 SE02 =2

ð6:26-3Þ

The field distortion caused by the finite thickness of the sheet is neglected. The ratio of the floating force to E0Q is 0.83, 0.75, and 0.5, for (a), (b), and (c), respectively. This means that the more remote the center of gravity of a conductor is from the ground, the closer the ratio is to 1.

6.27

Induced Charge and Floating Force for a Hemispherical Conductor on a Grounded Plane (in Sect. 6.26)

The configuration shown in Fig. 6.27-1 is the same as that for a hemispherical conductor existing under a uniform field E0 shown in Fig. 6.26-1(b). This field is given in the polar coordinates (r, h, u). From symmetry, the coordinate u need not be considered. It is not difficult to confirm that the following is the electric potential Ф in (r, h) coordinates;   R3 U ¼ 1  3 rE0 cos h r

ð6:27-1Þ

6.27

Induced Charge and Floating Force for a Hemispherical …

147

Fig. 6.27-1 Diagram for computing the induced charge and floating force

θ

This is because Ф satisfies Laplace’s equation as well as the boundary conditions of Ф = 0 on the sphere surface and a uniform field for a sufficiently large r. Differentiating Ф on the sphere surface gives the charge density r at r = R as r ¼ e0 E ¼ e0 @U=@r ¼ 3e0 E0 cos h

ð6:27-2Þ

This is the induced charge density at angle h on the hemispherical surface in Fig. 6.27-1. Integrating the charge density on the total hemisphere surface gives Q ¼ 3pe0 R2 E0

ð6:27-3Þ

The force exerted on a unit area is DF ¼ e0 En2 =2 ¼ r2 =ð2e0 Þ. Integrating the force in the upward direction DF cos h on the total hemisphere surface, we get F ¼ 2:25pe0 R2 E02

6.28

ð6:27-4Þ

Force on a Charged Spherical Shell Above a Grounded Plane (in Answer L3)

Figure L3-1(b) is electrically equivalent to Fig. L3-1(a) because the grounded plane has been replaced by a spherically symmetric image charge of the same radius with total charge −Q. Therefore, the electric field is the superposition of the symmetric fields caused by two single point charges Q and −Q located at the center of each sphere.

148

6 Supplementary Explanations

The force for Fig. L3-1(b) is calculated by superposing (or integrating) the force exerted on each small element of surface charge (charge density r) over the whole surface. Two important points should be considered: (a) The force exerted by the point charge on its own spherical shell is everywhere symmetric with only radial components remaining, and these cancel each other on the total sphere surface. As a result, we have to consider only the force exerted by the other (opposite) spherical or point charge. (b) As described above, the electric field E is the field resulting from charge Q (or −Q) located at the center of each sphere. At distance L, the field can be expressed as

E¼

1 Q  2; 4pe0 L

ð6:28-1Þ

where L2 ¼ ðW þ 2R  R cos hÞ2 þ ðR sin hÞ2 ¼ ðW þ 2RÞ2 þ R2  2RðW þ 2RÞ cos h and W is equal to 2H. Height H and radius R are given in Fig. L3-1. The charge element DQ on the annulus in the range Dh at angle h is DQ ¼ r R Dh  2pR Dh

ð6:28-2Þ

where r ¼ Q=ð4p R2 Þ. We have to consider the force in the z-direction only, because the other components (perpendicular to the z-direction) cancel out over the whole surface. Therefore, the resulting (attractive) force DF is expressed as DF ¼ DQ  E  2p R sin h  cos a

ð6:28-3Þ

where cos a ¼ ðW þ 2R  R cos hÞ=L. The final expression is Q2 ðW þ 2R  R cos hÞ sin h dh  8pe0 L3=2 2 Q ðW þ 2R  R cos hÞ sin h dh ¼ h i3=2 8p e0 ðW þ 2RÞ2 þ R2  2RðW þ 2RÞ cos h

DF ¼

ð6:28-4Þ

Force on a Charged Spherical Shell Above a Grounded Plane …

6.28

149

We can readily integrate this by converting h to t as t ¼  cos h. The indefinite integral form Z DF ¼ h

2

h

8