Electromagnetism 9781681744292, 9781681744285, 9781681744315

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Electromagnetism Problems and solutions

Electromagnetism Problems and solutions Carolina C Ilie State University of New York at Oswego, USA

Zachariah S Schrecengost State University of New York at Oswego, USA

Morgan & Claypool Publishers

Copyright ª 2016 Morgan & Claypool Publishers All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publisher, or as expressly permitted by law or under terms agreed with the appropriate rights organization. Multiple copying is permitted in accordance with the terms of licences issued by the Copyright Licensing Agency, the Copyright Clearance Centre and other reproduction rights organisations. Rights & Permissions To obtain permission to re-use copyrighted material from Morgan & Claypool Publishers, please contact [email protected]. ISBN ISBN ISBN

978-1-6817-4429-2 (ebook) 978-1-6817-4428-5 (print) 978-1-6817-4431-5 (mobi)

DOI 10.1088/978-1-6817-4429-2 Version: 20161101 IOP Concise Physics ISSN 2053-2571 (online) ISSN 2054-7307 (print) A Morgan & Claypool publication as part of IOP Concise Physics Published by Morgan & Claypool Publishers, 40 Oak Drive, San Rafael, CA, 94903 USA IOP Publishing, Temple Circus, Temple Way, Bristol BS1 6HG, UK

To my family, my mentors, and my students — CCI To my friends, family, and mentors — ZSS

Contents Preface

x

Acknowledgements

xii

About the authors

xiii

1

Mathematical techniques

1-1

1.1

Theory 1.1.1 1.1.2 1.1.3 1.1.4 1.1.5 1.1.6 1.1.7 1.1.8 1.1.9 1.1.10 1.1.11 1.1.12

1-1 1-1 1-1 1-2 1-2 1-2 1-3 1-3 1-4 1-4 1-4 1-4 1-4

2

Electrostatics

2-1

2.1

Theory 2.1.1 2.1.2 2.1.3 2.1.4 2.1.5

2-1 2-1 2-1 2-2 2-2 2-2

1.2

Dot and cross product Separation vector Transformation matrix Gradient Divergence Curl Laplacian Line integral Surface integral Volume integral Fundamental theorem for gradients Fundamental theorem for divergences (Gauss’s theorem, Green’s theorem, divergence theorem) 1.1.13 Fundamental theorem for curls (Stoke’s theorem, curl theorem) 1-4 1.1.14 Cylindrical polar coordinates 1-4 1.1.15 Spherical polar coordinates 1-5 1.1.16 One-dimensional Dirac delta function 1-5 1.1.17 Theory of vector fields 1-5 Problems and solutions 1-5 Bibliography 1-34

Coulomb’s law Electric field Gauss’s law Curl of E ⃗ Energy of a point charge distribution

vii

Electromagnetism

2.2

2.1.6 Energy of a continuous distribution 2.1.7 Energy per unit volume Problems and solutions Bibliography

3

Electric potential

3.1

Theory 3.1.1 Laplace’s equation 3.1.2 Solving Laplace’s equation 3.1.3 General solutions 3.1.4 Method of images 3.1.5 Potential due to a dipole 3.1.6 Multiple expansion 3.1.7 Monopole moment Problems and solutions Bibliography

3.2

3-1 3-1 3-1 3-1 3-4 3-5 3-6 3-6 3-6 3-6 3-29

4

Magnetostatics

4.1

Theory 4.1.1 Magnetic force 4.1.2 Force on a current carrying wire 4.1.3 Volume current density 4.1.4 Continuity equation 4.1.5 Biot–Savart law 4.1.6 Divergence of B ⃗ 4.1.7 Ampère’s law 4.1.8 Vector potential 4.1.9 Magnetic dipole moment 4.1.10 Magnetic field due to dipole moment Problems and solutions Bibliography

4.2

2-2 2-2 2-3 2-35

4-1 4-1 4-1 4-1 4-1 4-2 4-2 4-2 4-2 4-2 4-3 4-3 4-3 4-26

5

Electric fields in matter

5-1

5.1

Theory 5.1.1 Induced dipole moment of an atom in an electric field 5.1.2 Torque on a dipole due to an electric field 5.1.3 Force on a dipole

5-1 5-1 5-1 5-1

viii

Electromagnetism

5.2

5.1.4 Energy of a dipole in an electric field 5.1.5 Surface bound charge due to polarization P ⃗ 5.1.6 Volume bound charge due to polarization P ⃗ 5.1.7 Potential due to polarization P ⃗ 5.1.8 Electric displacement 5.1.9 Gauss’s law for electric displacement 5.1.10 Linear dielectrics 5.1.11 Energy in a dielectric system Problems and solutions Bibliography

6

Magnetic fields in matter

6.1

Theory 6.1.1 Torque on a magnetic dipole moment 6.1.2 Force on a magnetic dipole 6.1.3 H-field 6.1.4 Linear media 6.1.5 Surface bound current due to magnetization M⃗ 6.1.6 Volume bound current due to magnetization M⃗

6.2

Problems and solutions Bibliography

5-2 5-2 5-2 5-2 5-2 5-2 5-2 5-3 5-3 5-26 6-1

ix

6-1 6-1 6-1 6-1 6-2 6-2 6-2 6-2 6-14

Preface We wrote this book of problems and solutions having in mind the undergraduate student—sophomore, junior, or senior—who may want to work on more problems and receive immediate feedback while studying. The authors strongly recommend the textbook by David J Griffiths, Introduction to Electrodynamics, as a first source manual, since it is recognized as one of the best books on electrodynamics at the undergraduate level. We consider this book of problems and solutions a companion volume for the student who would like to work on more electrostatic problems by herself/himself in order to deepen their understanding and problems solving skills. We add brief theoretical notes and formulae; for a complete theoretical approach we suggest Griffiths’ book. Every chapter is organized as follows: brief theoretical notes followed by the problem text with the solution. Each chapter ends with a brief bibliography. We plan to write a second volume on electrodynamics, which will start with Maxwell’s equations and the conservation laws, and then discuss electromagnetic (EM) waves, potentials and fields, radiation, and relativistic electrodynamics. We follow here the notation of Griffiths, and use r ⃗ for the vector from a source ⃗ point r ⃗′ to the field point r ⃗ . Please note that rˆ = rr = ∣ rr⃗ ⃗ −− rr⃗ ′⃗ ′ ∣ and, as you see, this notation already greatly simplifies complex equations, but you need to be careful with your notation, in particular if you only use cursive or typed letters. Also, we use the same notation s for the distance to the z-axis in cylindrical coordinates as is used in Griffiths’ book. The chosen units are SI units—the international system. The reader should be aware that other books may employ either the Gaussian system (CGS) or the Heaviside–Lorentz (HL) system. The Coulomb force in each of the systems is as follows, SI system:

F⃗ =

1 q1q2 rˆ 4πε0 r 2

CGS:

F⃗ =

q1q2 rˆ r2

HL:

F⃗ =

1 q1q2 rˆ 4π r 2

Some of the problems are typical practice problems with the pedagogical role of improving understanding and problem solving skills. Several of the problems presented here appear in a variety of undergraduate textbooks on EM as they are classic examples; however, we felt it would be incomplete to omit these problems as

x

Electromagnetism

they are fundamental to the study of EM. We also present problems that are more general in nature, which may be a bit more challenging. We tried to maintain a balance between the two types of problems, and we hope that the readers will enjoy this variation and have as much thrill and excitement as we had while creating and solving these problems.

xi

Acknowledgements We want to thank to Dr Ilie’s students, Nicholas Jira, Vincent DeBiase, Ian Evans, and Andres Inga, who contributed to the editing (typing) of this book. We are particularly grateful to our illustrator, Julia D’Rozario, for making all of the figures. We thank Dr Ildar Sabirianov for providing useful suggestions. We thank the administration at SUNY Oswego and the office of Research and Individualized Student Experiences for overall support. We are grateful to Dr Peter Dowben, from the University of Nebraska at Lincoln, who thought that such a project has a niche. A thought of appreciation to Dr Charles Ebner, from the Ohio State University for his perfect Electrodynamics course. Also many thanks to our editors, Joel Claypool, Publisher at Morgan & Claypool Publishers, Jeanine Burke, Consulting Editor at the IOP Concise Physics e-book program, and Jacky Mucklow, Production Team Manager at the Institute of Physics. Lastly, we thank to our families and friends for their sense of humor, encouragement, and for keeping us sane and happy.

xii

About the authors Carolina C Ilie Carolina C Ilie is an Associate Professor with tenure at the State University of New York at Oswego. She taught Electromagnetic Theory for almost ten years and designed various problems for her students’ exams, group work, and quizzes. Dr Ilie obtained her PhD in Physics and Astronomy at the University of Nebraska at Lincoln, an MSc in Physics at Ohio State University and another MSc in Physics at the University of Bucharest, Romania. She received the President’s Award for Teaching Excellence in 2016 and the Provost Award for Mentoring in Scholarly and Creative Activity in 2013. She lives in Central New York with her spouse, also a physicist, and their two sons. Photograph courtesy of James Russell/SUNY Oswego Office of Communications and Marketing.

Zachariah S Schrecengost Zachariah S Schrecengost is a State University of New York alumnus. He graduated summa cum laude with a BS degree having completed majors in Physics, Software Engineering, and Applied Mathematics. He took the Advanced Electromagnetic Theory course with Dr Ilie and was thrilled to be involved in creating this book. He brings to the project both the fresh perspective of the student taking electrodynamics, as well as the enthusiasm and talent of an alumnus who is an electrodynamics and upper level mathematics aficionado. Mr Schrecengost works as a software engineer in Syracuse and is preparing to begin his graduate school studies in physics.

Julia R D’Rozario Julia R D’Rozario (illustrator) graduated from the State University of New York at Oswego in December 2016 where she completed a BS in Physics and a BA in Cinema and Screen Studies, and completed a minor in Astronomy by May 2016. She completed the Advanced Electromagnetic Theory course with Dr Ilie and has much experience of the arts through her career in film. Ms D’Rozario contributes her knowledge of electrodynamics and her talent in drawing using Inkscape software. Her future aim is to attend graduate school and continue to combine her passions for physics and cinema.

xiii

IOP Concise Physics

Electromagnetism Problems and solutions Carolina C Ilie and Zachariah S Schrecengost

Chapter 1 Mathematical techniques

There are a variety of mathematical techniques required to solve problems in electromagnetism. The aim of this chapter is to provide problems that will build confidence in these techniques. Concepts from vector calculus and curvilinear coordinate systems are the primary focus.

1.1 Theory 1.1.1 Dot and cross product Given vectors A⃗ = Ax xˆ + Ay yˆ + Az zˆ and B ⃗ = Bxxˆ + Byyˆ + Bzzˆ

A ⃗ ⋅ B ⃗ = Ax Bx + Ay By + Az Bz = AB cos θ xˆ yˆ zˆ ⃗ ⃗ A × B = Ax Ay Az Bx By Bz

with A ⃗ × B ⃗ = AB sin θ

where A = ∣A⃗ ∣ = Ax2 + A y2 + Az2 , B = ∣B ∣⃗ = between A⃗ and B ⃗ .

Bx2 + By2 + Bz2 , and θ is the angle

1.1.2 Separation vector This notation is outlined by David J Griffiths in his book Introduction to Electrodynamics (1999, 2013). Given a source point r ′⃗ and field point r ⃗ , the separation vector points from r ′⃗ to r ⃗ and is given by

r ⃗ = r ⃗ − r ⃗′ = (x − x′)xˆ + (y − y′)yˆ + (z − z′)zˆ

doi:10.1088/978-1-6817-4429-2ch1

1-1

ª Morgan & Claypool Publishers 2016

Electromagnetism

and the unit vector pointing from r ′⃗ to r ⃗ is

rˆ =

r ⃗ − r ⃗′ (x − x′)xˆ + (y − y′)yˆ + (z − z′)zˆ r⃗ = . = r ⃗ − r ⃗′ r (x − x′)2 + (y − y′)2 + (z − z′)2

As explained by Griffiths, this notation greatly simplifies later equations. 1.1.3 Transformation matrix Given vector A⃗ = Ax xˆ + Ay yˆ + Az zˆ in coordinate system K, the components of A⃗ in coordinate system K ′ are determined by rotational matrix R given by ⎛ Rxx Rxy Rxz ⎞ ⎜ ⎟ R = ⎜ Ryx Ryy Ryz ⎟ ⎜ ⎟ ⎝ Rzx Rzy Rzz ⎠ with

⎛ ⎞ ⎛ Ax ⎞ ⎜ Ax′ ⎟ ⎜ A ′ ⎟ = R ⎜ Ay ⎟ . ⎜⎜ ⎟⎟ ⎜ y⎟ ⎝ Az ⎠ ⎜ ′⎟ ⎝ Az ⎠ 1.1.4 Gradient Given a scalar function T , the gradients for various coordinate systems are given below. Cartesian

∇T =

∂T ∂T ∂T zˆ yˆ + xˆ + ∂x ∂y ∂z

Cylindrical

∇T =

1 ∂T ˆ ∂T ∂T sˆ + zˆ ϕ+ s ∂ϕ ∂z ∂s

Spherical

∇T =

1 ∂T ˆ 1 ∂T ˆ ∂T rˆ + θ+ ϕ r ∂θ r sin θ ∂ϕ ∂r

1.1.5 Divergence Given vector function v ⃗ , the divergences for various coordinate systems are given below. Cartesian

∇ ⋅ v⃗ =

∂vy ∂v ∂vx + + z ∂z ∂x ∂y

1-2

Electromagnetism

Cylindrical

∇ ⋅ v⃗ =

∂v 1 ∂ 1 ∂vϕ + z ( svs ) + ∂z s ∂s s ∂ϕ

Spherical

∇ ⋅ v⃗ =

∂ 1 ∂vϕ 1 ∂ 2 1 r vr + sin θ vθ + 2 r ∂r r sin θ ∂θ r sin θ ∂ϕ

( )

(

)

1.1.6 Curl Given vector function v ⃗ , the curls for various coordinate systems are given below. Cartesian

⎛ ∂v ⎛ ∂vy ∂vy ⎞ ⎛ ∂vx ∂v ⎞ ∂v ⎞ − z ⎟yˆ + ⎜ − x ⎟zˆ ∇ × v⃗ = ⎜ z − ⎟xˆ + ⎜ ⎝ ∂z ∂x ⎠ ∂y ⎠ ∂z ⎠ ⎝ ∂y ⎝ ∂x Cylindrical

⎛ 1 ∂vz ∂vϕ ⎞ ⎛ ∂vs ∂v ⎤ 1⎡ ∂ ∂v ⎞ − − z ⎟ϕˆ + ⎢ ( svϕ ) − s ⎥zˆ ∇ × v⃗ = ⎜ ⎟sˆ + ⎜ ⎝ ∂z ∂ϕ ⎦ ∂s ⎠ s ⎣ ∂s ∂z ⎠ ⎝ s ∂ϕ Spherical

∇ × v⃗ =

⎤ ∂ ∂vθ ⎤ 1 ⎡ 1 ∂vr 1 ⎡∂ − (rvϕ )⎥θˆ ⎢ (sin θ vϕ ) − ⎥rˆ + ⎢ ∂r ∂ϕ ⎦ ⎦ r sin θ ⎣ ∂θ r ⎣ sin θ ∂ϕ ∂v ⎤ 1⎡ ∂ + ⎢ ( rvθ ) − r ⎥ϕˆ ⎣ ∂θ ⎦ r ∂r

1.1.7 Laplacian Given a scalar function T , the Laplacians for various coordinate systems are given below. Cartesian

∇2 T =

∂ 2T ∂ 2T ∂ 2T + + ∂z 2 ∂y 2 ∂x 2

Cylindrical

∇2 T =

∂ 2T 1 ∂ ⎛ ∂T ⎞ 1 ∂ 2T ⎜s ⎟ + + ∂z 2 s ∂s ⎝ ∂s ⎠ s 2 ∂ϕ 2

Spherical

∇2 T =

∂ 2T 1 ∂ ⎛ ∂T ⎞ 1 ∂ ⎛ 2 ∂T ⎞ 1 ⎜r ⎟ + ⎜sin θ ⎟ + ∂θ ⎠ r 2 sin2 θ ∂ϕ 2 r 2 ∂r ⎝ ∂r ⎠ r 2 sin θ ∂θ ⎝

1-3

Electromagnetism

1.1.8 Line integral Given vector function v ⃗ and path P , a line integral is given by b⃗

∫a ⃗ P v ⃗ ⋅ dl⃗, where a ⃗ and b ⃗ are the end points, and dl⃗ is the infinitesimal displacement vector along P . In Cartesian coordinates dl⃗ = dx xˆ + dy yˆ + dz zˆ . 1.1.9 Surface integral Given vector function v ⃗ and surface S , a surface integral is given by

∫S

v ⃗ ⋅ da ⃗ ,

where da ⃗ is the infinitesimal area vector that has direction normal to the surface. Note that da ⃗ always depends on the surface involved. 1.1.10 Volume integral Given scalar function T and volume V , a volume integral is given by

∫V

T dτ ,

where dτ is the infinitesimal volume element. In Cartesian coordinates dτ = dx dy dz . 1.1.11 Fundamental theorem for gradients b⃗

∫a ⃗ P (∇T ) ⋅ dl⃗ = T ( b ⃗) − T ( a ⃗) 1.1.12 Fundamental theorem for divergences (Gauss’s theorem, Green’s theorem, divergence theorem)

∫V (∇ ⋅ v ⃗)dτ = ∮S v ⃗ ⋅ da ⃗ 1.1.13 Fundamental theorem for curls (Stoke’s theorem, curl theorem)

∫S (∇ × v ⃗) ⋅ da ⃗ = ∮P v ⃗ ⋅ dl⃗ 1.1.14 Cylindrical polar coordinates Here our infinitesimal quantities are

dl⃗ = ds sˆ + s dϕ ϕˆ + dz zˆ and

dτ = s ds dϕ dz . 1-4

Electromagnetism

1.1.15 Spherical polar coordinates Here our infinitesimal quantities are

dl⃗ = dr rˆ + r dθ θˆ + r sin θ dϕ ϕˆ and

dτ = r 2 sin θ dr dθ dϕ . 1.1.16 One-dimensional Dirac delta function The one-dimensional Dirac delta function is given by

⎧0 x ≠ a δ (x − a ) = ⎨ ⎩∞ x = a and has the following properties ∞

∫

δ(x − a )dx = 1

−∞ ∞

∫

f (x )δ(x − a )dx = f (a )

−∞

δ(kx ) =

1 δ(x ). k

1.1.17 Theory of vector fields If the curl of a vector field F ⃗ vanishes everywhere, then F ⃗ can be written as the gradient of a scalar potential V :

∇ × F ⃗ ↔ F ⃗ = −∇V. If the divergence of a vector vanishes everywhere, then F ⃗ can be expressed as the curl of a vector potential A⃗ :

∇ ⋅ F ⃗ = 0 ↔ F ⃗ = ∇ × A.⃗

1.2 Problems and solutions Problem 1.1. Given vectors A⃗ = 3xˆ + 9yˆ + 5zˆ and B ⃗ = xˆ − 7yˆ + 4zˆ , calculate A⃗ ⋅ B ⃗ and A⃗ × B ⃗ using vector components and find the angle between A⃗ and B ⃗ using both products.

1-5

Electromagnetism

Solution

A ⃗ ⋅ B ⃗ = (3xˆ + 9yˆ + 5zˆ ) ⋅ (xˆ − 7yˆ + 4zˆ ) = (3)(1) + (9)( −7) + (5)(4) = 3 − 63 + 20 A ⃗ ⋅ B ⃗ = −40 xˆ yˆ zˆ A⃗ × B ⃗ = 3 9 5 1 −7 4 = [(9)(4) − ( −7)(5)]xˆ + [(1)(5) − (3)(4)]yˆ + [(3)( −7) − (1)(9)]zˆ Aˆ × Bˆ = 71xˆ − 7yˆ − 30zˆ To find the angle θ between A⃗ and Bˆ we must first calculate A and B :

A = 32 + 9 2 + 5 2 =

115

B = 12 + ( −7)2 + 42 =

66 .

Using the dot product, we have

⎛ −40 ⎞ A ⃗ ⋅ B ⃗ = AB cos θ → θ = cos−1 ⎜ ⎟ ⎝ 115 66 ⎠ θ = 117.3°. Using the cross product, we have A ⃗ × B ⃗ = AB sin θ →

712 + ( −7)2 + ( −30)2 =

115 66 sin θ

θ = 62.7°. Note, however, that we can see that the angle between A⃗ and B ⃗ is greater than 90°. For any argument γ , −90° ⩽ sin−1(γ ) ⩽ 90°. Since the angle between A⃗ and B ⃗ is greater than 90°, we must adjust for this by subtracting our angle from 180°. Therefore, θ = 180° − 62.7° = 117.3° as expected.

1-6

Electromagnetism

Problem 1.2. The scalar triple product states A⃗ ⋅ (B ⃗ × C ⃗ ) = B ⃗ ⋅ (C ⃗ × A⃗ ). Prove this by expressing each side in terms of its components. Solution Starting with the left-hand side, the cross product is

xˆ yˆ zˆ ⃗ ⃗ B × C = Bx By Bz Cx Cy Cz

(

)

= (ByCz − BzCy )xˆ + BzCx − BxCz yˆ + (BxCy − ByCx )z. ˆ Now, dotting A⃗ with (B ⃗ × C ⃗ )

(

)

(

)

(

)

(

A ⃗ ⋅ B ⃗ × C ⃗ = Ax ByCz − BzCy + Ay BzCx − BxCz + Az BxCy − ByCx

)

= Ax ByCz − Ax BzCy + Ay BzCx − Ay BxCz + Az BxCy − Az ByCx

(

)

(

(

)

= Bx CyAz − CzAy + By CzAx − CxAz + Bz CxAy − CyAx

)

A ⃗ ⋅ B ⃗ × C ⃗ = B ⃗ ⋅ ⎡⎣ CyAz − CzAy xˆ + CzAx − CxAz yˆ + CxAy − CyAx zˆ⎤⎦ .

(

)

(

)

(

)

(

)

Note the term in brackets is precisely C ⃗ × A⃗ , therefore

(

)

(

A⃗ ⋅ B ⃗ × C⃗ = B ⃗ ⋅ C⃗ × A⃗

)

as desired. This procedure can easily be applied again to prove the final part of the triple product,

(

)

(

)

(

)

A⃗ ⋅ B ⃗ × C⃗ = B ⃗ ⋅ C⃗ × A⃗ = C⃗ ⋅ A⃗ × B ⃗ . Problem 1.3. Given source vector r ′⃗ = r cos θ xˆ + r sin θ yˆ and field vector r ⃗ = zzˆ , find the separation vector r ⃗ and the unit vector rˆ . Solution We have

r ⃗ = r ⃗ − r ⃗′ = zzˆ − r cos θ xˆ + r sin θ yˆ

(

)

r ⃗ = −r cos θ xˆ − r sin θ yˆ + zzˆ . To determine the unit vector rˆ , we must first find the magnitude of r ⃗ ,

r=

( −r cos θ )2 + ( −r sin θ )2 + z 2 =

1-7

(

)

r 2 cos2 θ + sin2 θ + z 2 =

r2 + z2 .

Electromagnetism

So

rˆ =

r⃗ −r cos θ xˆ − r sin θ yˆ + zzˆ = . r r2 + z2

Problem 1.4. Given A⃗ in coordinate system K , find the rotational matrix to give the components in system K ′.

Solution From the figures, we have

Ax′ = Ay ,

A y′ = Ax ,

Az′ = −Az .

We want to find the rotational matrix R that satisfies

⎛ ⎞ ⎛ Ax ⎞ ⎜ Ax′ ⎟ ⎜ A ′ ⎟ = R⎜ Ay ⎟ . ⎜⎜ ⎟⎟ ⎜ y⎟ ⎝ Az ⎠ ⎜ ′⎟ ⎝ Az ⎠ From our equations above

⎛ ⎞ ⎜ Ax′ ⎟ ⎛ 0 1 0 ⎞ ⎛ Ax ⎞ ⎜ A ′ ⎟ = ⎜ 1 0 0 ⎟ ⎜ Ay ⎟ . ⎟⎟ ⎜⎜ ⎟⎟ ⎜ y ⎟ ⎜⎜ ⎜ ′⎟ ⎝ 0 0 − 1⎠ ⎝ Az ⎠ ⎝ Az ⎠

1-8

Electromagnetism

Therefore,

⎛0 1 0 ⎞ ⎜ ⎟ R = ⎜ 1 0 0 ⎟. ⎜ ⎟ ⎝ 0 0 −1⎠ Problem 1.5. Find the gradient of the following functions: a) T = x 4 + y 2 + z 3 b) T = x 2 ln y z 3 c) T = x 2y + z 3 Solutions a) T = x 4 + y 2 + z 3

∇T =

∂T ∂T ∂T zˆ = 4x 3xˆ + 2yyˆ + 3z 2zˆ yˆ + xˆ + ∂x ∂y ∂z

b) T = x 2 ln y z 3

∇T =

x 2z 3 ∂T ∂T ∂T yˆ + 3x 2z 2 ln y zˆ zˆ = 2xz 3 ln y xˆ + yˆ + xˆ + y ∂x ∂y ∂z

c) T = x 2y + z 3

∇T =

∂T ∂T ∂T zˆ = 2xyxˆ + x 2yˆ + 3z 2zˆ yˆ + xˆ + ∂x ∂y ∂z

Problem 1.6. Find the divergence of the following functions: a) v ⃗ = xyxˆ − 2y 2 zyˆ + z 3zˆ b) v ⃗ = (x + y )xˆ + (y + z )yˆ + (z + x )zˆ Solutions a) v ⃗ = xyxˆ − 2y 2 zyˆ + z 3zˆ

∇ ⋅ v⃗ =

∂vy ∂v ∂vx + + z = y − 4yz + 3z 2 ∂z ∂x ∂y

b) v ⃗ = (x + y )xˆ + (y + z )yˆ + (z + x )zˆ

∇ ⋅ v⃗ =

∂vy ∂v ∂vx + + z =1+1+1=3 ∂z ∂x ∂y

Problem 1.7. Find the curl of the following functions: a) v ⃗ = xyxˆ − 2y 2 zyˆ + z 3zˆ b) v ⃗ = (x + y )xˆ + (y + z )yˆ + (z + x )zˆ c) v ⃗ = sin x xˆ + cos y yˆ

1-9

Electromagnetism

Solutions a) v ⃗ = xyxˆ − 2y 2 zyˆ + z 3zˆ

⎛ ∂v ⎛ ∂vy ∂vy ⎞ ⎛ ∂vx ∂v ⎞ ∂v ⎞ − x ⎟zˆ − z ⎟yˆ + ⎜ ∇ × v⃗ = ⎜ z − ⎟xˆ + ⎜ ⎝ ∂z ∂y ⎠ ∂x ⎠ ∂z ⎠ ⎝ ∂y ⎝ ∂x ⎡ 3 ⎡ ∂ −2y 2 z ⎤⎥ ∂ z3 ∂( xy ) ∂(z ) ⎢ ˆ+⎢ − − =⎢ x ⎥ ⎢ ∂z ∂z ∂x ∂y ⎣ ⎦ ⎣ ⎡ ∂ − 2y 2 z ⎤ ∂( xy ) ⎥ − + ⎢⎢ zˆ ∂x ∂y ⎥ ⎣ ⎦

(

(

(

)

( ) ⎤⎥yˆ ⎥ ⎦

)

)

= 0 + 2y 2 xˆ + (0 − 0)yˆ + (0 − x )zˆ ∇ × v ⃗ = 2y 2 xˆ − xzˆ

b) v ⃗ = (x + y )xˆ + (y + z )yˆ + (z + x )zˆ

⎛ ∂v ⎛ ∂vy ∂vy ⎞ ⎛ ∂vx ∂v ⎞ ∂v ⎞ − z ⎟yˆ + ⎜ − x ⎟zˆ ∇ × v⃗ = ⎜ z − ⎟xˆ + ⎜ ⎝ ∂z ∂x ⎠ ∂y ⎠ ∂z ⎠ ⎝ ∂y ⎝ ∂x ⎡ ∂(z + x ) ⎡ ∂(x + y ) ∂(z + x ) ⎤ ∂(y + z ) ⎤ − − =⎢ ⎥xˆ + ⎢ ⎥yˆ ⎣ ∂z ∂x ⎦ ∂z ⎦ ⎣ ∂y ⎡ ∂(y + z ) ∂(x + y ) ⎤ − +⎢ ⎥zˆ ∂y ⎦ ⎣ ∂x ∇ × v ⃗ = −xˆ − yˆ − zˆ

c) v ⃗ = sin x xˆ + cos y yˆ

⎛ ∂v ⎛ ∂vy ∂vy ⎞ ⎛ ∂vx ∂v ⎞ ∂v ⎞ − z ⎟yˆ + ⎜ − x ⎟zˆ ∇ × v⃗ = ⎜ z − ⎟xˆ + ⎜ ⎝ ⎠ ∂x ∂y ⎠ ∂z ⎠ ∂z ⎝ ∂y ⎝ ∂x ⎡ ∂(0) ⎡ ∂(sin x ) ∂(0) ⎤ ∂(cos y ) ⎤ − − =⎢ ⎥xˆ + ⎢ ⎥yˆ ⎣ ⎣ ∂y ∂x ⎦ ∂z ⎦ ∂z ⎡ ∂(cos y ) ∂(sin x ) ⎤ − +⎢ ⎥zˆ = 0 ⎣ ∂x ∂y ⎦

1-10

Electromagnetism

Problem 1.8. Prove ∇ × (∇T ) = 0. Solution

( )

∇ × ∇T =

xˆ

yˆ

zˆ

∂ ∂x

∂ ∂y

∂ ∂z

∂T ∂T ∂T ∂x ∂y ∂z ⎡ ∂ ⎛ ∂T ⎞ ⎡ ∂ ⎛ ∂T ⎞ ∂ ⎛ ∂T ⎞⎤ ∂ ⎛ ∂T ⎞⎤ ⎜ ⎟⎥yˆ =⎢ ⎜ ⎟ − ⎜ ⎟⎥xˆ + ⎢ ⎜ ⎟ − ⎣ ∂z ⎝ ∂x ⎠ ∂x ⎝ ∂z ⎠⎦ ⎣ ∂y ⎝ ∂z ⎠ ∂z ⎝ ∂y ⎠⎦ ⎡ ∂ ⎛ ∂T ⎞ ∂ ⎛ ∂T ⎞⎤ ⎜ ⎟⎥z ˆ +⎢ ⎜ ⎟− ⎣ ∂x ⎝ ∂y ⎠ ∂y ⎝ ∂x ⎠⎦ ∇ × (∇T ) = 0. Problem 1.9. Find the Laplacian of the following functions: a) T = x + y 2 + xz + 3 b) T = e x + sin y cos(2z ) c) T = sin x cos y d) v ⃗ = xyxˆ + z 2yˆ − 2zˆ Solutions a) T = x + y 2 + xz + 3

∇2 T =

∂ 2T ∂ 2T ∂ 2T =0+2+0=2 + + ∂z 2 ∂y 2 ∂x 2

b) T = e x + sin y cos(2z )

∇2 T =

∂ 2T ∂ 2T ∂ 2T = e x − sin y cos(2z ) − 4 sin y cos(2z ) + + ∂z 2 ∂y 2 ∂x 2

= e x − 5 sin y cos(2z ) c) T = sin x cos y

∇2 T =

∂ 2T ∂ 2T ∂ 2T = −sin x cos y − sin x cos y = −2 sin x cos y + + ∂z 2 ∂y 2 ∂x 2

1-11

Electromagnetism

d) v ⃗ = xyxˆ + z 2yˆ − 2zˆ

⎛ ∂ 2vy ⎛ ∂ 2v ∂ 2vy ∂ 2vy ⎞ ∂ 2vx ⎞ ∂ 2vx ˆ x + + + + ∇2 v ⃗ = ⎜ 2x + ⎜ ⎟yˆ ⎟ ∂z 2 ⎠ ∂y 2 ∂y 2 ∂z 2 ⎠ ⎝ ∂x ⎝ ∂x 2 ⎛ ∂ 2v ∂ 2vz ⎞ ∂ 2vz + + ⎜ 2z + ⎟zˆ 2 ∂z 2 ⎠ ∂y ⎝ ∂x ∇2 v ⃗ = (0 + 0 + 0)xˆ + (0 + 0 + 2)yˆ+(0 + 0 + 0)zˆ = 2yˆ Problem 1.10. Test the divergence theorem with v ⃗ = 2xyxˆ + y 2 z 3yˆ + (x 2z − 2y )zˆ and the volume below.

Solution The divergence theorem states

∫V

∇ ⋅ v ⃗ dτ =

∮S v ⃗ ⋅ da ⃗.

Starting with the left-hand side, we have the divergence

(

)

∇ ⋅ v ⃗ = 2y + 2yz 3 + x 2 = 2y z 3 + 1 + x 2 . We must split the volume into two pieces, (a) 0 ⩽ y ⩽ 1 and (b) 1 ⩽ y ⩽ 2. (a) 2 2

∫

∇ ⋅ v ⃗ dτ =

1

∫∫∫ 0 0 0

⎡2y z 3 + 1 + x 2⎤dy dx dz = 52 ⎣ ⎦ 3

(

1-12

)

Electromagnetism

(b) 2 2 4 − 2y

∫

∇ ⋅ v ⃗ dτ =

∫∫ ∫ 0 1

0

⎡2y z 3 + 1 + x 2⎤dy dx dz = 176 ⎣ ⎦ 15

(

)

So,

∫V

∇ ⋅ v ⃗ dτ =

52 176 436 . + = 3 15 15

Now we solve ∮ v ⃗ ⋅ da ⃗ , which must be evaluated over the six sides. S

(i) We must split this region into two sections (a) and (b), and da ⃗ = dy dz zˆ with x = 2. In (a), 0 ⩽ y ⩽ 1, 2

1

∫ v ⃗ ⋅ da ⃗ = ∫ ∫

2(2)y dy dz = 4.

0 0

In (b), 1 ⩽ y ⩽ 2 and 0 ⩽ z ⩽ 4 − 2y 2 4 − 2y

∫

v ⃗ ⋅ da ⃗ =

∫ ∫ 1

2(2)y dz dy =

0

1-13

16 . 3

Electromagnetism

(ii) Here, da ⃗ = −dy dz xˆ and x = 0, so v ⃗ ⋅ da ⃗ = 2(0)y( −dy dx )xˆ = 0. (iii) Here, da ⃗ = dx dy zˆ and z = 2 2

1

⎡ x 2(2) − 2y⎤ dy dx = 10 . ⎣ ⎦ 3

∫ v⃗ ⋅ da ⃗ = ∫ ∫ 0

0

(iv) Here, da ⃗ = −dx dy zˆ and z = 0 2

2

∫ v ⃗ ⋅ da ⃗ = ∫ ∫ 0

⎡ x 2(0) − 2y⎤( −dx dy ) = 8. ⎣ ⎦

0

(v) Here da ⃗ = dx dz yˆ and y = 0, so v ⃗ ⋅ da ⃗ = 02z 3( −dx dz ) = 0. (vi) Here, we have da ⃗ = dx dz′ nˆ where nˆ = nn⃗ . We can find n ⃗ by crossing vectors A⃗ = −yˆ + 2zˆ and B ⃗ = 2xˆ (the edges of the volume):

xˆ yˆ zˆ n ⃗ = A ⃗ × B ⃗ = 0 −1 2 = 4yˆ + 2zˆ . 2 0 0 So

n=

42 + 22 = 2 5

nˆ =

2 5 5 yˆ + zˆ . 5 5

and

We can also obtain dz′ by considering

1-14

Electromagnetism

5 2

so dz′ =

dz . Now

⎛2 5 5 ⎞ ⎛ 1 ⎞ 5 dx dz ⎜ yˆ + zˆ⎟ = ⎜ yˆ + zˆ⎟dx dz 5 ⎠ ⎝ 2 ⎠ 2 ⎝ 5

da ⃗ = and

z = 4 − 2y → y = 2 −

z . 2

So 2

2

∫ v ⃗ ⋅ da ⃗ = ∫ ∫ 0 2

0 2

∫∫

=

0

0

⎡ 2 3 1 2 ⎤ x z − 2y ⎥ dx dz ⎢⎣y z + ⎦ 2

(

)

⎧⎛ ⎛ z ⎞2 z ⎞⎤⎫ 42 1⎡ ⎨⎜ 2 − ⎟ z 3 + ⎢x 2z − 2⎜ 2 − ⎟⎥⎬ dx dz = . ⎝ 5 2⎠ 2⎣ 2 ⎠⎦⎭ ⎩⎝

Therefore

∮S v ⃗ ⋅ da ⃗ = 4 + 163

+

42 436 10 +8+ = 5 15 3

as expected. Problem 1.11. Test the curl theorem with v ⃗ = 5xy 2 xˆ + yz 2yˆ + 4x 2zzˆ and the surface below.

1-15

Electromagnetism

Solution The curl theorem states

∫S (∇ × v ⃗) ⋅ da ⃗ = ∮P v ⃗ ⋅ dl⃗. Starting with the left-hand side, the curl is given by

∇ × v⃗ =

xˆ

yˆ

zˆ

∂ ∂x

∂ ∂y

∂ ∂z

= −2yzxˆ − 8xzyˆ − 10xyzˆ

5xy 2 yz 2 4x 2z We also have da ⃗ = dy dz xˆ with 0 ⩽ z ⩽ − (y − 2)2 + 4. So 4 −(y − 2)2 + 4

∫S

( ∇ × v ⃗ ) ⋅ da ⃗ =

∫

∫

0

0

−2yz dz dy = −

1024 . 15

Now to solve ∮ v ⃗ ⋅ dl⃗ over the two paths (i) and (ii): P

(i) Here we have x = 0, z = 0, and dl⃗ = dy yˆ . So v ⃗ ⋅ dl⃗ = y(02)dy = 0. (ii) Here we have dl⃗ = dy yˆ + dz zˆ , x = 0, and z = − (y − 2)2 + 4 0

∫P

v ⃗ ⋅ dl⃗ =

∫P

yz 2dy + 4(0 2)z dz =

∫ 4

1-16

2 1024 . y⎡⎣ −(y − 2)2 + 4⎤⎦ dy = − 15

Electromagnetism

So,

∮P v ⃗ ⋅ dl⃗ = 0 + −1024 15

=

−1024 15

as expected. Problem 1.12. Test the gradient theorem with T = 3xz 2 − y 2 z and path z = y 2 and z = y3 from (0, 0, 0) → (0, 1, 1).

Solution The gradient theorem states b⃗

∫

()

()

∇T ⋅ dl⃗ = T b ⃗ − T a ⃗ .

a⃗

Starting with the right side, we have

T (0,1,1) − T (0,0,0) = 3(0)(12 ) − (12 )(1) − 0 = − 1. b⃗

Now to solve

∫

∇T ⋅ dl⃗ , the gradient of T is given by

a⃗

∇T =

∂ ∂ ∂ 3xz 2 − y 2 z xˆ + 3xz 2 − y 2 z yˆ + 3xz 2 − y 2 z zˆ ∂z ∂y ∂x

(

)

(

(

)

)

∇T = 3z 2xˆ − 2yz yˆ + 6xz − y 2 zˆ .

1-17

(

)

Electromagnetism

Here, dl⃗ = dy yˆ + dz zˆ with x = 0. So

∇T ⋅ dl⃗ = −2yz dy + ⎡⎣6(0)( z ) − y 2 ⎤⎦dz = −2yz dy − y 2 dz . For path (i), we have z = y 2 → dz = 2y dy . So

( )

∇T ⋅ dl⃗ = −2y y 2 dy − y 2 (2y dy ) = −4y 3 dy and b⃗

∫

1

∇T ⋅ dl⃗ =

a⃗

∫

−4y 3dy = −1

0

as expected. For path (ii), we have z = y3 → dz = 3y 2 dy . So

( )

(

)

∇T ⋅ dl⃗ = −2y y 3 dy − y 2 3y 2 dy = −5y 4dy and b⃗

∫

1

∇T ⋅ dl⃗ =

a⃗

∫

− 5y 4dy = −1

0

also as expected. Problem 1.13. Verify the following integration by parts given f = xy 2 z and A⃗ = z 2xˆ + 4xyyˆ − x 2zzˆ and the surface below,

∫S f ( ∇ × A⃗ ) ⋅ da ⃗ = ∫S

⎡A ⃗ × ∇f ⎤ ⋅ da ⃗ + ⎣ ⎦

( )

1-18

∮P fA⃗ ⋅ dl⃗.

Electromagnetism

Solution Starting with the left-hand side

∇ × A⃗ =

xˆ

yˆ

zˆ

∂ ∂x

∂ ∂y

∂ ∂z

= [2z − ( −2xz )] yˆ + 4yzˆ = 2z(x + 1)yˆ + 4yzˆ .

z 2 4xy −x 2z Now

(

)

f ∇ × A ⃗ = 2xy 2 z 2(x + 1)yˆ + 4xy 3zzˆ . Here we have da ⃗ = dx′ dz nˆ where nˆ = xˆ + yˆ and n = from

2 so nˆ =

2 2

xˆ +

2 2

yˆ . Also

we have

dx′ =

2 dx

with y = 1 − x .

Now

da ⃗ =

⎛ 2 2 ⎞ 2 dx dz ⎜ xˆ + yˆ ⎟ = dx dz xˆ + yˆ . 2 ⎠ ⎝ 2

(

)

Therefore, 1

1

∫S f ( ∇ × A⃗ ) ⋅ da ⃗ = ∫ ∫ 0

0

1

1

∫∫

=

0

⎡ 2xy 2 z 2(x + 1)yˆ + 4xy 3zzˆ⎤ ⋅ (xˆ + yˆ )dx dz ⎣ ⎦

2x(1 − x )2 z 2(x + 1)dx dz =

0

Next, we will solve the ∮ fA⃗ ⋅ dl⃗ term for the four segments. P

1-19

7 . 90

Electromagnetism

Segment (i)

z = 0 → f = xy 2 (0) = 0. Segment (ii)

x = 0 → f = (0)y 2 z = 0. Segment (iii)

dl⃗ = dx xˆ + dy yˆ , z = 1,

and

y = 1 − x → dy = − dx .

Segment (iv)

y = 0 → f = x(0 2)z = 0. So

f A ⃗ ⋅ dl⃗ = xy 2 z 2dx + 4xy dy = x(1 − x )2 ⎡⎣ 1 − 4x(1 − x )⎤⎦dx

(

)

(

)

and 1

∮P Now to solve the

fA ⃗ ⋅ dl⃗ =

∫ 0

1 . x(1 − x )2 ⎡⎣ 1 − 4x(1 − x )⎤⎦dx = 60

∫S [A × ∇f ] ⋅ da ⃗ term. First, we have ∇f = y 2 zxˆ + 2xyzyˆ + xy 2 zˆ .

1-20

Electromagnetism

So

xˆ yˆ zˆ 2 A × ( ∇f ) = z 4xy −x 2z y 2 z 2xyz xy 2

(

)

(

)

(

)

= 4x 2y 3 + 2x 3yz 2 xˆ + −x 2y 2 z 2 − xy 2 z 2 yˆ + 2xyz 3 − 4xy 3z zˆ . As before, da ⃗ = dx dz (xˆ + yˆ ). So 1 1

∫S [A × (∇f )] ⋅ da ⃗ = ∫ ∫

⎡⎣ 4x 2(1 − x )3 + 2x 3(1 − x )z 2

0 0

−x 2(1 − x )2 z 2 − x(1 − x )2 z 2⎤⎦dx dz 11 . ∫S [A × (∇f )] ⋅ da ⃗ = 180 So

11 ∫S [A × (∇f )] ⋅ da ⃗ + ∮P fA⃗ ⋅ dl⃗ = 180

+

1 7 = 60 90

as expected. Problem 1.14. Find the divergence and curl of the following functions: a) v ⃗ = r 2rˆ + cos θ sin ϕ θˆ + sin θ cos ϕ ϕˆ b) v ⃗ = s cos ϕ sˆ + cos ϕ sin ϕ ϕˆ +z sin ϕ zˆ Solutions a) v ⃗ = r 2rˆ + cos θ sin ϕ θˆ + sin θ cos ϕ ϕˆ

∇ ⋅ v⃗ =

∂ 1 ∂vϕ 1 ∂ 2 1 r vr + sin θ vθ + 2 r ∂r r sin θ ∂θ r sin θ ∂ϕ

( )

(

)

=

∂ ∂ 1 1 ∂ 4 1 (sin θ cos ϕ) r + sin θ cos θ sin ϕ + 2 r ∂r r sin θ ∂θ r sin θ ∂ϕ

=

1 1 sin ϕ −sin2 θ + cos2 θ + ( −sin ϕ) 4r 3 + 2 r r sin θ r

( )

( )

(

)

(

)

1-21

Electromagnetism

= 4r +

∇ ⋅ v ⃗ = 4r +

∇ × v⃗ =

sin ϕ sin ϕ 1 − 2 sin2 θ − r sin θ r

(

)

sin ϕ csc θ − 2 sin θ − 1 r

(

)

∂v ⎤ 1 ⎡∂ sin θ vϕ − θ ⎥rˆ ⎢ ∂ϕ ⎦ r sin θ ⎣ ∂θ

(

+

)

⎤ ∂v ⎤ ∂ 1⎡ ∂ 1 ⎡ 1 ∂vr − ( rvϕ )⎥θˆ + ⎢ ( rvθ ) − r ⎥ϕˆ ⎢ ⎣ ∂θ ⎦ ∂r r ⎣ sin θ ∂ϕ r ∂r ⎦

⎤ ∂ 1 ⎡∂ sin2 θ cos ϕ − (cos θ sin ϕ)⎥rˆ ⎢ ⎦ ∂ϕ r sin θ ⎣ ∂θ

(

=

)

+

⎤ ∂ 1⎡ 1 ∂ 2 r − (r sin θ cos ϕ)⎥θˆ ⎢ ⎦ ∂r r ⎣ sin θ ∂ϕ

+

∂ 2 ⎤ˆ 1⎡ ∂ r ⎥ϕ ⎢ (r cos θ sin ϕ) − ∂ϕ ⎦ r ⎣ ∂r

=

∇ × v⃗ =

( )

( )

sin θ cos ϕ ˆ 1 (2 sin θ cos θ cos ϕ − cos θ cos ϕ)rˆ − θ r sin θ r cos θ sin ϕ ˆ + ϕ r

cos θ cos ϕ sin θ cos ϕ ˆ cos θ sin ϕ ˆ θ+ ϕ (2 − csc θ )rˆ − r r r

b) v ⃗ = s cos ϕ sˆ + cos ϕ sin ϕ ϕˆ + z sin ϕ zˆ

∇ ⋅ v⃗ =

=

∂v 1 ∂ 1 ∂vϕ + z ( svs ) + ∂z s ∂s s ∂ϕ ∂ 1 ∂ 2 1 ∂ s cos ϕ + (cos ϕ sin ϕ) + (z sin ϕ) ∂z s ∂s s ∂ϕ

(

= 2 cos ϕ +

)

1 −sin2 ϕ + cos2 ϕ + sin ϕ s

(

∇ ⋅ v ⃗ = 2 cos ϕ + sin ϕ +

)

cos2 ϕ − sin2 ϕ s

1-22

Electromagnetism

⎛ 1 ∂vz ∂vϕ ⎞ ⎛ ∂vs ∂v ⎤ 1⎡ ∂ ∂v ⎞ − − z ⎟ϕˆ + ⎢ ( svϕ ) − s ⎥zˆ ∇ × v⃗ = ⎜ ⎟sˆ + ⎜ ⎝ ∂z ∂ϕ ⎦ ∂s ⎠ s ⎣ ∂s ∂z ⎠ ⎝ s ∂ϕ ⎡1 ∂ ⎤ ⎡∂ ⎤ ∂ ∂ (z sin ϕ) − (cos ϕ sin ϕ)⎥sˆ + ⎢ (s cos ϕ) − (z sin ϕ)⎥ϕˆ =⎢ ⎣ ∂z ⎦ ⎣ s ∂ϕ ⎦ ∂s ∂z +

=

∇ × v⃗ =

⎤ 1⎡ ∂ ∂ (s cos ϕ)⎥zˆ ⎢ (s cos ϕ sin ϕ) − s ⎣ ∂s ∂ϕ ⎦

1 z cos ϕ sˆ + (cos ϕ sin ϕ + s sin ϕ)zˆ s s sin ϕ z cos ϕ sˆ + (cos ϕ + s )zˆ s s

Problem 1.15. Find the gradient and Laplacian of: a) T = r 2(cos θ sin ϕ + sin θ cos ϕ ) b) T = z 2 sin ϕ − s cos2 ϕ Solutions a) T = r 2(cos θ sin ϕ + sin θ cos ϕ )

∇T =

1 ∂T ˆ 1 ∂T ˆ ∂T rˆ + θ+ ϕ r ∂θ r sin θ ∂ϕ ∂r

= 2r(cos θ sin ϕ + sin θ cos ϕ)rˆ + +

1 2 r ( −sin θ sin ϕ + cos θ cos ϕ)θˆ r

1 r 2(cos θ cos ϕ − sin θ sin ϕ)ϕˆ r sin θ

= 2r(cos θ sin ϕ + sin θ cos ϕ)rˆ + r(cos θ cos ϕ − sin θ sin ϕ)θˆ +

r (cos θ cos ϕ − sin θ sin ϕ)ϕˆ sin θ

∇T = 2r sin(θ + ϕ)rˆ + r cos(θ + ϕ)θˆ +

r cos(θ + ϕ)ϕˆ . sin θ

Note we could have written T as T = r 2 sin(θ + ϕ ) and then computed the gradient.

∇2 T =

⎛ ∂ 2T ⎞ 1 ∂ ⎛ ∂T ⎞ 1 ∂ ⎛ 2 ∂T ⎞ 1 ⎜r ⎟ + ⎜ sin θ ⎟ + ⎜ ⎟ ∂θ ⎠ r 2 sin2 θ ⎝ ∂ϕ 2 ⎠ r 2 ∂r ⎝ ∂r ⎠ r 2 sin θ ∂θ ⎝

1-23

Electromagnetism

=

1 ∂⎡ 3 1 ∂ ⎡ 2 ⎤ ⎤ ⎣ 2r sin(θ + ϕ)⎦ + 2 ⎣ r sin θ cos(θ + ϕ)⎦ 2 r ∂r r sin θ ∂θ 1 ∂ ⎡ 2 ⎤ + 2 2 ⎣ r cos(θ + ϕ)⎦ r sin θ ∂ϕ

= 6 sin(θ + ϕ) + +

1 [cos θ cos(θ + ϕ) − sin θ sin(θ + ϕ)] sin θ

1 −sin(θ + ϕ) sin2 θ

(

∇2 T = 5 sin(θ + ϕ) +

)

sin(θ + ϕ) cos θ . cos(θ + ϕ) − sin2 θ sin θ

b) T = z 2 sin ϕ − s cos2 ϕ

∇T =

1 ∂T ˆ ∂T ∂T sˆ + zˆ ϕ+ s ∂ϕ ∂z ∂s

1 ∂ 2 ∂ 2 z sin ϕ − s cos2 ϕ sˆ + z sin ϕ − s cos2 ϕ ϕˆ ∂s s ∂ϕ ∂ 2 + z sin ϕ − s cos2 ϕ zˆ ∂z 1 = −cos2 ϕ sˆ + ⎡⎣ z 2 cos ϕ − 2s cos ϕ( −sin ϕ)⎤⎦ϕˆ + 2z sin ϕ zˆ s

(

=

)

(

∇T = − cos2 ϕ sˆ +

∇2 T =

(

)

)

cos ϕ 2 z + 2s sin ϕ ϕˆ + 2z sin ϕ zˆ s

(

)

∂ 2T 1 ∂ ⎛ ∂T ⎞ 1 ∂ 2T ⎜s ⎟ + + ∂z 2 s ∂s ⎝ ∂s ⎠ s 2 ∂ϕ 2

∂ ⎛ ∂T ⎞ ∂T ∂T = −s cos2 ϕ → ⎜ s ⎟ = −cos2 ϕ = −cos2 ϕ → s ∂s ⎝ ∂s ⎠ ∂s ∂s ∂ 2T ∂T = −z 2 sin ϕ + 2s −sin2 ϕ + cos2 ϕ = z 2 cos ϕ + 2s cos ϕ sin ϕ → ∂ϕ 2 ∂ϕ

(

∂ 2T ∂T = 2 sin ϕ = 2z sin ϕ → ∂z ∂z

∇2 T = −

cos2 ϕ 2 z2 + −sin2 ϕ + cos2 ϕ − 2 sin ϕ + 2 sin ϕ s s s

(

)

1-24

)

Electromagnetism

∇2 T =

⎛ cos2 ϕ 2 z2 ⎞ − sin2 ϕ + ⎜ 2 − 2 ⎟sin ϕ ⎝ s s s ⎠

Problem 1.16. Test the divergence theorem with v ⃗ = r cos ϕ rˆ + r cos θ sin θ θˆ + r sin ϕ ϕˆ and the volume below (the upper half of the sphere of radius R with a cone of radius a = R cut out). 3

Solution The divergence theorem states

∫V

∮S v ⃗ ⋅ da ⃗.

∇ ⋅ v ⃗ dτ =

Starting with the left-hand side, the divergence is

∇ ⋅ v⃗ =

=

∂ 1 ∂vϕ 1 ∂ 2 1 + + θ r v sin v r θ r 2 ∂r r sin θ ∂θ r sin θ ∂ϕ

( )

(

)

∂ 1 ∂ 1 ∂ 3 1 (r sin ϕ) r cos ϕ + r sin2 θ cos θ + 2 r ∂r r sin θ ∂θ r sin θ ∂ϕ

(

= 3 cos ϕ +

)

(

)

cos ϕ 1 2 sin θ cos2 θ − sin3 θ + sin θ sin θ

(

)

∇ ⋅ v ⃗ = 3 cos ϕ + 2 cos2 θ − sin2 θ +

cos ϕ . sin θ

For the volume,

0 ⩽ r ⩽ R,

⎛ 1 ⎞ π ⎛a⎞ π π tan−1⎜ ⎟ = tan−1⎜ = → ⩽θ⩽ , ⎟ ⎝ R⎠ 6 2 ⎝ 3⎠ 6

1-25

0 ⩽ ϕ ⩽ 2π .

Electromagnetism

So R

π 2

2π

∫V ∇ ⋅ v ⃗ dτ = ∫ ∫ ∫ 0

∫V

∇ ⋅ v ⃗ dτ = −

π 6

0

⎛ cos ϕ ⎞ 2 ⎜ 3 cos ϕ + 2 cos2 θ − sin2 θ + ⎟ r sin θ dϕ dθ dr ⎝ sin θ ⎠

(

)

3 3 πR . 12

Now for the right-hand side, we have three surfaces: the bottom (i), the outer shell (ii), and the inner part where the cone is cut out (iii). We have

v ⃗ = r cos ϕ rˆ + r sin ϕ ϕˆ + r cos θ sin θ θˆ. π For (i), we have da ⃗ = r dr dϕ θˆ and θ = 2 . So

v ⃗ ⋅ da ⃗ = r 2 cos

π π sin dr dϕ = 0 2 2

and

∫

v ⃗ ⋅ da ⃗ = 0.

(i )

For (ii), we have r = R and da ⃗ = r 2 sin θ dθ dϕ rˆ = R2 sin θ dθ dϕ rˆ . So

v ⃗ ⋅ da ⃗ = R3 cos ϕ sin θ dθ dϕ and 2π

∫

v ⃗ ⋅ da ⃗ = R3

(ii )

For (iii), we have θ =

∫∫ 0

π 6

π 2

cos ϕ sin θ dθ dϕ = 0.

π 6

and da ⃗ = −r sin θ dr dϕ θˆ = − 12 r dr dϕ θˆ . So

1 3 2 π π v ⃗ ⋅ da ⃗ = − r 2 cos sin = − r 2 6 6 8 and

∫ (iii )

3 v ⃗ ⋅ da ⃗ = − 8

R 2π

∫∫ 0

0

1-26

r 2 dϕ dr = −

3 πR3. 12

Electromagnetism

Therefore,

∮S v ⃗ ⋅ da ⃗ = − 123 πR3 as expected. Problem 1.17. Test the curl theorem with v ⃗ = s 2z sˆ + sin ϕ cos ϕ ϕˆ + zs cos ϕ zˆ and half of a cylindrical shell with radius R and height h.

Solution The curl theorem states

∫S ( ∇ × v ⃗) ⋅ da ⃗ = ∮P v ⃗ ⋅ dl⃗. Starting with the left-handed side, we have

da ⃗ = s dϕ dz sˆ = R dϕ dz sˆ . Since we are dotting da ⃗ with ∇ × v ⃗ , we only need the sˆ component of the curl:

⎡1 ∂ ⎤ ⎛ 1 ∂vz ∂vϕ ⎞ ∂ − [∇ × v ⃗ ] s = ⎜ (zs cos ϕ) − (sin ϕ cos ϕ)⎥sˆ ⎟sˆ = ⎢ ∂z ∂z ⎠ ⎣ s ∂ϕ ⎦ ⎝ s ∂ϕ = −z sin ϕ sˆ . So

( ∇ × v ⃗) ⋅ da ⃗ = −Rz sin ϕ dϕ dz. 1-27

Electromagnetism

We have

0⩽ϕ⩽π

and

0⩽z⩽h

so h

π

∫S ( ∇ × v ⃗) ⋅ da ⃗ = ∫ ∫ 0

−Rz sin ϕ dϕ dz = −h 2R .

0

For the left-hand side, we have four curves

with

v ⃗ = s 2z sˆ + sin ϕ cos ϕ ϕˆ + z cos ϕ zˆ . For curve (i), dl⃗ = dϕ ϕˆ , z = 0, and s = R . So

v ⃗ ⋅ dl⃗ = sin ϕ cos ϕ dϕ and π

∫

sin ϕ cos ϕ dϕ = 0.

0

For curve (ii), dl⃗ = dz zˆ , ϕ = π , and s = R . So

v ⃗ ⋅ dl⃗ = zs cos ϕ dz = zR cos π dz = −zR dz and h

∫ 0

1 −zR dz = − h 2R . 2

1-28

Electromagnetism

For curve (iii), dl⃗ = dϕ ϕˆ , z = h, and s = R . So

v ⃗ ⋅ dl⃗ = sin ϕ cos ϕ dϕ and 0

∫

sin ϕ cos ϕ dϕ = 0.

π

For curve (iv), dl⃗ = dz zˆ , ϕ = 0, and s = R . So

v ⃗ ⋅ dl⃗ = zs cos ϕ dz = zR cos(0)dz = zR dz and 0

∫ h

1 zR dz = − h 2R2 . 2

So,

∮ v ⃗ ⋅ dl⃗ = − 12 h2R − 12 h2R = − h2R P

as expected. Problem 1.18. Test the gradient theorem using T = sz 2 sin ϕ and the half helix path (radius R , height h).

1-29

Electromagnetism

Solution The gradient theorem states

∫P ∇T ⋅ dl⃗ = T ( b ⃗) − T ( a ⃗). Starting with the right-hand side ⎛ π ⎞ ⎛ ⎛ π⎞ π ⎞ π T b ⃗ − T a ⃗ = T ⎜ R , , h⎟ − T ⎜ R , − , 0⎟ = Rh 2 sin − R(0)2 sin⎜ − ⎟ = h 2R . ⎝ 2 ⎠ ⎝ ⎠ ⎝ 2⎠ 2 2

()

()

Now, the gradient is

∇T =

1 ∂T ˆ ∂T ∂T sˆ + zˆ = z 2 sin ϕ sˆ + z 2 cos ϕ ϕˆ + 2sz sin ϕ zˆ . ϕ+ s ∂ϕ ∂z ∂s

We also have s = R and l⃗ = s dϕ ϕˆ + dz zˆ = R dϕ ϕˆ + dz zˆ. So ∇T ⋅ dl⃗ = Rz 2 cos ϕ dϕ + 2Rz sin ϕ dz . We need a way to relate z and ϕ. Note that as ϕ increases, z increases linearly. So, using the equation of line

z − z0 = γ (ϕ − ϕ0) , when z = 0 and ϕ =

when z = h and ϕ =

π − 2,

π , 2

⎛ π⎞ z = γ⎜ ϕ + ⎟, ⎝ 2⎠ ⎛π π⎞ h h = γ⎜ + ⎟ → γ = , ⎝2 ⎠ 2 π

so

z=

h h ϕ− 2 π

and

dz =

h dϕ . π

Using our expressions for z and dz , we have

⎡ ⎛h ⎛h ⎛ h ⎞⎤ h ⎞2 h⎞ ∇T ⋅ dl⃗ = ⎢R⎜ ϕ + ⎟ cos ϕ + 2R⎜ ϕ + ⎟ sin ϕ ⎜ ⎟⎥dϕ . ⎝π ⎝ π ⎠⎦ 2⎠ 2⎠ ⎣ ⎝π So π 2

b⃗

∫

∇ T ⋅ d l⃗ =

a⃗

∫ −π 2

⎡ ⎛h ⎛h ⎛ h ⎞⎤ h ⎞2 h⎞ ⎢R⎜ ϕ + ⎟ cos ϕ + 2R⎜ ϕ + ⎟ sin ϕ ⎜ ⎟⎥dϕ = h 2R ⎝ π ⎠⎦ ⎝π 2⎠ 2⎠ ⎣ ⎝π

as expected.

1-30

Electromagnetism

Problem 1.19. Evaluate the following integrals: 3

a)

∫

(2x 2 − x + 4)δ (x − 2)dx

1 1

b)

∫

(x 2 + 4)δ (x − 2)dx

−1 6

c)

∫

3x

sin( 2 )δ (x − π )dx

2 2

d)

∫

(2x 3 + 1)δ (4x )dx

−2 ∞

e)

∫

x 2δ (2x + 1)dx

−∞ a

f)

∫

δ (x − b )dx

0

Solutions a) 3

∫ ( 2x 2 − x + 4)δ(x − 2)dx. 1

Since 2 ∈ (1, 3) and f (x ) = 2x 2 − x + 4, we have 3

∫ ( 2x 2 − x + 4)δ(x − 2)dx = f (2) = 2(2)2 − 2 + 4 = 10. 1

b)

1

∫ ( x 2 + 4)δ(x − 2)dx. −1

Since 2 ∉ ( −1, 1), we have 1

∫ ( x 2 + 4)δ(x − 2)dx = 0. −1

c)

6

∫ 2

⎛ 3x ⎞ sin ⎜ ⎟δ(x − π )dx . ⎝ 2 ⎠

1-31

Electromagnetism

3x

Since π ∈ (2, 6) and f (x ) = sin( 2 ), we have 6

⎛ 3x ⎞ ⎛ 3π ⎞ sin ⎜ ⎟δ(x − π )dx = f (π ) = sin ⎜ ⎟ = −1. ⎝ 2 ⎠ ⎝2⎠

∫ 2

d) 2

∫ ( 2x3 + 1)δ(4x)dx. −2

Since 0 ∈ ( −2, 2) and f (x ) = 2x 3 + 1, we have 2

∫ ( 2x3 + 1)δ(4x)dx = −2

e)

1 1 1 2(0)3 + 1) = . f (0) = ( 4 4 4

∞

∫

x 2δ(2x + 1)dx .

−∞

This can be rewritten as ∞

∫

∞ 2

x δ(2x + 1)dx =

−∞

Since

∫ −∞

1 −2

⎡ ⎛ 1 ⎞⎤ x δ⎢2⎜x + ⎟⎥dx = ⎣ ⎝ 2 ⎠⎦ 2

∞

∫

x2

−∞

1 ⎛ 1⎞ δ⎜x + ⎟dx . ⎝ 2 2⎠

∈ ( −∞ , ∞) and f (x ) = x 2 , we have ∞

∫

x2

−∞

1 ⎛ 1⎞ 1 ⎛ 1 ⎞2 1 1 ⎛ 1⎞ ⎜− ⎟ = . δ⎜x + ⎟dx = f ⎜− ⎟ = 2 ⎝ 2⎠ 2 ⎝ 2⎠ 8 2 ⎝ 2⎠

f)

a

∫

δ(x − b)dx .

0

Here we have a

∫ 0

δ(x − b)dx =

{

1 if 0 < b < a . 0 otherwise

Problem 1.20. Suppose we have two vector fields F1⃗ = y 2 zˆ and F2⃗ = xxˆ + yyˆ + zzˆ . Calculate the divergence and curl of each. Which can be written as the gradient of a scalar and which can be written as the curl of a vector? Find a scalar and a vector potential.

1-32

Electromagnetism

Solution For F1⃗ , we have

⎛∂ ∂( y 2 ) ∂ ∂ ⎞ =0 ∇ ⋅ F1⃗ = ⎜ xˆ + yˆ + zˆ⎟ ⋅ ( y 2 zˆ ) = ∂y ∂z ⎠ ∂z ⎝ ∂x and

xˆ yˆ zˆ ∂ ∂ ∂ ∇ × F1⃗ = = 2yxˆ . ∂x ∂y ∂z 0 0 y2 For F2⃗ , we have

⎛∂ ∂ ∂ ⎞ yˆ + zˆ⎟ ⋅ (xxˆ + yyˆ + zzˆ ) = 1 + 1 + 1 = 3 ∇ ⋅ F2⃗ = ⎜ xˆ + ∂y ∂z ⎠ ⎝ ∂x and

xˆ yˆ zˆ ∂ ∂ ∂ ∇ × F2⃗ = = (0 − 0)xˆ + (0 − 0)yˆ + (0 − 0)zˆ = 0. ∂x ∂y ∂z x y z Since ∇ ⋅ F1⃗ = 0, F1⃗ can be expressed as F1⃗ = ∇ × A⃗ . We can find A⃗ by considering

xˆ yˆ zˆ ∂ ∂ ∂ ∇ × A⃗ = ∂x ∂y ∂z 0 0 y2 ⎛ ∂Ay ⎛ ∂A ∂Ay ⎞ ⎛ ∂Ax ∂Az ⎞ ∂Ax ⎞ ⎟yˆ + ⎜ − − =⎜ z − ⎟zˆ . ⎟xˆ + ⎜ ⎝ ∂z ∂x ⎠ ∂y ⎠ ∂z ⎠ ⎝ ∂x ⎝ ∂y By inspection:

∂Ay ∂Az − = 0, ∂y ∂z

∂Az ∂Ax = 0, − ∂x ∂z

∂Ay ∂Ax = y2 . − ∂y ∂x

This is satisfied by

ˆ A ⃗ = y 2 xy, which is just one example. Since ∇ × F2⃗ = 0, F2⃗ can be expressed as F2⃗ = −∇V . We can find V by considering

1-33

Electromagnetism

⎛ ∂V ∂V ∂V ⎞ zˆ⎟ . yˆ + F2⃗ = −⎜ xˆ + ∂y ∂z ⎠ ⎝ ∂x By inspection:

x=−

∂V , ∂x

y=−

∂V , ∂y

z=−

∂V . ∂z

This is satisfied by

⎛ x2 y2 z2 ⎞ V = −⎜ + + ⎟ ⎝2 2 2⎠ which is again just one example.

Bibliography Byron F W and Fuller R W 1992 Mathematics of Classical and Quantum Physics (New York: Dover) Griffiths D J 1999 Introduction to Electrodynamics 3rd edn (Englewood Cliffs, NJ: Prentice Hall) Griffiths D J 2013 Introduction to Electrodynamics 4th edn (New York: Pearson) Halliday D, Resnick R and Walker J 2010 Fundamentals of Physics 9th edn (New York: Wiley) Halliday D, Resnick R and Walker J 2013 Fundamentals of Physics 10th edn (New York: Wiley) Purcell E M and Morin D J 2013 Electricity and Magnetism 3rd edn (Cambridge: Cambridge University Press) Rogawski J 2011 Calculus: Early Transcendentals 2nd edn (San Francisco, CA: Freeman)

1-34

IOP Concise Physics

Electromagnetism Problems and solutions Carolina C Ilie and Zachariah S Schrecengost

Chapter 2 Electrostatics

Electrostatics is the topic of this chapter. Coulomb’s law, Gauss’s law, and the energy of various charge distributions are a few ways of understanding the electric field. The methods employed will make use of the specific degrees of symmetry. The mathematical skills obtained in chapter 1 will be applied here to analyze different charge distributions in Cartesian, spherical, or cylindrical coordinates.

2.1 Theory 2.1.1 Coulomb’s law The force on a point charge q due to a charge Q , separated by a distance r , is given by JG 1 Q F = q rˆ, 4πεo r 2 where ε0(= 8.85 × 10−12

C2 ) Nm2

is the permittivity of free space.

2.1.2 Electric field In general, for a volume charge density ρ(r ⃗ ), the electric field at r ⃗ is given by JG 1 ρ(r ⃗′) rˆ dτ′ . E (r ⃗ ) = 4πεo V r 2

∫

For a surface charge density σ (r ⃗ ), the electric field is given by

JG E (r ⃗ ) =

doi:10.1088/978-1-6817-4429-2ch2

1 4πεo

∫S σr(r2⃗′) rˆ da′.

2-1

ª Morgan & Claypool Publishers 2016

Electromagnetism

For a linear charge density λ(r ⃗ ), the electric field is given by JG 1 λ(r ⃗′) rˆ dl′ . E (r ⃗ ) = 4πεo P r 2

∫

2.1.3 Gauss’s law

JG For an electric field E and surface S , Gauss’s law states JG JG q E ⋅ da = enc , ε0 S

∮

where the enclosed charge is

qenc = In differential form

∫V ρ dτ.

JG ρ ∇⋅ E = , εo

where ρ is the volume charge density. JG 2.1.4 Curl of E

JG

∮P E

JG ⋅ dl⃗ = 0 →  ∇ × E = 0,

JG where E is an electrostatic field.

2.1.5 Energy of a point charge distribution The energy required to assemble n charges q1, q2, …, qn is given by ⎛ ⎞ n ⎜ n 1 qj ⎟ 1 n 1 ⎟ = ∑q V (ri ⃗ ), W = ∑ qi ⎜ ∑ 2 i = i ⎜⎜ j = i 4πεo rij ⎟⎟ 2 i = 1 i ⎝ j ≠i ⎠ where V (ri ⃗ ) is the potential at charge qi and r ij is the distance between charges qi and qj . 2.1.6 Energy of a continuous distribution

W=

1 2

∫ ρV (r ⃗) dτ =   ε2o ∫

all space

2.1.7 Energy per unit volume

W ε = o E2 volume 2

2-2

E 2 dτ .

Electromagnetism

2.2 Problems and solutions Problem 2.1. Given the charge distribution below, find the force on charge q1 = q with q2 = 3q , q3 = −2q , and q4 = q .

Solution The force on q1 from q2 is given by JG 1 q1q2 1 q1q2 JG rˆ = r , F21 = 2 4πεo r 4πεo r 3 where

JG r = 3xˆ + 4yˆ

and

r =

32 + 42 = 5.

So

JG F21 =

1 3q 2 q2 ⎛ 9 12 ⎞ ⎜ ˆ ˆ (3 4 ) + = x y xˆ + yˆ ⎟ . 3 4πεo 5 4πεo ⎝ 125 125 ⎠

The force on q1 from q3 is given by JG F31 =

1 q1q3 rˆ, 4πεo r 2

where

r=4 and

ˆ rˆ = y.

2-3

Electromagnetism

So

JG F31 =

1 ⎛ 2q 2 ⎞ q2 1 yˆ . ⎜ − 2 ⎟yˆ = − 4πεo ⎝ 4 ⎠ 4πεo 8

The force on q1 from q4 is given by JG F41 =

1 q1q4 rˆ, 4πεo r 2

where

r=2 and

ˆ rˆ = x. So

JG F41 =

1 q2 q2 1 rˆ = xˆ . 2 4πεo 2 4πεo 4

Therefore,

JG JG JG JG ⎛ 12 q 2 ⎡⎛ 9 1⎞ 1⎞ ⎤ + ⎟xˆ + ⎜ − ⎟yˆ ⎥ F1 = F21 + F31 + F41 = ⎢⎜ ⎝ 125 4πεo ⎣⎝ 125 4⎠ 8⎠ ⎦ JG q 2 ⎛ 161 29 ⎞ ⎜ F1 = xˆ − yˆ ⎟ . 4πεo ⎝ 500 1000 ⎠ Problem 2.2. Given a charged sheet with surface charge density σ = ky (where k is a constant) and sides of length 2d , find the electric field z above the center of the sheet.

2-4

Electromagnetism

The electric field is given by

JG E =

1 4πεo

∫ rσ2 rˆ da.

The horizontal components cancel so we only have the zˆ -component: z rˆ → cos θ zˆ = r zˆ .

Also we have da = dx dy and r 2 = x 2 + y 2 + z 2 . Note that the piece of the sheet in each quadrant of the xy -plane contributes the same amount to the total field. Therefore,

JG JG E = 4Equad =

kzzˆ = πεo

d

∫ 0

4 4πεo

d d

∫∫ 0 0

⎡ y⎢ ⎢ y2 + z2 ⎣

(

kyzzˆ

(

kzzˆ dx dy = 3/2 πεo x2 + y2 + z2

)

⎤ ⎥ dy = kzdzˆ πεo y 2 + z 2 + x 2 ⎥⎦ x=0 x= d

x

)

d d

∫∫ 0 0

y dx dy 3/2  (x + y + z 2 ) 2

2

d

∫

y 0 (

y 2

+ z2

)

y2 + z2 + d 2

Let

u2 = y2 + z2 so

2u du = 2y dy → u du = y dy . Evaluating u at the endpoints yields

u 2(y = 0) = z 2 → u = z u 2 (y = d ) = d 2 + z 2 → u =

d 2 + z2 .

Now

JG kzdzˆ E = πεo

=

z 2+ d 2

∫ z

du u u2 + d 2

kzdzˆ ⎡⎢ 1 ⎛ d + ln⎜ πεo ⎢⎣ d ⎝

d 2 + u 2 ⎞⎤⎥ ⎟ u ⎠⎥⎦ u=

u=z

z 2+d 2

⎤ ⎡ 2 2 z2 + d 2 ⎥ JG kz ⎢ d + z + d ln E = ⎥zˆ . 2 2 πεo ⎢⎢ + + z d 2 d z ⎥⎦ ⎣

(

)

(

)

2-5

⎛ 2 2 ⎜ d+ d +z kzzˆ ⎜ z ln = πεo ⎜⎜ d + 2d 2 + z 2 ⎜ ⎝ z2 + d 2

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

dy .

Electromagnetism

Problem 2.3. Find the electric field d above a cylinder of radius R , height h, and volume density ρ (ignoring edge effects).

Solution We have

JG E =

1 4πεo

∫ rρ2 rˆ dτ.

Note our horizontal components cancel, so rˆ → cos θ zˆ with

cos θ =

d+h−z . r

Also

dτ = s ds dϕ dz and

r 2 = s 2 + (d + h − z )2 . Therefore,

JG E =

ρ 4πεo

2π

R

h

∫ ∫ ∫ 0

0

0

(d + h − z )s zˆ ⎡ s 2 + (d + h − z )2 ⎤3/2 ⎣ ⎦

dz ds dϕ

2πρ ⎡ 2 R + d 2 − R2 + (d + h)2 + h⎤⎦zˆ 4πεo ⎣ JG ρ ⎡ 2 E = R + d 2 − R2 + (d + h)2 + h⎤⎦zˆ . 2εo ⎣ =

2-6

Electromagnetism

Note if R ≫ d and R ≫ h, the field reduces to JG ρh E = zˆ , 2εo which is the field given by an infinite sheet of surface charge σ = ρh. Problem 2.4. Given the bottom hemisphere of a spherical shell of radius R , thickness d , and volume charge density ρ, find the electric field z above the center (above the open part, ignoring edge effects).

Solution The electric field is given by JG 1 E = 4πεo

∫ rρ2 rˆ dτ.

We can see from

that

R ⩽ r ⩽ R + d,

0 ⩽ ϕ ⩽ 2π ,

2-7

π ⩽ θ ⩽ π. 2

Electromagnetism

Also, dτ = r 2 sin θ dr dϕ dθ . From the law of cosines,

r 2 = z 2 + r 2 − 2rz cos θ . Since the horizontal components cancel, rˆ becomes

rˆ → cos γ zˆ =

z+b zˆ , r

where b is given by

b → b = −r cos θ . r

cos(π − θ ) = So

z − r cos θ zˆ . r

rˆ → Therefore,

JG E =

R+d

ρ 4πεo

ρzˆ = 2εo

2π

π

∫ ∫ ∫ π 2

R

R+d

0

π

(z

2

+ r 2 − 2rz cos θ

r 2(z − r cos θ )sin θ zˆ

∫ ∫

(z

π 2

R

r 2(z − r cos θ ) sin θ zˆ

2

+ r 2 − 2rz cos θ

3/2

)

3/2

)

dϕ dθ dr

dθ dr .

Let u = cos θ and du = −sin θ dθ

⎛ π⎞ u⎜θ = ⎟ = 0 ⎝ 2⎠ u (θ = π ) = − 1 JG ρzˆ E = 2εo

0

R +d

r 2(z − ru )zˆ

∫ ∫ R

ρzˆ = 2εo z 2

( z 2 + r 2 − 2rzu )3/2

−1 R +d

∫ R

⎛ ⎜r 2 − ⎝

⎡ ρzˆ ⎢ r 3 = − 2εoz 2 ⎢⎣ 3 JG ρzˆ E = 6εoz 2

⎞ ⎟ dr r2 + z2 ⎠ r3

r 2 + z 2 ( r 2 − 2z 2 ) ⎤ ⎥ ⎥⎦ 3

r=R +d

r=R

{ (R + d ) −

− R3 +

3

(R

du dr

2

+ z2

(R + d )2 + z 2 ⎡⎣ (R + d )2 − 2z 2⎤⎦

)( R 2-8

2

− 2z 2

)}.

Electromagnetism

JG Problem 2.5. Given the electric field E = k [2xzxˆ + z 2yˆ + (x 2 + 2yz )zˆ ] (with constant k ) find the following: a) The charge density ρ. b) The charge enclosed by a cylinder of height h, radius R , and base on the xy -plane center at the origin (below). c) The charge enclosed by an upper hemisphere of radius R centered at the origin.

Solutions a) The charge density ρ. Gauss’s law states

JG ρ ∇⋅E = . εo So

JG ρ = εo  ∇ ⋅ E = kεo(2y + 2z ) = 2kεo(y + z ).

b) The charge enclosed by a cylinder of height h, radius R , and base on the xy -plane center at the origin. We have

qenc =

∫V ρ dτ,

2-9

Electromagnetism

with

ρ = 2kεo(y + z ). We can transform ρ into cylindrical coordinates using x = s cos ϕ, y = s sin ϕ, and z = z. So R

qenc = 2kεo

h

2π

∫∫∫ 0

0

(s sin ϕ + z )s dϕ dz ds = πkεoh 2R2 .

0

c) The charge enclosed by an upper hemisphere of radius R centered at the origin. Again

qenc =

∫V ρ dτ

with

ρ = 2kεo(y + z ), but now y = r sin ϕ sin θ and z = r cos θ . So R 2π

qenc = 2kεo

π 2

∫∫∫ 0

0

r(sin ϕ sin θ + cos θ )r 2 sin θ dθ dϕ dr =

0

kεoπR 4 . 2

Problem 2.6. Given a charge q located in the center of a spherical shell of radius R and surface charge σ = k sin θ (with constant k ), find the electric field inside and outside the shell.

Solution We will use Gauss’s law to find the field JG JG q E ⋅ da = enc , εo S

∮

2-10

Electromagnetism

where

JG

∮S E

JG ⋅ da =

∮S E da = E ∮S da = E 4πr 2.

For r < R , we have

qenc = q. So

E 4πr 2 =

JG 1 q q rˆ . →E = 4πεo r 2 εo

For r > R , we have 2π π

qenc = q +

∫ σ da = q + k ∫ ∫ R2 sin2 θ dθ dϕ. 0

0

2 2

qenc = q + kR π . So

E 4πr 2 =

q kR2π 2 + εo εo

and

JG E =

1 q + kR2π 2 rˆ . 4πεo r2

Problem 2.7. Given a line of charge carrying λ surrounded by a cylindrical shell with inner radius a , outer radius b, and charge density ρ = ks 2 , find the electric field in the regions s < a , a < s < b, and b < s .

Solution Gauss’s law states

JG

∮S E

JG q ⋅ da = enc , εo

2-11

Electromagnetism

where

JG

∮S E

JG ⋅ da =

∮S E da = E ∮S

da = E 2πsl .

For s < a , we have

qenc = λl . So

E 2πsl =

λl εo

and

JG E =

λ sˆ . 2πεos

For a < s < b, we have l

qenc = λl +

2π

s

∫ ρ dτ = λ l + ∫ ∫ ∫ 0

0

a

⎡ ⎤ kπ 4 k (s′)2 s′ ds′ dϕ′ dz′ = l⎢λ + s − a 4 ⎥. ⎣ ⎦ 2

(

)

So

JG JG q E ⋅ da = enc εo S

∮

⎡ ⎤ kπ 4 l⎢λ + s − a4 ⎥ ⎣ ⎦ 2 → E 2πsl = εo

(

)

and

(

)

JG 2λ + kπ s 4 − a 4 sˆ . E = 4πεos For b < s , we have l

qenc = λl +

2π

b

∫ ρ dτ = λ l + ∫ ∫ ∫ 0

0

a

⎡ ⎤ kπ 4 k (s′)2 s′ ds′ dϕ′ dz′ = l⎢λ + b − a 4 ⎥. ⎣ ⎦ 2

(

So

JG

∮S E

JG q ⋅ da = enc εo

⎡ ⎤ kπ 4 l⎢λ + b − a4 ⎥ ⎣ ⎦ 2 → E 2πsl = εo

(

2-12

)

)

Electromagnetism

and

(

)

JG 2λ + kπ b 4 − a 4 sˆ . E = 4πεos Problem 2.8. Which of the following is a possible electrostatic field? JG a) E = k (yzxˆ + xzyˆ + x 2zˆ ) JG b) E = k (xxˆ + yyˆ + zzˆ ) JG c) E = k [2xzxˆ + z 2yˆ + (x 2 + 2yz )zˆ ] where k is a constant with the appropriate units for the given field. For the possible electric field, find the electric potential JG using the origin as your reference point. Check your answer by verifying that E = −∇V . Solutions JG a) E = k yzxˆ + xzyˆ + x 2zˆ

(

xˆ JG ∇×E =

yˆ

)

zˆ

∂ ∂ ∂ ∂x ∂y ∂z yz xz x 2

⎡∂ ⎤ ⎡∂ ⎤ ⎡∂ ∂ ∂ 2⎤ ∂ =⎢ x2 − (xz )⎥xˆ + ⎢ (yz ) − x )⎥yˆ + ⎢ (xz ) − (yz )⎥zˆ ( ⎣ ∂z ⎦ ∂y ⎣ ∂y ⎦ ⎣ ∂x ⎦ ∂x ∂z

( )

= (0 − x )xˆ + (y − 2x )yˆ + (z − z )zˆ = −xxˆ + ( −2x + y )yˆ . JG Since ∇ × E ≠ 0, this is not a possible electric field. JG b) E = k (xxˆ + yyˆ + zzˆ )

JG ∇×E =

xˆ yˆ zˆ ∂ ∂ ∂ ∂x ∂y ∂z x y z

⎡∂ ⎤ ⎡∂ ⎤ ⎡∂ ⎤ ∂ ∂ ∂ (y )⎥xˆ + ⎢ (x ) − (z )⎥yˆ + ⎢ (y ) − (x )⎥zˆ = 0. = ⎢ (z ) − ⎣ ∂z ⎣ ∂y ⎣ ∂x ∂y ⎦ ∂x ⎦ ∂z ⎦ JG Since ∇ × E = 0, this is a possible electric field. Let us find the electric potential by integrating along the path given by

2-13

Electromagnetism

The potential is given by r⃗

V=−

∫

JG E ⋅ dl ⃗ .

O

where

JG E ⋅ dl⃗ = k (xxˆ + yyˆ + zzˆ ) ⋅ (dxxˆ + dyyˆ + dzzˆ ) = k (x dx + y dy + z dz ). Note along (i) we only have dx , along (ii) we only have dy , and along (iii) we only have dz . Therefore, taking the origin O=(0, 0, 0) as our reference point, the potential will be given by

r⃗

V=−

∫ O

⎛ ⎞ ⎜ x ⎟ y z JG ⎜ ⎟ ⃗ E ⋅ dl = −k ⎜ x′dx′ + y′dy′ + z′dz′ ⎟ . ⎜ 0 ⎟ 0 0 ⎜       ⎟ ⎝ (i) ⎠ (ii) (iii)

∫

∫

∫

So,

V (r ) = −

k 2 x + y2 + z2 . 2

(

2-14

)

Electromagnetism

We can check this using JG E = − ∇V

⎧∂ ⎡ k ⎤ ⎤ ∂ ⎡ k 2 − ( x + y 2 + z 2 )⎥yˆ = − ⎨ ⎢ − ( x 2 + y 2 + z 2 )⎥xˆ + ⎢ ⎦ ⎦ ∂y ⎣ 2 ⎩ ∂x ⎣ 2 +

⎤ ⎫ ∂⎡ k 2 2 2 ⎢⎣ − ( x + y + z )⎥⎦zˆ⎬ ∂z ⎭ 2

2kx 2ky 2kz xˆ + yˆ + zˆ 2 2 2 JG = k (xxˆ + yyˆ + zzˆ ) = E . =

JG c) E = k [2xzxˆ + z 2yˆ + (x 2 + 2yz )zˆ ]

xˆ yˆ JG ∂ ∂ ∇×E = ∂x ∂y 2xz z 2 x 2

zˆ ∂ ∂z + 2yz

⎡∂ ⎡∂ ⎤ ∂ 2 ∂ 2⎤ = ⎢ ( x 2 + 2yz ) − z )⎥xˆ + ⎢ (2xz ) − x + 2yz )⎥yˆ ( ( ⎣ ∂z ⎦ ⎣ ∂y ⎦ ∂x ∂z ⎡∂ ⎤ ∂ + ⎢ ( z 2) − (2xz )⎥zˆ ⎣ ∂x ⎦ ∂y = k ⎡⎣ (2z − 2z )xˆ + (2x − 2x )yˆ + (0 − 0)zˆ⎤⎦ = 0. JG Since ∇ × E = 0, this is a possible electric field. Let us find the electric potential by integrating along the same path as before. The potential is given by r⃗

V=−

∫

JG E ⋅ dl ⃗

O

where

JG E ⋅ dl⃗ = k ⎡⎣ 2xzxˆ + z 2yˆ + x 2 + 2yz zˆ⎤⎦ ⋅ (dxxˆ + dyyˆ + dzzˆ )

(

)

= k ⎡⎣ 2xzdx + z 2dy + x 2 + 2yz dz⎤⎦ .

(

)

Note along (i) we only have dx with y = 0 and z = 0, along (ii) we only have dy with x = 1 and z = 0, and along (iii) we only have dz with x = 1 and y = 1. Therefore, taking the origin O = (0, 0, 0) as our reference point, the potential will be given by

2-15

Electromagnetism

r⃗

V =−

∫ O

⎛ ⎞ ⎜ x ⎟ y z JG ⎜ ⎟ E ⋅ dl ⃗ = − k ⎜ 2x′z dx′ + z 2dy′ + (2yz′ + x 2 )dz′ ⎟ ⎜ 0 ⎟ 0  

0         ⎟ ⎜  ⎝ ⎠ (i) (ii) (iii)

∫

∫

∫

⎛ ⎞ ⎜ x ⎟ y z ⎜ ⎟ 2 2 = − k⎜ 2x′(0)dx′ + 0 d y′ + (2(1)z′+1 )dz′ ⎟ ⎜ 0 ⎟ 0  

0        ⎟ ⎜  ⎝ ⎠ (i) (ii) (iii)

∫

∫

(

∫

)

V (r ) = − k yz 2 + x 2z . We can check this using

JG E = −∇V ⎧∂ ⎫ ∂⎡ ∂ ⎡ 2 2 ⎤ˆ 2 2 ⎤ ⎬ − + ˆ − + + = − ⎨ ⎡⎣ −k yz 2 + x 2z ⎤⎦xˆ + k yz x z y k yz x z z ⎦ ⎦ ⎭ ∂z ⎣ ∂y ⎣ ⎩ ∂x JG = k ⎡⎣ (2xz )xˆ + z 2 yˆ + x 2 + 2yz zˆ⎤⎦ = E .

(

)

( )

(

(

)

(

)

)

Problem 2.9. Find the electric field and the electric potential inside and outside a thin spherical shell of radius R that carries a uniform surface charge σ . Set the reference point at infinity.

Solution Let us find the electric field everywhere by using Gauss’s law, given by JG JG q E ⋅ da = enc , ε0 S

∮

2-16

Electromagnetism

where

JG

∮S E

JG ⋅ da =

∮S E da = E ∮S da = E 4πr 2.

For r < R, we have our Gaussian surface given by

where r is the radius of the Gaussian sphere with radius smaller than R. Note that

qenc = 0. So,

JG E 4πr 2 = 0 → E = 0.

For r > R, we have our Gaussian surface given by

Now we have

qenc = σ 4πR2 . So,

E 4πr 2 =

σ 4πR2 ε0

2-17

Electromagnetism

and

JG σR2 ˆ r. E = ε0r 2 Now let us calculate the electric potential everywhere taking the reference point at infinity. We will use r

V=−

∫

JG E ⋅ dl ⃗ ,

∞

where

dl⃗ = dr rˆ + r dθ θˆ + r sin θ dϕ ϕˆ . For r > R ,

therefore, r⃗

V =−

∫ ∞

JG E ⋅ dl ⃗ = −

r⃗

∫ ∞

σR 2 rˆ ⋅ dr rˆ + r dθ θˆ + r sin θ dϕ ϕˆ = − ε0r 2

(

)

2

V=

σR . ε0r

For r < R

2-18

r

∫ ∞

σR 2 dr ′ ε0r′2

Electromagnetism

r⃗

V=−

∫

JG E ⋅ dl ⃗ = −

∞

R

∫ ∞

σR 2 dr − ε0r

r

∫

0 dr ′ =

R

σR 2 σR −0= = const. ε0R ε0

Note that the potential inside the shell is constant, as the electric field is zero. Problem 2.10. Calculate the electric field and the electric potential inside and outside a solid sphere of radius R having a uniform charge distribution ρ. Use infinity as your JG reference point. Then obtain the gradient of the potential everywhere and check that E = −∇V . Plot the potential versus distance from the center of the sphere.

Solution Starting with the electric field, we use Gauss’s law, given by JG JG q E ⋅ da = enc , ε0 S

∮

where

JG

∮S E

JG ⋅ da =

∮S E da = E ∮S

da = E 4πr 2 .

For r > R, we have our Gaussian surface given by

2-19

Electromagnetism

Here we simply have

qenc = ρVsp =

ρ4πR3 , 3

where Vsp is the volume of the sphere. So

E 4πr 2 =

ρ4πR3 3ε0

and

JG ρR3 rˆ . E = 3ε0r 2 For r < R , our Gaussian surface becomes

Now,

qenc = ρVenc =

ρ4πr 3 , 3

where Venc is the enclosed volume. So

E 4πr 2 =

ρ4πr 3 3ε0

and

JG ρr E = rˆ . 3ε0

2-20

Electromagnetism

The plot of electric field is given by

Now we can calculate the electric potential. This is done using r⃗

V=−

JG E ⋅ dl ⃗

∫ ∞

with infinity as the reference point. For r > R

with r⃗

V=−

∫ ∞

JG E ⋅ dl ⃗ = −

r

∫ ∞

2-21

ρR3 ρR3 dr ′ = . 2 3ε0r′ 3ε0r

Electromagnetism

For r < R

with r⃗

V=−

∫ ∞

JG E ⋅ dl ⃗ = −

R

∫ ∞

ρR2 dr − 3ε0r 2

r

∫ R

ρr′ ρR2 ρr 2 dr ′ = . − 3ε0 2ε0 6ε0

JG We can check using E = −∇V . For r > R ,

JG ∂ ⎛ ρR2 ρr 2 ⎞ ρr − E = −∇V = − ⎜ rˆ ⎟ rˆ = ∂r ⎝ 2ε0 6ε0 ⎠ 3ε0 and for r < R ,

JG ∂ ⎛ ρR3 ⎞ ρR3 rˆ E = −∇V = − ⎜ ⎟ rˆ = ∂r ⎝ 2ε0r ⎠ 3ε0r 2 both of which are in agreement with what we found from Gauss’s law. Problem 2.11. Calculate the electric field and the electric potential for a sphere of radius R that carries a charge density ρ = kr 2 , where k is a constant.

2-22

Electromagnetism

Solution Starting with the electric field, we use Gauss’s law, given by

JG

∮S E

JG q ⋅ da = enc , ε0

where

JG

∮S E

JG ⋅ da =

∮S E da = E ∮S

da = E 4πr 2 .

For r > R, we have our Gaussian surface given by

and our enclosed charge is given by 2π

qenc =

∫ ρ dτ = ∫ 0

R

π

dϕ

∫

sin θ dθ

0

∫ 0

Therefore,

E 4πr 2 =

4πkR5 5ε0

so

JG kR5 rˆ . E = 5ε0r 2

2-23

kr 2r 2dr = 4πk

R5 . 5

Electromagnetism

For r < R , our Gaussian surface becomes

Now, 2π

qenc =

∫

ρ dτ =

∫ 0

R

π

dϕ

∫

sin θ dθ

0

∫

kr′2 r′2 dr′ =

0

so

E 4πr 2 =

4πkr 5 5ε 0

and

JG kr 3 E = rˆ . 5ε0 Now we can calculate the electric potential. This is done using r⃗

V=−

∫

JG E ⋅ dl ⃗

∞

with infinity as the reference point. For r > R

2-24

4πkr 5 5

Electromagnetism

with r⃗

V=−

∫

JG E ⋅ dl ⃗ = −

∞

r

∫ ∞

kR5 kR5 dr ′ = . 2 5ε0r′ 5ε0r

For r < R

with r⃗

V =−

JG E ⋅ dl ⃗ = −

∫ ∞

= V=

R

kR5 dr ′ − 5ε0r′2

∫ ∞

k 4R 4 − r 4 + R 4 20ε0

(

r

∫ R

kr′3 kR5 kr 4 kR 4 dr ′ = −0− + 5ε0 5ε0R 20ε0 20ε0

)

kR 4 ⎛ r4 ⎞ ⎜5 − 4 ⎟ . 20ε0 ⎝ R ⎠

Problem 2.12 A long cylinder of radius a carries a charge density ρ = ks 2 , where k is a constant and s is the distance from the axis of the cylinder. Find the electric field and the electric potential everywhere. Take the reference point at a distance b from the axis (b > a ).

2-25

Electromagnetism

Solution Starting with the electric field, we use Gauss’s law, given by JG JG q E ⋅ da = enc . ε0 S

∮

Note the left-hand side is always given by JG JG E ⋅ da = E da = E

∮S

∮S

∮S

da = E 2πsl .

For s > a , we have our Gaussian surface given by

with enclosed charge given by 2π

qenc =

∫ ρ dτ = ∫ 0

a

l

dϕ

∫

dz

0

∫

ks 2s ds =

0

Therefore,

E 2πsl =

πk la 4 2ε 0

and

JG ka 4 sˆ . E = 4ε0s For s < a , our Gaussian surface becomes

2-26

πk la 4 . 2

Electromagnetism

Now, 2π

qenc =

∫ ρ dτ = ∫

s

l

dϕ

∫

0

dz

0

∫

ks′2 s′ ds′ =

0

πk ls 4 2

so

πk ls 4 2ε 0

E 2πsl = and

JG ks 3 E = sˆ . 4ε0 Now we can calculate the electric potential. This is done using r⃗

V=−

∫JG

JG E ⋅ dl ⃗

b

with b as the reference point. For s > a ,

with r⃗

V=−

∫JG b

JG E ⋅ dl ⃗ = −

s

∫ b

ka 4 ka 4 s ka 4 ds ′ = − (ln s − ln b) = − ln . 4ε0s′ 4ε 0 4ε 0 b

2-27

Electromagnetism

For s < a ,

with

V =−

∫

JG E ⋅ dl ⃗ = −

a

∫ b

V=

(

ka 4 ds − 4ε0s

s

∫ a

ks′3 ka 4 a k (s 4 − a 4 ) ds ′ = − ln − 16ε0 4ε 0 4ε 0 b

)

k a4 − s4 ka 4 b ln + . 4ε 0 a 16ε0

Problem 2.13. Verify the electrostatic boundary condition using the charge distribution in problem 2.9. Solution The electrostatic boundary condition is given by JG JG σ ˆ Eabove − E below = n. ε0 From problem 2.9, our electric fields are

⎧ 0 r R ⎩ ε0r At r = R , we have

JG σ Eabove = rˆ ε0 and

JG E below = 0.

Therefore,

JG JG σ σ Eabove − E below = rˆ − 0 = nˆ , ε0 ε0 where nˆ is normal to the sphere which has the direction of rˆ .

2-28

Electromagnetism

Problem 2.14. Find the work required to assemble the charge distribution below.

Solution We can denote the following q1 = 3q , q2 = −2q, q3 = −q, q4 = q . Starting with q1, W = 0. Moving in q2 , we have

W2 =

⎛q ⎞ 1 6q 2 1 q2 3 2 q2⎜ 1 ⎟ = − =− . 4πε0 ⎝ r12 ⎠ 4πε0 a 2 4πε0 a

Moving in q3, we have

W3 =

⎛q q ⎞ 1 q ⎛ 3q 2q ⎞ q 2 ⎛ 2 2 −3 ⎞ ⎟. ⎜ q3⎜ 1 + 2 ⎟ = − − ⎜ ⎟= 4πε0 ⎝ r13 4πε0 ⎝ 2a r 23 ⎠ 2 a ⎠ 4πε0 ⎝ 2a ⎠

Moving in q4 , we have

W4 =

=

⎛q q q ⎞ 1 q ⎛ 3q 2q q ⎞ q4⎜ 1 + 2 + 3 ⎟ = − − ⎜ ⎟ 4πε0 ⎝ r14 2a r 24 r34 ⎠ 4πε0 ⎝ a 10 a 2⎠ q 2 ⎛ 3 10 − 5 2 − 10 ⎞ ⎟. ⎜ 4πε0 ⎝ 10a ⎠

Therefore,

W = W2 + W3 + W4 =

2 2 −3 3 10 − 5 2 − 10 ⎞ q2 ⎛ 3 2 ⎟ ⎜− + + 4πε0 ⎝ 2a 10a a ⎠

⎡ ⎤ q 2 ⎢ 3 10 − 25 2 + 1 ⎥ W= ⎥. 4πε0 ⎢ 10a ⎣ ⎦

(

)

Problem 2.15. Find the energy stored in a spherical shell of inner radius a and outer radius b with a charge distribution ρ = kr 2 .

2-29

Electromagnetism

Solution The work is given by

W= 

ε0 2

∫

E 2 dτ

all space

so we need to find the field in all three regions. For r < a , we have qenc = 0. So E = 0. For a < r < b, we have

JG

∮S E

JG q ⋅ da = enc ε0

with r

qenc =

∫ ρ dτ = 4π ∫

k (r′)2 (r′)2 dr′ =

a

4πk 5 (r − a5) 5

and

JG

JG ⋅ da = E 4πr 2 .

∮S E So

E=

k (r 5 − a 5) . 5ε0r 2

For r > b, we have b

qenc =

∫ ρ dτ = 4π ∫

k (r′)2 (r′)2 dr′ =

a

2-30

4πk 5 (b − a5). 5

Electromagnetism

So

(

k b5 − a 5 E=

5ε0r

2

).

Now the work is given by

⎧ ε0 ⎪ W = ⎨4π 2⎪ ⎩ W=

b

∫ a

⎡ k r5 − a5 ⎢ ⎢ 5ε0r 2 ⎣

(

) ⎤⎥ r 2

⎥ ⎦

∞ 2

dr + 4π

∫ b

⎡ k b5 − a 5 ⎢ ⎢ 5ε0r 2 ⎣

(

) ⎤⎥ r 2

⎥ ⎦

2

⎫ ⎪ dr ⎬ ⎪ ⎭

k 2π 5a 9 − 9a 5b 4 + 4b9 . 45ε0

(

)

Problem 2.16. Given a charge density ρ = ke−r , with k a constant, find the radius of a sphere that maximizes the energy per unit volume. Solution The energy per unit volume is given by

W ε = 0 E 2, volume 2 where the field is given by

JG

∮S E

JG q ⋅ da = enc ε0

with r

qenc =

∫ ρ dτ = 4π ∫

2

( )

k e− r ′ r′

(

)

dr′ = 4πk e−r 2e r − r 2 − 2r − 2 .

0

Also,

JG

∮S E

JG ⋅ da = E 4πr 2 .

So

(

k e−r 2e r − r 2 − 2r − 2 E=

ε0r

2

).

The energy per unit volume contained in a sphere of radius r is given by

(

2

)

k 2 e−2r 2e r − r 2 − 2r − 2 W ε = 0E2 = . volume 2 2ε0r 4

2-31

Electromagnetism

To maximize this, we have

d ⎛⎜ W ⎞⎟ =0 dr ⎝ volume ⎠ so

(

)(

−k 2 e−2r 2e r − r 2 − 2r − 2 4e r − r 3 − 2r 2 − 4r − 4 ε0r

5

) = 0.

Since r ≠ 0, k ≠ 0, and e−2r ≠ 0, we have

2e r − r 2 − 2r − 2 = 0 → r = 0 and

4e r − r 3 − 2r 2 − 4r − 4 = 0 → r = 0,

r = 1.45123.

But r ≠ 0, so a sphere of radius r = 1.45123 has the maximum energy per unit volume. Problem 2.17. A metal sphere of radius R and charge q is surrounded by two concentric metal shells. a) Obtain the surface charge density σ at R , a , b, c , and d . b) Calculate the potential at the center of the sphere by taking infinity as the reference point.

Solutions a) Obtain the surface charge density σ at R , a , b, c , and d . For r = R , the sphere is metallic, therefore all charge q is distributed on the surface of the sphere. This gives a surface charge density of

2-32

Electromagnetism

σ=

q 4πR2

By influence and due to the sphere with charge q , the inner shell is redistributing the electric charge such that the surface with radius a has charge –q and the surface with radius b has charge +q . Similarly for the outer shell. Therefore, the surface charge densities are

σa = −

q 4πa 2

σb =

q 4πb 2

σc = −

q 4πc 2

σd =

q . 4πd 2

b) Calculate the potential at the center of the sphere by taking infinity as the reference point Taking our reference point at infinity, the electric potential at the center is given by 0 JG V = − ∫ E ⋅ dl⃗ ∞ d

=

∫ ∞

V=

q dr − 4πε0r 2

c

∫ d

b

0 dr −

∫ c

q dr − 4πε0r 2

a

∫

R

0 dr −

b

∫ a

q dr − 4πε0r 2

0

∫

0 dr

R

q ⎛1 1 1 1 1⎞ ⎜ + − + − ⎟. ⎝ 4πε0 d b c R a⎠

Problem 2.18. Calculate the capacitance of the spherical shell capacitor of radii a (inner) and b (outer) shown below.

Solution The electric field is due to the inner charge, so

E=

Q . 4πε0r 2

2-33

Electromagnetism

The electric potential is then b

V=−

∫

JG E ⋅ dl ⃗ =

a

a

∫ b

Q ⎛1 1⎞ Q (b − a ) Q ⎜ − ⎟= dr = . 2 ⎝ ⎠ 4πε0r 4πε0 a b 4πε0 ab

We can find capacitance using

V=

Q Q →C= . V C

So,

C=

Q 4πε0ab . = Q (b − a ) b−a 4πε0 ab

Problem 2.19. Calculate the capacitance of a cylindrical capacitor of length L with two metal cylinders of radii a (inner) and b (outer) shown below. Ignore the edge effects, obtain the capacitance per unit length.

Solution Let us consider that the charge on the inner cylinder is Q (at the radius a ). The electric field is obtained from Gauss’s law

JG

∮S E

JG q ⋅ da = enc , ε0

where

JG

∮S E

JG ⋅ da =

∮S E da = E ∮S

and the enclosed charge is

qenc = Q.

2-34

da = E 2πsL

Electromagnetism

Therefore,

JG E =

Q sˆ . 2πε0sL

The potential difference between the two cylinders is then b

V (b ) − V (a ) = −

∫

JG E ⋅ dl ⃗ = −

a

b

∫ a

Q b Q ds = ln . 2πε0sL 2πε0L a

Therefore, the capacitance is given by

C=

Q 2πε0L = . b V ln a

Bibliography Byron F W and Fuller R W 1992 Mathematics of Classical and Quantum Physics (New York: Dover) Griffiths D J 1999 Introduction to Electrodynamics 3rd edn (Englewood Cliffs, NJ: Prentice Hall) Griffiths D J 2013 Introduction to Electrodynamics 4th edn (New York: Pearson) Halliday D, Resnick R and Walker J 2010 Fundamentals of Physics 9th edn (New York: Wiley) Halliday D, Resnick R and Walker J 2013 Fundamentals of Physics 10th edn (New York: Wiley) Jackson J D 1998 Classical Electrodynamics 3rd edn (New York: Wiley) Rogawski J 2011 Calculus: Early Transcendentals 2nd edn (San Fransisco, CA: Freeman)

2-35

IOP Concise Physics

Electromagnetism Problems and solutions Carolina C Ilie and Zachariah S Schrecengost

Chapter 3 Electric potential

Chapter 3 contains different methods for obtaining the electric potential. We will focus on calculating the potential as finding the field is a straightforward calculation once the potential has been determined. Laplace’s equation is solved using different methods, depending on the type of charge distribution and on the symmetry of the problem. The method of images, separation of variables, and multipole (in particular dipole) expansions are discussed using appropriate examples.

3.1 Theory 3.1.1 Laplace’s equation Cartesian

∇2 T =

∂ 2T ∂ 2T ∂ 2T =0 + + 2 2 ∂z 2 ∂y ∂x

Cylindrical

∇2 T =

∂ 2T 1 ∂ ⎛ ∂T ⎞ 1 ∂ 2T ⎜s ⎟ + =0 + ∂z 2 s ∂s ⎝ ∂s ⎠ s 2 ∂ϕ 2

Spherical

∇2 T =

∂ 2T 1 ∂ ⎛ ∂T ⎞ 1 ∂ ⎛ 2 ∂T ⎞ 1 ⎜r ⎟ + ⎜sin θ ⎟ + =0 ∂θ ⎠ r 2 sin2 θ ∂ϕ 2 r 2 ∂r ⎝ ∂r ⎠ r 2 sin θ ∂θ ⎝

3.1.2 Solving Laplace’s equation As an introduction to solving problems using Laplace’s equation, we will outline the solutions in Cartesian and spherical coordinates. Laplace’s equation can be solved by the method separation of variables when we know the boundary conditions. The doi:10.1088/978-1-6817-4429-2ch3

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ª Morgan & Claypool Publishers 2016

Electromagnetism

general solutions will be outlined below, but seeing how they are derived is important. We will leave them in a general form and problems in this chapter will provide examples of using the boundary conditions to solve for the constants. Two-dimensional Cartesian coordinates Let us look at a general two-dimensional case where Laplace’s equation is given by

∇2 V =

∂ 2V ∂ 2V = 0. + ∂y 2 ∂x 2

We look for a solution of the type V (x , y ) = X (x )Y (y ) and we replace the desired solution in Laplace’s equation which becomes

Y (y )

∂ 2Y (y ) ∂ 2X (x ) ( ) X x =0 + ∂y 2 ∂x 2

and in a simpler form

Y

∂ 2Y ∂ 2X X = 0. + ∂y 2 ∂x 2

We want to separate the variables, which can easily be done by dividing the equation by X (x )Y (y ) = V (x , y )

1 ∂ 2X 1 ∂ 2Y = 0. + X ∂x 2 Y ∂y 2 Note that the first term depends only on x and the second term depends only on y . This means that each of the two terms must be constant, and the two constants must be equal in magnitude but opposite in sign. So

1 ∂ 2X =F X ∂x 2

1 ∂ 2Y = −F . Y ∂y 2

We choose F positive, and we can rewrite the equations as

1 d2X = k2 X dx 2

1 d2Y = −k 2. Y dy 2

Note that the initial partial differential equation was replaced by two ordinary differential equations. Rearranging yields

d2X = k 2X dx 2

d2Y = −k 2Y . dy 2

The two equations have the following solutions

X (x ) = Ae kx + B e−kx

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and

Y (y ) = C sin(ky ) + D cos(ky ). Going back to the electric potential, V becomes

V (x , y ) = X (x )Y (y ) = Ae kx + B e−kx ⎡⎣ C sin(ky ) + D cos(ky )⎤⎦ .

(

)

The next step is to apply the boundary conditions in order to obtain the constants A, B , C , and D and to (usually) impose some constraints on k . Two-dimensional spherical coordinates Here we will assume azimuthal symmetry (no dependence on ϕ) where Laplace’s equation is given by

∇2 V =

∂ ⎛ ∂V ⎞ 1 ∂ ⎛ 2 ∂V ⎞ 1 ⎜r ⎟ + ⎜sin θ ⎟ = 0. 2 2 ∂θ ⎠ r ∂r ⎝ ∂r ⎠ r sin θ ∂θ ⎝

We look for a solution which has a radial component and an angular component

V (r , θ ) = R(r )Θ(θ ). Note that here R is the function of r , and not merely the radius of the sphere. We plug our solution in the previous equation and we obtain

Θ(θ )

∂Θ(θ ) ⎞ ∂ ⎛ 2 ∂R(r ) ⎞ R(r ) ∂ ⎛ ⎜r ⎟+ ⎜sin θ ⎟ = 0. ∂θ ⎠ ∂r ⎝ ∂r ⎠ sin θ ∂θ ⎝

We want to use the method of separation of variables so we will divide the previous equation by V (r, θ ) = R(r )Θ(θ ),

dΘ(θ ) ⎞ 1 d ⎛ 2 dR(r ) ⎞ 1 d⎛ ⎜r ⎟+ ⎜sin θ ⎟ = 0. dθ ⎠ dr ⎠ Θ(θ )sin θ dθ ⎝ R(r ) dr ⎝ Note that each term is only a function of a single variable so we were able to replace the partial derivates with ordinary derivates. Now we have one term in R(r ) and another term in Θ(θ ), so we have separated the variables. Therefore, each term must be constant. For well know reasons (more apparent in quantum mechanics), we choose the constant as following

1 d ⎛ 2 dR(r ) ⎞ ⎜r ⎟ = l (l + 1) dr ⎠ R(r ) dr ⎝ dΘ(θ ) ⎞ 1 d⎛ ⎜sin θ ⎟ = − l (l + 1). Θ(θ )sin θ dθ ⎝ dθ ⎠ Now let us analyze each of the equations and find the solution.

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The radial equation

d ⎛ 2 dR(r ) ⎞ ⎜r ⎟ = l (l + 1)R(r ) dr ⎝ dr ⎠ has the general solution, with A and B constants

B . r l +1

R(r ) = Ar l + The angular equation

dΘ(θ ) ⎞ d⎛ ⎜sin θ ⎟ = −l (l + 1)Θ(θ )sin θ dθ ⎠ dθ ⎝ is not at all trivial. The solutions constitute Legendre polynomials with the variable cos θ . Legendre polynomials are a special class of polynomials. So the solution to the angular equation is

Θ(θ ) = Pl (cos θ ), where the general form is given by Rodrigues formula

Pl (x ) =

2 1 ⎛ d ⎞l 2 ⎜ ⎟ x −1 . l ⎝ 2 l! dx ⎠

(

)

Therefore, the separable solution of the Laplace equation, considering azimuthal symmetry, is

⎛ B ⎞ V (r , θ ) = R(r )Θ(θ ) = ⎜Ar l + l +1 ⎟Pl (cos θ ) ⎝ r ⎠ and the general solution is the linear combination of the separable solutions ∞

V (r , θ ) =

⎛

∑⎜⎝Al r l + l =0

Bl ⎞ ⎟Pl (cos θ ). r l +1 ⎠

3.1.3 General solutions Cartesian

∂ 2V ∂ 2V = 0 → V (x , y ) = (Ae kx + B e−kx)⎡⎣ C sin(ky ) + D cos(ky )⎤⎦ + ∂y 2 ∂x 2 Spherical

1 ∂ ⎛ 2 ∂V ⎞ 1 ∂ ⎛ ∂V ⎞ ⎜r ⎟ + ⎜sin θ ⎟ = 0 2 2 ∂θ ⎠ r ∂r ⎝ ∂r ⎠ r sin θ ∂θ ⎝

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∞

V (r , θ ) =

⎛

∑ ⎜⎝Al r l + l =0

Bl ⎞ ⎟Pl (cos θ ), r l +1 ⎠

where Pl are Legendre polynomials given by the Rodrigues formula

Pl (x ) =

l 1 ⎛ d ⎞l 2 ⎜ ⎟ x −1 . l ⎝ 2 l! dx ⎠

(

)

Note

P0(x ) = 1 P1(x ) = x P2(x ) =

3x 2 − 1 2

P3(x ) =

5x 3 − 3x 2

P4(x ) =

35x 4 − 30x 2 + 3 . 8

Cylindrical

∂ 2V 1 ∂ ⎛ ∂V ⎞ 1 ∂ 2V ⎜s ⎟ + =0 + ∂z 2 s ∂s ⎝ ∂s ⎠ s 2 ∂ϕ 2 ∞

V (s , ϕ) = a0 + b0 ln(s ) +

∑ {s k⎡⎣ ak cos(kϕ) + bk sin(kϕ)⎤⎦ k=1

+ s−k⎡⎣ ck cos(kϕ) + dksin(kϕ)⎤⎦ .

}

3.1.4 Method of images The method of images is a very useful technique for calculating the electric field and the electric potential for problems with symmetry. By using the uniqueness theorem we know that the electric field is uniquely determined at any point in space, thus we can replace an apparently difficult problem with another problem in which we use the initial charge(s) and also the ‘images’ of the charges. The boundary conditions need to be fulfilled. Typically, this involves the condition for a zero electric potential for a grounded conductor and the condition for zero potential very far away from the system of charges. The potential can be determined in the permitted region, which is in general the region of the real charge. The region of the image charges is the ‘forbidden’ region; the potential cannot be calculated there. The best way to learn this method is by solving problems and checking the examples.

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3.1.5 Potential due to a dipole

V (r ⃗ ) ≅

JG 1 qd cos θ 1 p · rˆ = 4π ε 0 4π ε 0 r 2 r2

3.1.6 Multiple expansion ∞

V (r ⃗ ) =

1 1 ∑ 4π ε 0 n = 0 r n + 1

V (r ⃗ ) =

1 ⎡1 ⎢ 4π ε 0 ⎣ r +

1 r3

∫ (r′)nPn(cos θ′) ρ(r ⃗′)dτ′

∫ ρ(r ′⃗ )dτ′ + r12 ∫ r′ cos θ′ ρ(r ⃗′)dτ′ ⎛

⎞

⎤

∫ (r′)2⎜⎝ 32 cos2 θ′ − 12 ⎟⎠ ρ(r ⃗′)dτ′ + ⋯⎥⎦.

3.1.7 Monopole moment n

Q=

∑qi i=0

3.2 Problems and solutions Problem 3.1. Solve the Laplace equation in spherical and cylindrical coordinates for the cases where V is only dependent on one coordinate at a time. Solution In spherical coordinates

∇2 V =

∂ 2V 1 ∂ ⎛ ∂V ⎞ 1 ∂ ⎛ 2 ∂V ⎞ 1 ⎜r ⎟ + ⎜sin θ ⎟ + = 0. ∂θ ⎠ r 2 sin2 θ ∂ϕ 2 r 2 ∂r ⎝ ∂r ⎠ r 2 sin θ ∂θ ⎝

If V only depends on r

1 d ⎛ 2 dV ⎞ ⎜r ⎟ = 0, r 2 dr ⎝ dr ⎠ which means

r2

dV = C. dr

So

V=C

∫ r−2 dr → V (r) = C ( −r−1 + A) = k − Cr . 3-6

Electromagnetism

If V only depends on θ

dV ⎞ 1 d⎛ ⎜sin θ ⎟ = 0, dθ ⎠ r sin θ dθ ⎝ 2

which mean

sin θ

dV = C. dθ

So

V =C

∫ sin1 θ dθ = C ∫ csc θ dθ = C ( ln csc θ − cot θ

V (θ ) = k + C ln csc θ − cot θ . If V only depends on ϕ

1 d2V = 0, r 2 sin2 θ dϕ 2 which means

dV = C. dϕ So

V (ϕ) = k + Cϕ . In cylindrical coordinates

∂ 2V 1 ∂ ⎛ ∂V ⎞ 1 ∂ 2V ⎜s ⎟ + = 0. + ∂z 2 s ∂s ⎝ ∂s ⎠ s 2 ∂ϕ 2 If V only depends on s

1 d ⎛ dV ⎞ ⎜s ⎟ = 0, s ds ⎝ ds ⎠ which means

s

dV = C. ds

So

V =C

∫ s−1 ds = C ( ln s

V (s ) = k + C ln s .

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)

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Electromagnetism

If V only depends on ϕ,

1 d2V = 0, s 2 dϕ 2 which means

dV = C, dϕ which is the same as ϕ dependence in spherical coordinates. So

V (ϕ) = k + Cϕ . If V only depends on z

d2V = 0, dz 2 which means

dV = C, dz which is the same form as ϕ dependence. So

V (z ) = k + Cz . Problem 3.2. In two-dimensional Cartesian coordinates, the general solution to the Laplace equation is

V (x , y ) = Ae kx + B e−kx ⎡⎣ C sin(ky ) + D cos(ky )⎤⎦ .

(

)

Verify that this does in fact satisfy the Laplace equation. Solution Here, the Laplace equation is

∂ 2V ∂ 2V = 0. + ∂y 2 ∂x 2 So we need to compute

∂ 2V ∂x 2

and

∂ 2V . ∂y 2

We have

∂V = Ak e kx − Bk e−kx ⎡⎣ C sin(ky ) + D cos(ky )⎤⎦ ∂x

(

)

and

∂ 2V = ∂x 2

( Ak e

2 kx

+ Bk 2 e−kx ⎡⎣ C sin(ky ) + D cos(ky )⎤⎦

)

= k 2 Ae kx + B e−kx ⎡⎣ C sin(ky ) + D cos(ky )⎤⎦ .

(

)

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Electromagnetism

Also,

∂V = Ae kx + B e−kx ⎡⎣ Ck cos(ky ) − Dk sin(ky )⎤⎦ ∂y

(

)

and

∂ 2V = Ae kx + B e−kx ⎡⎣ −Ck 2 sin(ky ) − Dk 2 cos(ky )⎤⎦ ∂y 2

(

)

= − k 2 Ae kx + B e−kx ⎡⎣ C sin(ky ) + D cos(ky )⎤⎦ .

(

)

Putting this all together, we have

∂ 2V ∂ 2V = k 2 Ae kx + B e−kx ⎡⎣ C sin(ky ) + D cos(ky )⎤⎦ + ∂y 2 ∂x 2 − k 2 Ae kx + B e−kx ⎡⎣C sin(ky ) + D cos(ky )⎤⎦

(

)

(

)

∂ 2V ∂ 2V =0 + ∂y 2 ∂x 2 as expected. Problem 3.3. A charge q is placed at a distance d from an infinite grounded conducting plane. Using the method of images, find the electric potential. Which is the ‘forbidden’ region, for which we cannot calculate the potential?

Solution We replace the previous problem with a completely different problem: charge q at (0, 0, d ) and its image, charge −q situated at (0, 0, −d). The grounded conducting plane disappeared. This new problem is shown below

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Electromagnetism

The potential needs to satisfy the following boundary conditions: a) V = 0 for z = 0 (grounded plane in the initial problem). b) V → 0 for a point far from charge q

x 2 + y 2 + z 2 ≫ d 2. The electric potential due to both point charges is:

V (x , y , z ) =

q 4π ε 0 x 2 + y 2 + ( z − d ) 2

+

−q 4π ε 0 x 2 + y 2 + ( z + d ) 2

.

Let us check the boundary conditions: a) For z = 0 it is easy to see that V = 0. b) For x 2 + y 2 + z 2 ≫ d 2 , V → 0.

x 2 + y 2 + (z − d ) 2 ≅

x 2 + y 2 + (z + d ) 2 ,

and

It is important to note that the only region for which we are able to obtain the electric potential is the region in space above the grounded conducting plane, i.e. the semi-space where charge q is located. For z < 0, we are not able to obtain the electric potential.

Problem 3.4. A charge q is placed in an opened grounded conducting parallelepiped at (a, b, c ), where a, b, c are positive. Using the method of images, obtain the electric potential in the region of the charge q (for which x > 0, y > 0, z > 0).

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Solution We have the one real charge q at (a, b, c ) and seven image charges as following: −q at ( −a, −b, −c ), ( −a, b, c ), (a, −b, c ), and (a, b, −c ); q at (a, −b, −c ), ( −a, b, −c ), ( −a, −b, c ). The electric potential is given by

V (x , y , z ) ⎛ 1 ⎜ q = + ⎜ 2 4π ε 0 ⎝ ( x − a ) + ( y − b ) 2 + ( z − c ) 2 +

q 2

2

(x + a ) + (y − b ) + (z + c ) +

+

2

−q (x + a )2 + (y + b )2 + (z + c )2 −q (x − a )2 + (y − b )2 + (z + c )2

+

+

+

q 2

(x − a ) + (y + b )2 + (z + c )2 q 2

(x + a ) + (y + b )2 + (z − c )2 −q (x − a )2 + (y + b )2 + (z − c )2 ⎞ ⎟. ⎟ (x + a )2 + (y − b )2 + (z − c )2 ⎠ −q

If we check the limit x = 0, y = 0, z = 0 successively, we obtain a zero potential for all the three sides of the parallelepiped, where the grounded conductors were in the equivalent problem. Also, for the points far away from the point (a, b, c ) in the ‘eighth’ part of space for which x ≫ 0, y ≫ 0, z ≫ 0, the potential becomes zero as well. Again, note that this is the electric potential only for this part of space, accessible for investigation using the method of images. Problem 3.5. Let us imagine that we have n charges placed as following: q1 = −q at (0, 0, d ), q2 = 2q at (0, 0, 2d ), …, qn = ( −1) nnq at (0, 0, nd ) above a grounded, conducting xy -plane (shown below). Obtain the electric potential using the method of images.

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Solution The image charges will be equal in magnitude, of different sign, and situated symmetrically with the xy -plane. In the new problem we eliminate the grounded, conducting plane, but we use the xy -plane for geometrical purposes.

For each charge qn = ( −1) nnq we have the image charge qn′ = ( −1) n+1nq located at (0, 0, −nd ). The electric potential is, therefore,

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Electromagnetism

⎛ −1 q ⎜ + V (x , y , z ) = ⎜ 2 2 4πε0 ⎝ x + y + (z − d )2 +

+

( −1)2 2 x 2 + y 2 + ( z − 2d ) 2 ( −1)k k x 2 + y 2 + (z − kd )2

+⋯ +

+

+

( −1)nn x 2 + y 2 + (z − nd )2

1 2

2

x + y + (z + d )2 ( −1)2+12 x 2 + y 2 + ( z + 2d ) 2

+ ⋯

( −1)k+1k x 2 + y 2 + (z + kd )2 +

⎞ ⎟. ⎟ x 2 + y 2 + (z + nd )2 ⎠ ( −1)n+1n

It is easy to see that V = 0 for z = 0 and also V = 0 for a point very far from charge x 2 + y 2 + z 2 ≫ (nd )2 . Problem 3.6. A conducting sphere of radius R , centered at the origin, is grounded. Find the potential outside the sphere, if a point charge +q is placed at a distance d from the sphere, d > R . Use the method of images.

Solution We replace our problem with the grounded, conducting sphere of radius R and the charge +q at distance d > R with a different problem. The sphere, the charge +q and the image charge q′, situated at (0, 0, a), with a < R , is given by

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Electromagnetism

We need V = 0 everywhere on the sphere of radius R . Note that we can only find the electric potential outside the sphere. Consider the point P depicted above; here the electric potential at P is given by

V (r ⃗ ) =

1 ⎛q 1 ⎛ q q′ ⎞ q′ ⎞ JG + ⎜ + ⎟= ⎜ JG ⎟ . 4π ε0 ⎝ r1 r⃗ − a ⎠ r 2 ⎠ 4π ε 0 ⎝ r ⃗ − d

Using the law of cosines, we have

r 12 = d 2 + r 2 − 2dr cos θ and

r 22 = a 2 + r 2 − 2ar cos θ . We can rewrite V as

V (r ⃗ ) =

1 ⎛ q ⎜ + 2 2 4π ε0 ⎝ d + r − 2dr cos θ

⎞ ⎟. a 2 + r 2 − 2ar cos θ ⎠ q′

The potential should be zero for r = R

V (R ) =

1 ⎛ q ⎜ + 2 2 4π ε0 ⎝ d + R − 2dR cos θ

⎞ ⎟ = 0. a 2 + R2 − 2aR cos θ ⎠ q′

So

q d 2 + R2 − 2Rd cos θ

=

− q′ a 2 + R2 − 2aR cos θ

.

We need to obtain both q′ and a , so we need two equations. We choose two convenient values for θ , θ = 0 and θ = π . For θ = 0, cos θ = 1, so

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Electromagnetism

q

=

d 2 + R2 − 2Rd

− q′ a 2 + R2 − 2aR

→

q (d − R )2

=

− q′ (R − a )2

.

Choosing the positive square root,

q − q′ = d−R R−a and solving for the image charge, we obtain

q′ = −q

R−a . d−R

For θ = π , cos θ = −1, so

q − q′ . = d+R R+a By substituting q′, we have

q (R − a )q . = d+R (d − R )(a + R ) Solving for a , we obtain

a=

R2 . d

Using this, we can find our image charge

⎛ R2 ⎞ ⎜R − ⎟q ⎝ d ⎠ R(d − R )q q′ = − =− (d − R )d d−R q′ = −

R q. d

Now we have the electric potential, since we obtained the image charge and its position. Problem 3.7. Given two infinitely long grounded plates at y = 0 and y = a connected by the metal strip at x = −b with constant potential –V0 and x = b with constant potential V0. Find the potential inside the pipe.

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Electromagnetism

Solution This is independent of z so we have

∂ 2V ∂ 2V =0 + 2 ∂y 2 ∂x with boundary conditions (i) V (y = 0) = 0 (ii) V (y = a ) = 0 (iii) V (x = b ) = V0 (iv) V (x = −b ) = −V0 . Our general solution is given by

V (x , y ) = Ae kx + B e−kx ⎡⎣ C sin(ky ) + D cos(ky )⎤⎦ .

(

)

From boundary condition (i)

(

)

V (x , 0) = Ae kx + B e−kx (D ) = 0 ⇒ D = 0. So our solution becomes

V (x , y ) = Ae kx + B e−kx ⎡⎣ C sin(ky )⎤⎦ .

(

)

From boundary condition (ii)

V (x , y ) = Ae kx + B e−kx ⎡⎣ C sin(ka )⎤⎦ = 0

(

)

we have

k=

nπ , a

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Electromagnetism

where n is an integer. By symmetry

V (x , y ) = − V ( − x , y )

( Ae

kx

)

(

)

+ B e−kx [C sin(ka )] = − Ae−kx + B e kx [C sin(ka )] Ae kx + B e−kx = −Ae−kx + B e kx

(

)

(

)

A e kx + e−kx = −B e kx + e−kx . So

A = −B. Absorbing our constants, the solution becomes

(

)

V (x , y ) = C e kx + e−kx sin(ky ) = C sinh(kx ) sin(ky ) and in general ∞

V (x , y ) =

⎛ nπx ⎞ ⎛ nπy ⎞ ⎟sin⎜ ⎟. a ⎠ ⎝ a ⎠

∑Cn sinh⎜⎝ n=1

Now to find Cn we take ∞

V (b , y ) =

⎛ nπb ⎞ ⎛ nπy ⎞ ⎟ = V ⎟ sin⎜ 0 a ⎠ ⎝ a ⎠

∑Cn sinh⎜⎝ n=1

so

⎛ nπb ⎞ ∑Cn sinh⎜⎝ ⎟⎠ a n=1 ∞

a

∫ 0

⎛ nπy ⎞ ⎛ n′πy ⎞ ⎟ sin⎜ ⎟dy = sin⎜ ⎝ a ⎠ ⎝ a ⎠

a

∫ 0

⎛ n′πy ⎞ ⎟dy . V0 sin⎜ ⎝ a ⎠

Note that when n ≠ n′, a

∫ 0

⎛ nπy ⎞ ⎛ n′πy ⎞ ⎟ sin⎜ ⎟dy = 0 sin⎜ ⎝ a ⎠ ⎝ a ⎠

and when n = n′, a

∫ 0

⎛ nπy ⎞ ⎛ n′πy ⎞ a ⎟ sin⎜ ⎟dy = . sin⎜ ⎝ a ⎠ ⎝ a ⎠ 2

Therefore,

⎧ 0 n is even ⎪ ⎛ nπb ⎞ a V0a ⎟ = [1 − cos(nπ )] = ⎨ 2V0a Cn sinh⎜ . ⎝ a ⎠2 nπ n is odd ⎪ ⎩ nπ

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So

Cn =

4V0 1 nπ sinh nπb

( ) a

for n = 1, 3, 5, … and our potential is given by ∞ 4V0 1 sinh V (x , y ) = ∑ π n=1,3,5 n sinh

( ) sin⎛ nπy ⎞. n πx a n πb a

( )

⎜

⎟

⎝ a ⎠

Problem 3.8. Suppose a thin spherical shell of radius R has potential 3 V (θ ) = V0(1 − 2 sin2 θ ) specified at the surface. Find the potential inside and outside the sphere. Solution Our general solution is given by ∞

V (r , θ ) =

⎛

∑⎜⎝Al r l + l =0

Bl ⎞ ⎟Pl (cos θ ). r l +1 ⎠

Inside: Here, we must have Bl = 0 so the potential does not blow up at the origin. So our potential becomes ∞

V (r , θ ) =

∑Al r l Pl (cos θ ). l =0

At the surface, we have

⎛

∞

V (R , θ ) =

∑Al Rl Pl (cos θ ) = V0⎜⎝1 − l =0

⎞ 3 sin2 θ ⎟ . ⎠ 2

Note that

⎛ 3 cos2 θ − 1 ⎞ ⎡ ⎤ ⎛ ⎞ 3 3 V0⎜1 − sin2 θ ⎟ = V0⎢1 − (1 − cos2 θ )⎥ = V0⎜ ⎟ = V0P2(cos θ ). ⎣ ⎦ ⎝ ⎠ ⎝ ⎠ 2 2 2 So ∞

∑Al Rl Pl (cos θ ) = V0P2(cos θ ). l =0

This means we only have the l = 2 term,

A2 R2P2(cos θ ) = V0P2(cos θ ) → A2 =

V0 . R2

Therefore,

⎛ r ⎞2 ⎛ 3 cos2 θ − 1 ⎞ V (r , θ ) = A2 r 2P2(cos θ ) = V0⎜ ⎟ ⎜ ⎟. ⎝R⎠ ⎝ ⎠ 2

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Electromagnetism

Outside: Here, we must have Al = 0 so the potential does not blow up as r → ∞. So our potential becomes ∞

B

∑ r l +l 1 Pl (cos θ ).

V (r , θ ) =

l =0

At the surface, we have

⎛ ⎞ 3 Bl P (cos θ ) = V0⎜1 − sin2 θ ⎟ . l +1 l ⎝ ⎠ 2 R l =0 ∞

V (R , θ ) =

∑

Again, the right-hand side is V0P2(cos θ ), so

B2 P2(cos θ ) = V0P2(cos θ ) → B2 = V0R3. R2 +1 Therefore,

V (r , θ ) =

⎛ R ⎞3⎛ 3 cos2 θ − 1 ⎞ B2 ⎜ ⎟ ⎜ (cos ) P V θ = ⎟. 2 0 ⎝r⎠⎝ ⎠ 2 r3

Problem 3.9. A spherical shell of radius R has surface charge σ0(θ ) = sin θ sin 3θ smeared on its surface. Find the potential inside and outside the sphere. Solution Our general solution is given by ∞

V (r , θ ) =

⎛

∑⎜⎝Al r l + l =0

Bl ⎞ ⎟Pl (cos θ ). r l +1 ⎠

Inside we must have Bl = 0, otherwise V → ∞ as r → 0. So our potential becomes ∞

Vin(r , θ ) =

∑Al r l Pl (cos θ ). l =0

Outside we must have Al = 0, otherwise V → ∞ as r → ∞. So our potential becomes ∞

Vout(r , θ ) =

B

∑ r l +l 1 Pl (cos θ ). l =0

At the surface they must be equal, so

Vin(R , θ ) = Vout(R , θ ) ∞

∞

B

∑Al Rl Pl (cos θ ) = ∑ Rl +l 1 Pl (cos θ ) l=0

l=0

Bl = Al R2l +1.

3-19

Electromagnetism

We must also have

⎛ ∂Vout ∂Vin ⎞ 1 ⎜ ⎟ = − σ0(θ ), − ⎝ ∂r ⎠ ∂r r=R ϵ0 where

∂Vout = ∂r

∞

B

∑ − (l + 1) r l +l2 Pl (cos θ ) l=0

and

∂Vin = ∂r

∞

∑lAl r l −1Pl (cos θ ). l =0

Thus,

⎛ ∂Vout ∂Vin ⎞ ⎜ ⎟ = − ⎝ ∂r ∂r ⎠ r=R

∞

B

∑ − (l + 1) Rl +l 2 Pl (cos θ ) − lAl Rl −1Pl (cos θ ). l =0

Substitution of Bl yields

⎛ ∂Vout ∂Vin ⎞ ⎜ ⎟ = − ⎝ ∂r ∂r ⎠ r=R

∞

∑(2l + 1)Al Rl −1Pl (cos θ ) = l =0

1 σ0(θ ). ϵ0

Since Legendre polynomials are orthogonal, when l ≠ l ′ we have π

∫

Pl (cos θ )Pl ′(cos θ )sin θ dθ =

0

2 . 2l + 1

It follows that

1 Al = 2ϵ0Rl −1

π

∫

σ0(θ )Pl (cos θ )sin θ dθ .

0

Note that σ0(θ ) can be rewritten as

σ0(θ ) = sin θ sin 3θ = sin θ (sin 2θ cos θ + cos 2θ sin θ ) = sin θ ⎡⎣ (2 sin θ cos θ )cos θ + sin θ (2 cos2 θ − 1)⎤⎦ = 2 sin2 θ cos2 θ + sin2 θ (2 cos2 θ − 1) = 2(1 − cos2 θ )cos2 θ + (1 − cos2 θ )(2 cos2 θ − 1) = 2 cos2 θ − 2 cos4 θ + 2 cos2 θ − 2 cos4 θ − 1 + cos2 θ = −4 cos4 θ + 5 cos2 θ − 1.

3-20

Electromagnetism

We can find α , β , and γ such that

−4 cos4 θ + 5 cos2 θ − 1 = αP4(cos θ ) + βP2(cos θ ) + γP0(cos θ ). So,

−4 cos4 θ + 5 cos2 θ − 1 = α It follows that α = − 32 ,β= 35

1 Al = 2ϵ0Rl −1

π

22 , 21

35 cos4 θ − 30 cos2 θ + 3 3 cos2 θ − 1 +β + γ. 8 2 2 and γ = − 15 and we can now solve for Al .

⎡ 32 22 P2(cos θ )Pl (cos θ )sin θ ⎢⎣ − P4(cos θ )Pl (cos θ )sin θ + 21 35

∫ 0

−

⎤ 2 P0(cos θ )Pl (cos θ )sin θ ⎥dθ . ⎦ 15

−

2 32 32 ⎛ 35 cos4 θ − 30 cos2 θ + 3 ⎞ . ⎜ ⎟ sin θ dθ = − ⎠ 315ϵ0R3 35 ⎝ 8

If l = 4, we have

1 A4 = 2ϵ0R 4−1

π

∫ 0

If l = 2, we have

1 A2 = 2ϵ0R2−1

π

∫ 0

2 22 22 ⎛ 3 cos2 θ − 1 ⎞ . ⎜ ⎟ sin θ dθ = ⎠ 105ϵ0R 21 ⎝ 2

If l = 0, we have

1 A0 = 2ϵ0R 0−1

π

∫ 0

−

2R 2 . sin θ dθ = − 15ϵ0 15

We can now find B4 , B2 , and B0 ,

B4 = A4 R2(4)+1 = − B2 = A2 R2(2)+1 =

32R6 315ϵ0

22R 4 105ϵ0

B0 = A0 R2(0)+1 = −

2R2 . 15ϵ0

Therefore, inside we have

Vin(r , θ ) = A0 + A2 r 2P2(cos θ ) + A4 r 4P4(cos θ ).

3-21

Electromagnetism

So

R⎡ 2 22 ⎛⎜ r ⎞⎟2 ⎛ 3 cos2 θ − 1 ⎞ ⎢− + ⎜ ⎟ ⎠ 105ϵ0 ⎝ R ⎠ ⎝ 2 ϵ0 ⎣ 15

Vin(r , θ ) =

−

32 ⎛⎜ r ⎞⎟4⎛ 35 cos4 θ − 30 cos2 θ + 3 ⎞⎤ ⎜ ⎟⎥ ⎠⎦ 315ϵ0 ⎝ R ⎠ ⎝ 8

and outside we have

Vout(r , θ ) =

B B0 B + 2+21 P2(cos θ ) + 4+41 P4(cos θ ). 1 r r r

So

Vout(r , θ ) 22 ⎛ R ⎞2 ⎛ 3 cos2 θ − 1 ⎞ 32 ⎛ R ⎞4⎛ 35 cos4 θ − 30 cos2 θ + 3 ⎞⎤ R2 ⎡ 2 ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎢− = + ⎟− ⎟⎥ . ⎠ 315 ⎝ r ⎠ ⎝ ⎠⎦ 105 ⎝ r ⎠ ⎝ 2 8 ϵ0r ⎣ 15 Problem 3.10. An infinitely long cylindrical shell of radius R is held at a potential V0(ϕ ) = α cos(4ϕ ). Find the potential inside and outside the shell. Solution Our general solution in cylindrical coordinates is given by ∞ ⎧ V (s , ϕ) = a 0 + b0 ln(s ) + ∑⎨s k⎡⎣ ak cos(kϕ) + bk sin(kϕ)⎤⎦ k =1⎩ ⎪

⎪

⎫ + s −k⎡⎣ ck cos(kϕ) + dk sin(kϕ)⎤⎦ ⎬ . ⎭ ⎪

⎪

Inside: Here, we must have b0 = ck = dk = 0, otherwise the potential would blow up at the center. Our potential becomes ∞

V (s , ϕ ) = a 0 +

∑s k⎡⎣ ak cos(kϕ) + bk sin(kϕ)⎤⎦. k =1

At the surface, we have ∞

V (R , ϕ ) = a 0 +

∑ Rk⎡⎣ ak cos(kϕ) + bk sin(kϕ)⎤⎦ = α cos(4ϕ). k=1

Note we have a 0 = 0, bk = 0, and ak = 0, except for a 4 . So α R 4a 4 cos(4ϕ) = α cos(4ϕ) → a 4 = 4 . R Therefore, for s ⩽ R , we have

⎛ s ⎞4 V (s , ϕ) = α⎜ ⎟ cos(4ϕ). ⎝R⎠

3-22

Electromagnetism

Outside: Here, we must have b0 = ak = bk = 0, otherwise the potential would blow up as s → ∞. Also, since we must have V → 0 as s → ∞, a 0 = 0. Our potential becomes ∞

V (s , ϕ ) =

∑s −k⎡⎣ ck cos(kϕ) + dk sin(kϕ)⎤⎦. k =1

At the surface, we have ∞

V (R , ϕ ) =

∑R−k⎡⎣ ck cos(kϕ) + dk sin(kϕ)⎤⎦ = α cos(4ϕ). k =1

Note we have dk = 0 and ck = 0, except for c4 . So

R−4c4 cos(4ϕ) = α cos 4ϕ → c4 = αR 4. Therefore, for s ⩾ R , we have

⎛ R ⎞4 V (s , ϕ) = α⎜ ⎟ cos(4ϕ). ⎝s⎠ Problem 3.11. Given an infinitely long cylindrical shell of radius R and surface charge σ0(ϕ ) = α cos(2ϕ ) + β sin(3ϕ ), find the potential inside and outside the cylinder. Solution Our general solution in cylindrical coordinates is given by ∞

V (s , ϕ) = a 0 + b0 ln(s ) +

⎧

∑⎨s k⎡⎣ ak cos(kϕ) + bk sin(kϕ)⎤⎦ ⎪

k =1⎩ ⎪

⎫ + s −k⎡⎣ ck cos(kϕ) + dk sin(kϕ)⎤⎦ ⎬ . ⎭ ⎪

⎪

Inside, we must have b0 = ck = dk = 0, otherwise the potential would blow up at the center. Our potential becomes ∞

Vin(s , ϕ) = a 0 +

∑s k⎡⎣ ak cos(kϕ) + bk sin(kϕ)⎤⎦. k =1

Outside we must have b0 = ak = bk = 0, otherwise the potential would blow up as s → ∞. Also, since we must have V → 0 as s → ∞, a 0 = 0. Our potential becomes ∞

Vout(s , ϕ) =

∑s −k⎡⎣ ck cos(kϕ) + dk sin(kϕ)⎤⎦. k =1

At the surface, we have

⎛ ∂Vout ∂Vin ⎞ 1 ⎜ ⎟ = − σ0(ϕ), − ⎝ ∂s ∂s ⎠ s=R ϵ0

3-23

Electromagnetism

where

∂Vout = ∂s

∞

∑ −ks −k−1⎡⎣ ck cos(kϕ) + dk sin(kϕ)⎤⎦ k=1

and

∂Vin = ∂s

∞

∑ks k−1⎡⎣ ak cos(kϕ) + bk sin(kϕ)⎤⎦. k =1

Thus,

⎛ ∂Vout ∂Vin ⎞ ⎜ ⎟ = − ⎝ ∂s ∂s ⎠ s=R

∞

∑ − kR−k−1⎡⎣ ck cos(kϕ) + dk sin(kϕ)⎤⎦ k =1

− kR k−1⎡⎣ ak cos(kϕ) + bk sin(kϕ)⎤⎦ =−

α cos(2ϕ) + β sin(3ϕ) . ϵ0

From this, we can see that ck = ak = 0, except when k = 2, and dk = bk = 0, except when k = 3. This means

(

)

(

)

2 cos(2ϕ) R−3c2 + Ra2 + 3 sin(3ϕ) R−4d3 + R2b3 =

α cos(2ϕ) β sin(3ϕ) + . ϵ0 ϵ0

Separating out the sine and cosine term, we have

(

)

2 R−3c2 + Ra2 =

⎛ α ⎞ α → c2 = R3⎜ − Ra2⎟ ⎝ 2ϵ 0 ⎠ ϵ0

and

3(R−4d3 + R2b3) =

⎛ β ⎞ β → d3 = R 4⎜ − R2b3⎟ . ⎝ 3ϵ 0 ⎠ ϵ0

Since V is continuous, we have

Vout(R , ϕ) = Vin(R , ϕ) R−2c2 cos(2ϕ) + R−3d3 sin(3ϕ) = a 0 + R2a2 cos(2ϕ) + R3b3 sin(3ϕ). We can see that a 0 = 0. Also, considering the sine and cosine terms separately, we have

R−2c2 = R2a2 .

3-24

Electromagnetism

Substitution of c2 yields

⎛ α ⎞ − Ra2⎟ = R2a2 R−2R3⎜ ⎝ 2ϵ 0 ⎠ so

a2 =

α . 4Rϵ0

Also

R−3d3 = R3b3. Substitution of d3 yields

⎛ β ⎞ − R2b3⎟ = R3b3 R−3R 4⎜ ⎝ 3ϵ0 ⎠ So

b3 =

β . 6R2ϵ0

Therefore,

⎛ α α ⎞ αR3 −R c2 = R3⎜ ⎟= ⎝ 2ϵ 0 4Rϵ0 ⎠ 4ϵ 0 and

⎛ β β ⎞ βR 4 d3 = R 4⎜ − R2 2 ⎟ = . ⎝ 3ϵ 0 6R ϵ 0 ⎠ 6ϵ0 Combining everything, the potential inside is

Vin(s , ϕ) = s 2

=

β α cos(2ϕ) + s 3 2 sin(3ϕ) 4Rϵ0 6R ϵ0

⎤ β ⎛ s ⎞3 R ⎡ α ⎛⎜ s ⎞⎟2 cos(2ϕ) + ⎜ ⎟ sin(3ϕ)⎥ ⎢ ϵ0 ⎣ 4 ⎝ R ⎠ 6⎝R⎠ ⎦

and outside is

Vout(s , ϕ) = s −2

=

βR 4 αR3 cos(2ϕ) + s −3 sin(3ϕ) 4ϵ 0 6ϵ0

⎤ β ⎛ R ⎞3 R ⎡ α ⎛ R ⎞2 ⎢ ⎜ ⎟ cos(2ϕ) + ⎜ ⎟ sin(3ϕ)⎥ . ϵ0 ⎣ 4 ⎝ s ⎠ 6⎝ s ⎠ ⎦

3-25

Electromagnetism

Problem 3.12. The electric potential varies as 1 for a monopole, as 12 for a dipole, as r r 1 for a quadrupole, and as 14 for an octopole. How will the electric potential depend 3 r

r

on r for a mutipole with n charges (n being a k power of 2, n = 2k )? Solution Number of charges

Potential

Monopole

n = 20 = 1; k = 0

Dipole

n = 21 = 2; k = 1

1 r k +1 1 V ∼ k +1 r 1 V ∼ k +1 r 1 V ∼ k +1 r 1 V ∼ k +1 r

Quadrupole

n = 22 = 4; k = 2

Octopole

n = 23 = 8; k = 3

Multipole

n = 2k ; k

V∼

= = = =

1 r 1 r2 1 r3 1 r4

Problem 3.13. Let us consider an electric dipole with charges q and –q situated at distance d from each other, shown below. Calculate the electric potential at a point P in the far approximation r ≫ d .

Solution The total electric potential is obtained by superposition 1 ⎛q −q ⎞ V (r ⃗ ) = V1 + V2 = ⎜ + ⎟. 4πε0 ⎝ r1 r2 ⎠ From the law of cosines

⎛ d ⎞2 d r 12 = ⎜ ⎟ + r 2 − 2 r cos θ ⎝2⎠ 2 and

⎛ d ⎞2 d r 22 = ⎜ ⎟ + r 2 − 2 r cos(π − θ ). ⎝2⎠ 2 3-26

Electromagnetism

Note that cos(π − θ ) = −cos θ . We can rewrite our potential as

⎛ q ⎜ V= ⎜ 4πε0 ⎜ ⎝

1 d2 4

⎞ ⎟ ⎟ 2 + r + rd cos θ ⎟⎠ 1

−

d2 4

2

+ r − rd cos θ

or

⎛ q ⎜ ⎜ V= 4πε0 ⎜ r ⎝ 1+ When r ≫ d,

d2 4r 2

1 d2 4r 2

1

− −

d r

cos θ

r 1+

d2 4r 2

+

d r

⎞ ⎟ ⎟. cos θ ⎟⎠

is very small and can be ignored. If we consider x =

can use the binomial theorem and obtain 1

x 2

1

x . 2

(1 + x )− 2 ≅ 1 − and

(1 − x )− 2 ≅ 1 + From this, we have

1 r 1−

d r

=

⎞ 1⎛ d ⎜1 + cos θ ⎟ ⎝ ⎠ r 2r

=

⎞ 1⎛ d ⎜1 − cos θ ⎟ . ⎠ r⎝ 2r

cos θ

and

1 r 1+

d r

cos θ

Therefore,

V=

=

⎞ 1⎛ ⎞⎤ q ⎡1 ⎛ d d cos θ ⎟ − ⎜1 − cos θ ⎟⎥ ⎢ ⎜1 + ⎠ r⎝ ⎠⎦ 4πε0 ⎣ r ⎝ 2r 2r ⎤ q ⎡ d d cos θ − 1 + cos θ ⎥ ⎢⎣1 + ⎦ 4πε0r 2r 2r

qd cos θ . 4πε0r 2 JG JG Taking the dipole moment p = qd , we have JG p · rˆ . V= 4πε0r 2 V=

3-27

d r

cos θ ≪ 1, we

Electromagnetism

Problem 3.14. Find the electric field of the dipole in problem 3.13, centered at the JG origin with the dipole moment p in the z-direction. Solution From problem 3.13, the electric potential is given by JG qd p · rˆ p cos θ cos . V (r ⃗ ) = = θ = 2 2 4πε0r 4πε0r 4πε0r 2 We can find the field from the potential using JG E = −∇V. We need to use the gradient in spherical coordinates

Er = −

∂V 2p cos θ = ∂r 4πε0r 3

Eθ = −

p sin θ 1 ∂V = r ∂θ 4πε0r 3

Eϕ = −

1 ∂V = 0. r sin θ ∂ϕ

Therefore, the electric field due to the dipole is JG p 2 cos θ rˆ + sin θ θˆ . Edipole(r , θ ) = 4πε0r 3

(

)

Problem 3.15. Two point charges +4q and −q are separated by a distance d. The first charge is placed at (0, 0, d ) and the second one at the origin. Find: (a) the monopole moment; (b) the dipole moment; (c) the electric potential in spherical coordinates for r ≫ d . Include only the monopole and dipole contributions.

3-28

Electromagnetism

Solution (a) Monopole moment:

Q = 4q − q = 3q . (b) Dipole moment:

JG p =

2

∑qiri ⃗ = −q(0, 0, 0) + 4q(0, 0, d ) = 4qdzˆ . i=1

(c) The electric potential:

V (r ⃗ ) =

=

⎞ 1 ⎛1 1 1 ⎛ 3q 1 JG ⎞ ⎜⎜ ∑qi + ∑qi ri ′ cos θi′ + ⋯⎟⎟ = ⎜ + 2 p · rˆ⎟ 2 ⎠ 4πε0 ⎝ r i r r i ⎠ 4πε0 ⎝ r 1 ⎛ 3q p cos θ ⎞ 1 ⎛ 3q 4qd cos θ ⎞ ⎜ ⎟= ⎜ ⎟. + + 2 ⎠ 4πε0 ⎝ r ⎠ 4πε0 ⎝ r r r2

Bibliography Byron F W and Fuller R W 1992 Mathematics of Classical and Quantum Physics (New York: Dover) Griffiths D J 1999 Introduction to Electrodynamics 3rd edn (Englewood Cliffs, NJ: Prentice Hall) Griffiths D J 2013 Introduction to Electrodynamics 4th edn (New York: Pearson) Halliday D, Resnick R and Walker J 2010 Fundamentals of Physics 9th edn (New York: Wiley) Halliday D, Resnick R and Walker J 2013 Fundamentals of Physics 10th edn (New York: Wiley) Jackson J D 1998 Classical Electrodynamics 3rd edn (New York: Wiley) Rogawski J 2011 Calculus: Early Transcendentals 2nd edn (San Francisco, CA: Freeman)

3-29

IOP Concise Physics

Electromagnetism Problems and solutions Carolina C Ilie and Zachariah S Schrecengost

Chapter 4 Magnetostatics

This chapter introduces magnetic fields in a vacuum and the methods for calculating the magnetic field. Magnetic fields are intrinsically determined by electric charges in motion. We imagine these small currents as magnetic dipoles. From the general Biot–Savart law, to the more straightforward Ampère’s law applicable to configurations with higher degree of symmetry, the suggested problems constitute good practice in magnetostatics.

4.1 Theory 4.1.1 Magnetic force A charge q moving with velocity υ⃗ in a magnetic field B ⃗ experiences a force given by

Fm⃗ = qυ ⃗ × B ⃗. 4.1.2 Force on a current carrying wire The force on a current carrying wire in a magnetic field B ⃗ is

Fm⃗ =

∫ I ( dl⃗ × B ⃗).

4.1.3 Volume current density The current density of a current I ⃗ is

J⃗ =

dI ⃗ da⊥

and the current density of a charge density ρ moving at velocity υ⃗ is

J ⃗ = ρυ ⃗ .

doi:10.1088/978-1-6817-4429-2ch4

4-1

ª Morgan & Claypool Publishers 2016

Electromagnetism

4.1.4 Continuity equation The divergence of the charge density J ⃗ is related to the charge density ρ by

∇ ⋅ J⃗ = −

∂ρ . ∂t

4.1.5 Biot–Savart law The magnetic field due to current distributions is given by

dl′⃗ × rˆ r2

B ⃗( r ⃗ ) =

μ0I 4π

B ⃗( r ⃗ ) =

μ0 4π

∫

K⃗ (r ′⃗ ) × rˆ da′ r2

B ⃗( r ⃗ ) =

μ0 4π

∫

J ⃗(r ′⃗ ) × rˆ dτ ′. r2

∫

4.1.6 Divergence of B ⃗ Given magnetic field B,⃗ we have

∇ ⋅ B ⃗ = 0. 4.1.7 Ampère’s law Given magnetic field B,⃗ we have

∇ × B ⃗ = μ 0 J ⃗. By applying Stoke’s law, we also have

∮S B ⃗ ·

dl⃗ = μ0Ienc,

where

Ienc =

∫ J ⃗ · da ⃗ .

4.1.8 Vector potential The vector potential due to current distributions is given by μI 1 ′⃗ A ⃗ (r ⃗ ) = 0 dl 4π r

∫

A ⃗ (r ⃗ ) =

μ0 4π

∫

K⃗ (r ′⃗ ) d a′ r

A ⃗ (r ⃗ ) =

μ0 4π

∫

J (⃗ r ′⃗ ) dτ ′. r

4-2

Electromagnetism

Also,

B ⃗ = ∇ × A⃗ and

∇2 A ⃗ = μ0J ⃗. 4.1.9 Magnetic dipole moment The magnetic dipole moment due to a current I is

m⃗ =

∫ I da ⃗ .

4.1.10 Magnetic field due to dipole moment Given magnetic dipole moment m⃗ , the magnetic field is

⃗ = Bdip

μ0m 2 cos θ rˆ + sin θ θˆ . 4πr 3

(

)

4.2 Problems and solutions Problem 4.1. A proton travels through a uniform magnetic and electric field. The magnetic field is B ⃗ = ayˆ , where a is a positive constant. If at one moment the velocity of the proton is υ⃗ = bzˆ , where b is a positive constant, what is the force acting on the proton if the electric field is E ⃗ = −cxˆ ? Solution

F ⃗ = q E ⃗ + υ ⃗ × B ⃗ = q⎡⎣ −cxˆ + (bzˆ ) × (ayˆ )⎤⎦ = q⎡⎣ − cxˆ + ab( −xˆ ) ⎤⎦ = (cxˆ + abxˆ )

(

)

F ⃗ = −q(c + ab)xˆ .

Problem 4.2. A particle of charge q enters a region of uniform magnetic field B ⃗ (out of the page, in the z-direction) with an initial velocity υ⃗ (in the x-direction). The particle is deflected a distance y above the initial direction. If the region has a width of x , find the sign of the charge and the deflected distance y as a function of q , υ, B , and x.

4-3

Electromagnetism

Solution

F ⃗ = qυ ⃗ × B ⃗ Since the charge is deflected as shown, the charge is negative (determined from the right-hand rule). In the x -direction we have no force, and therefore no acceleration

x = υt. In the y -direction

ma = q υB sin 90. So

a=

q υB m

and

y = yo + υoyt +

a yt 2 2

=

a yt 2 2

=

q υBt 2 . 2m

By substituting the time

t=

x υ

we obtain

x2 2 2 υ 2 = q Bx = q Bx , 2m 2mυ 2p

q υB y=

where p is the momentum of the particle. Problem 4.3. The current density in a wire of circular cross section of radius R is dependent on the distance from the axis, given by J ⃗ = ks 2zˆ , where k is a constant. Find a) the total current in the wire and b) the current density if the current in a) is uniformly distributed.

4-4

Electromagnetism

Solution a) Given current density J ⃗ = ks 2zˆ , the current is R

2π

I=

∫ J ⋅⃗ da ⃗ = ∫ ( ks zˆ) ⋅ ( s dϕ ds zˆ) = ∫ 2

dϕ

0

∫

ks 3ds = 2πk

0

s4 4

R

= 0

πkR 4 . 2

b) If this current was uniformly distributed, the current density is simply

J=

I 1 πkR 4 kR2 . = = area 2 πR2 2

Problem 4.4. a) In the famous experiment of J J Thompson, he measured the charge to mass

q m

ratio of the catode rays. Find q when you know B, R , and υ, and that B ⃗ is m perpendicular to υ⃗ . b) He also had the beams going in a region with perpendicular electric field and magnetic field and ‘tuned’ them such that the electrons left the region with unchanged direction. If the speed of the electrons is υ and the magnetic field is B,⃗ what should be the value of the electric field? Solution a) The magnitude of the magnetic force is given by

Fm⃗ = q υ ⃗ × B ⃗, where we have υ⃗ ⊥ B ⃗ . Also

⃗ Fm⃗ = Fcentripetal . So,

Fm = q υB and

Fcent =

mυ 2 . R

q υB =

mυ 2 R

Therefore from

we have

q υ . = m BR

4-5

Electromagnetism

b) Setting the magnetic and electric forces as equal, we have

Fm⃗ = Fe⃗ → qυ ⃗ × B ⃗ = qE ⃗ . Dividing by q and expressing this in terms of magnitudes, we have

υB sin 90 = E so

E = υB.

Problem 4.5. Find the magnetic field at: a) The center of a circular wire loop of radius R carrying current I.

b) The center of a wire loop that consists of half a loop of radius R and half a square loop of side 2R , carrying current I.

Solution a) The Biot–Savart law states

4-6

Electromagnetism

B⃗ =

μ0I 4π

∫ dlr×2 r , ⃗

ˆ

where

dl⃗ = R dϕ ⃗ = R dϕ ϕˆ and

rˆ = −rˆ . Since r = R , the Biot–Savart law becomes

B⃗ =

μ0IR 4πR2

2π

∫ 0

μI ϕˆ × rˆ dϕ = 0 zˆ 2R

(out of page).

b) From part a), we can determine the field contribution due to the circular part is

Bc⃗ =

μ0I zˆ , 4R

which is half that of the full loop. As for the square, we consider the field R above the wire. We have

So

B⃗ =

μ0I 4π

∫ dl r×2 r , ′⃗

ˆ

where dl′⃗ × rˆ points in the zˆ -direction (out of the page). Also,

dl′ sin γ = dl′ cos θ and

l′ = R tan θ → dl′ =

4-7

R dθ cos2 θ

Electromagnetism

and

1 cos2 θ = . 2 r R2

r 2 = l′2 + R2 → Therefore,

μI B⃗ = 0 4π

θ2

∫ θ1

⎛ cos2 θ ⎞⎛ R ⎞ μ0I ⎟cos θ dθ = (sin θ2 − sin θ1). ⎜ ⎟⎜ 2 2 ⎝ R ⎠⎝ cos θ ⎠ 4πR π

So for each R-lengthed segment (i) and (ii), we have θ1 = 0 and θ2 = 4 , and for π π the 2R -lengthed segment, we have θ1 = − 4 and θ2 = 4 . So

⎛μI μ I ⎡ ⎛π ⎞ ⎛ π ⎞⎤ π ⎞ B ⃗ = Bc⃗ + 2⎜ 0 sin zˆ⎟ + 0 ⎢sin⎜ ⎟ − sin⎜ − ⎟⎥zˆ ⎝ ⎠ ⎝ 4 ⎠⎦ ⎣ ⎝ 4πR ⎠ 4 4πR 4 =

μ0I π +2 2 zˆ 4πR

(

)

(out of page).

Problem 4.6. Consider a cylindrical shell of radius R and length L , carrying σ and rotating at ω. Find the magnetic field d from the end of the shell (on the axis).

Solution Here we have

B⃗ =

μ0 4π

∫ K r×2 r da, ⃗

ˆ

where

da = R dz dϕ and 0 ⩽ z ⩽ L . The surface charge is given by

K⃗ = συ ⃗ = σωRϕˆ .

4-8

Electromagnetism

From the figure above, we have l′ = L − z and

r=

(l′ + d )2 + R2 =

(L − z + d )2 + R 2 .

Note that the field cancels such that the zˆ -component is the only component that survives. So 2 ⎡ K⃗ × rˆ ⎤ = σωR sin θ zˆ = σωR R zˆ = σωR zˆ . ⎣ ⎦z r r

Putting everything together, we have

B⃗ =

=

μ0 4π

2π

L

∫ ∫ 0

0

μ0σωR3zˆ 2

σ ωR 2 R ⎡ (L − z + d )2 + R2⎤3/2 ⎣ ⎦ L

∫ 0

zˆ dz dϕ

dz . ⎡ (L − z + d )2 + R2⎤3/2 ⎣ ⎦

Therefore,

B⃗ =

μ0σωR ⎡⎢ d+L – 2 ⎢⎣ R2 + (d + L )2

⎤ ⎥zˆ . R2 + d 2 ⎥⎦ d

Problem 4.7. A hemisphere of radius R and charge density ρ is rotating at ω. Find the magnetic field d above the center.

4-9

Electromagnetism

Solution Here we have

B⃗ =

μ0 4π

∫ J r×2 r dτ, ⃗

ˆ

where

dτ = r 2 sin θ dr dϕ dθ . From the figure below, we can see that

r 2 = d 2 + r 2 − 2dr cos θ .

Also,

J ⃗ = ρυ ⃗ = ρωr sin θ ϕˆ .

4-10

Electromagnetism

Note that the field cancels such that the zˆ -component is the only component that survives. So

⎡ J ⃗ × rˆ ⎤ = ρωr sin θ ⎛⎜ r sin θ ⎞⎟zˆ . ⎣ ⎦z ⎝ r ⎠ Therefore

μ B⃗ = 0 4π

2π

R

ρωr 2 sin2 θ r 2 sin θ zˆ

∫ ∫ ∫ π 2

0

μ ρω = 0 2

B⃗ =

π

π

2

+ r − 2dr cos θ

R

∫ ∫ π 2

(d

0

2

0

)

r 4 sin3 θ zˆ

(d

2

2

+ r − 2dr cos θ

)

3 2

3 2

dr dθ dϕ

dr dθ

μ0ρω ⎡ R2 + d 2 −2R 4 + d 2R2 − 12d 4 + 2R5 + 5d 3R2 + 12d 5⎤⎦zˆ . 30d 3 ⎣

(

)

Problem 4.8. A spherical shell of radius R , carrying σ and rotating at ω, is centered at the origin. Find the velocity a loop of wire, carrying λ with radius a centered at the origin, required to cancel the magnetic field at the center.

Solution First, we have

B⃗ =

μ0 4π

∫ K r×2 r da, ⃗

where

r=R

4-11

ˆ

Electromagnetism

rˆ = −rˆ K⃗ = συ ⃗ = σωR sin θ ϕˆ da = R2 sin θ dθ dϕ and

K⃗ × rˆ = σωR sin2 θ zˆ . Putting this together, we have

Bs⃗ =

Bs⃗ =

μ0 4π

π

2π

∫ ∫ 0

0

μ σωR zˆ σωR sin2 θ R2 sin θ zˆ dϕ dθ = 0 2 R 2

π

∫

sin3 θ dθ

0

2μ0σωR zˆ . 3

Now a line of charge λ rotating at υ⃗ ‘looks’ like a wire carrying current I ⃗ = λυ ⃗= λωl aϕˆ . From problem 4.5(a), we know this produces magnetic field

Bl⃗ =

μ0λωl a μ λωl zˆ = 0 zˆ . 2a 2

We want Bl⃗ + Bs⃗ = 0, so

μ0λωl 2μ σωR λωl 2σωR . + 0 =0→ =− 2 3 2 3 Therefore,

ωl = −

4ωσR . 3λ

Problem 4.9. A long straight wire carries a steady current I. Obtain the magnetic field at a distance s from the wire.

4-12

Electromagnetism

Solution Here, we apply Ampère’s law, with an Amperian loop that is a circle centered on the wire and in a plane perpendicular to the wire. Ampère’s law is

∮S B ⋅⃗ dl⃗ = μ0Ienc. Noticing B ⃗ dl⃗ , it follows that B is a constant at a certain distance s from the wire. So the left-hand side is given by

∮S B ⋅⃗ dl⃗ = ∮S B dl = B ∮S dl = B 2πs. The enclosed current is just simply given by

Ienc = I So

B=

μ0I . 2πs

Since the magnetic field is tangent on the circle at every point

B⃗ =

μ0I ˆ ϕ. 2πs

Problem 4.10. An electric current flows through a long cylinder wire of radius a . Find the magnetic field inside and outside the wire, and plot it, in the following cases, where k is a constant with the appropriate units: a) I = constant (steady current). b) Current density J is proportional to the distance from the axis: J = ks . c) J = ks 2 . Solution a) Here we have constant I . For s > a , our Amperian loop is given by

4-13

Electromagnetism

Ampère’s law states

∮S B ⋅⃗ dl⃗ = μ0Ienc In all cases, the left-hand side yields

∮S B ⋅⃗ dl⃗ = ∮S B dl = B ∮S dl = B 2πs. Here we simply have

Ienc = I So

B 2πs = μ0I → B =

μ0I . 2πs

Applying the right-hand rule to a current coming out of the page, we have

B⃗ =

μ0I ˆ ϕ. 2πs

For s < a , our Amperian loop is inside the wire, at radius s .

Again we have

∮S B ⋅⃗ dl⃗ = μ0Ienc. Since I is uniform, the current density is constant,

J=

I I = enc2 2 πa πs

so

Ienc = I

4-14

s2 . a2

Electromagnetism

Therefore

μ0sI μ sI → B ⃗ = 0 2 ϕˆ . 2 2πa 2πa

B=

The plot of the magnetic field is given below.

b) Here we have J = ks . For s > a , the Amperian loop is the same as part a) for s > a . We still have the left hand side of Ampère’s law given by

∮S B ⋅⃗ dl⃗ = B 2πs. Now, since J is not constant, we need to integrate in order to find the enclosed current, so

Ienc =

a

2π

a

∫ J da = ∫

∫

∫

Js ds

0

dϕ =

0

2π 2

ks ds

0

∫ 0

ks 3 d ϕ = 2π 3

a

= 0

2πka 3 . 3

Therefore,

B 2πs =

μ02πka 3 μ ka 3 →B= 0 3 3s

with

B⃗ =

μ0ka 3 ˆ ϕ. 3s

For s < a , we again have the same Amperian loop as part a). Here, our enclosed current is given by s

Ienc =

∫ J da = ∫ 0

2π

ks′ s′ ds′

∫ 0

4-15

ks′3 d ϕ = 2π 3

s

= 0

2πks 3 . 3

Electromagnetism

Therefore,

B 2πs =

μ02πks 3 μ ks 2 →B= 0 3 3

with

μ0ks 2 ˆ ϕ. 3

B⃗ =

The plot of the magnetic field is given below.

c) Now we have J = ks 2 . For s > a , the enclosed current is given by a

Ienc =

∫ J da = ∫

2π

Js ds

0

∫

a

∫

d ϕ = 2π

0

ks 2 s ds = 2π

0

ks 4 4

a

= 2π 0

ka 4 . 4

Therefore,

B 2πs = μ02π

μ ka 4 ka 4 →B= 0 4 4s

with

B⃗ =

μ0ka 4 ˆ ϕ. 4s

For s < a , we have s

Ienc =

∫

J da =

∫ 0

2π

Js′ ds′

∫

s

dϕ = 2π

∫

0

ks′2 s′ ds′′ =

0

Therefore,

B=

μ0ks 3 4

4-16

2πks′4 4

s

= 0

2πks 4 . 4

Electromagnetism

with

B⃗ =

μ0ks 3 ˆ ϕ. 4

The plot of magnetic field is given below.

Problem 4.11. Use Ampère’s law to obtain the magnetic field inside and outside a N solenoid of n = L , where N is the number of turns, and L is the length of the solenoid. The solenoid is carrying the current I . Solution Let us choose two Amperian loops given by

Starting with the outside loop (loop A), the magnetic field does not have any radial Br component or Bϕ . Ampère’s law is given by

∮S B ⋅⃗ dl⃗ = μ0Ienc. 4-17

Electromagnetism

Since we have Ienc = 0, B (c ) = B (d ), but since B → 0 for large distances, B = 0 outside the solenoid. For loop B, the left hand side of Ampère’s law is given by

∮S B ⋅⃗ dl⃗ = ∮S B dl = BL. The sides perpendicular on the solenoid yield zero dot product, as the magnetic field is oriented parallel to the solenoid’s axis in the z -direction (by the right-hand rule). The enclosed current is given by

Ienc = InL = IN. Substituting these quantities into Ampère’s law yields

BL = μ0InL → B = μ0In =

μ0IN L

with

B ⃗ = μ0Inzˆ =

μ0IN z. ˆ L

Problem 4.12. A current carrying empty cylinder of inner radius a and outer radius b has a current density J , which is proportional to the distance from the axis; J = ks, k constant. Find the magnetic field in all regions.

Solution There are three significant regions: s < a , a < s < b, and s > b. The easiest to find is the field for s < a ,

4-18

Electromagnetism

where the enclosed current is zero. Therefore,

∮S B ⋅⃗ dl⃗ = 0, so B ⃗ = 0. For a < s < b, we have

Using Ampère’s law,

∮S B ⋅⃗ dl⃗ = μ0Ienc. We have the left-hand side is given by

∮S B ⋅⃗ dl⃗ = B 2πs, with enclosed current given by s

Ienc =

∫

J ⃗ · da ⃗ =

∫ a

2π

Js′ ds′

∫ 0

s

d ϕ = 2π

∫ a

4-19

2πks′3 ks′ s′ ds′ = 3

s

a

(

2πk s 3 − a 3 =

3

)

Electromagnetism

(

μok s 3 − a 3

B=

)

3s

with

B⃗ =

(

μok s 3 − a 3 3s

) ϕˆ .

For s > b, we have

Again,

∮S B ⋅⃗ dl⃗ = B 2πs with b

Ienc =

∫

J ⃗ ⋅ da ⃗ =

∫ a

2π

Js ds

∫

b

dϕ = 2π

0

∫

ks s ds = 2π

a

ks 3 3

b

a

(

2πk b3 − a 3 =

3

).

Therefore,

(

k b3 − a 3 B=

)

3s

with

B⃗ =

(

k b3 − a 3 3s

) ϕˆ .

Problem 4.13. Find the vector potential d above a spinning disk of radius R, with angular velocity ω and carrying σ .

4-20

Electromagnetism

Solution We have

A⃗ =

μ0 4π

⃗

∫ Kr da,

where

K⃗ = συ ⃗ = σωrϕˆ . From

we also have

r=

r2 + d 2

and

da = 2πr dr . So

μ A⃗ = 0 4π

R

∫ 0

μ σωϕˆ dr = 0 2 r2 + d 2

σωr 2πrϕˆ

R

∫ 0

μ σω ⎡ = 0 ⎣⎢R R2 + d 2 − d 2 ln R + 4

(

4-21

r2 r2 + d 2

dr

⎤ R2 + d 2 + d 2 ln d ⎦⎥ϕˆ

)

Electromagnetism

A⃗ =

⎞⎤ ⎟⎥ϕˆ . R2 + d 2 ⎠⎥⎦

⎛ μ0σω ⎡ ⎢R R2 + d 2 + ln⎜ 4 ⎢⎣ ⎝R +

d

Problem 4.14. What current density produces vector potential A⃗ = sin ϕ zˆ ? Solution First, check ∇ ⋅ A⃗ = 0:

∇ ⋅ A ⃗ = ∇ ⋅ (sin ϕ zˆ ) =

∂ (sin ϕ) = 0. ∂z

Now, B ⃗ = ∇ × A⃗ and ∇ × B ⃗ = μ0 J ,⃗ so

B ⃗ = ∇ × A ⃗ = ∇ × (sin ϕ zˆ ) =

⎤ 1⎡ ∂ cos ϕ sˆ . ⎢ (sin ϕ)⎥sˆ = ⎦ s ⎣ ∂ϕ s

Also,

μ0J ⃗ = ∇ × B ⃗ = ∇ ×

⎛ cos ϕ ⎞ 1 ⎡ ∂ ⎛ cos ϕ ⎞⎤ sin ϕ ⎜ ⎟⎥z ˆ= sˆ⎟ = ⎢ − ⎜ zˆ . ⎝ s ⎠ ⎝ ⎠ s ⎣ ∂ϕ s ⎦ s2

Therefore,

J⃗ =

sin ϕ zˆ . μ0s 2

This can be checked using the product rule

(

)

(

)

∇ × ∇ × A ⃗ = μ0J ⃗ = ∇ ∇ ⋅ A ⃗ − ∇2 A ⃗ . Since ∇ ⋅ A⃗ = 0, ∇2 A⃗ = μ0 J ⃗ where

−∇2 A ⃗ = −

1 ∂2 1 sin ϕ (sin ϕ)zˆ = − 2 ( −sin ϕ)zˆ = zˆ . 2 s ∂ϕ s s2

So

J⃗ =

1 sin ϕ −∇2 A ⃗ = zˆ μ0 μ0s 2

(

)

as expected. Problem 4.15. Find the vector potential inside and outside a wire of radius R that is carrying current density J ⃗ = kszˆ , where k is a constant.

4-22

Electromagnetism

Solution We can find the field inside by

∮S B ⃗ ⋅ dl⃗ = μ0Ienc, where

∮S B ⃗ ⋅ dl⃗ = B 2πs and s

Ienc =

∫

∫

J da =

2πks 3 . 3

ks′2πs′ ds′ =

0

So,

∮S B ⃗ ⋅ dl⃗ = B 2πs = μ0 2π3ks

3

and

B⃗ =

μ0ks 2 ˆ ϕ. 3

We must have that A⃗ depends only on s and is in the direction of the current. So A⃗ = A(s )zˆ and ∇ × A⃗ = B ⃗ . Note

∇ × A⃗ = −

μ ks 2 ∂A ˆ ϕ = B ⃗ = 0 ϕˆ . 3 ∂s

Therefore,

dA = −

μ0k 2 s 3

and

⎛ μ ks 3 ⎞ A⃗ = ⎜− 0 + C ⎟zˆ . 9 ⎝ ⎠ We will express this as

A⃗ = −

μ0k 3 s − α 3 zˆ . 9

(

)

Outside, our total current is 3

Itot = Ienc =

∫ J da = 2πkR 3 4-23

.

Electromagnetism

From

∮S B ⋅⃗ dl⃗ = μ0Ienc we have

B 2πs = μ0

2πkR3 3

and

B⃗ =

μ0kR3 ˆ ϕ. 3s

Again,

dA = − B ds = −

μ0kR3 ds 3s

so

⎛ μ kR3 ⎞ A⃗ = ⎜− 0 ln s + C ⎟zˆ . 3 ⎝ ⎠ We will express this as

A⃗ = −

μ0kR3 ⎛ s ⎞ ⎜ln ⎟zˆ . 3 ⎝ β⎠

Since A⃗ is continuous at R ,

−

μ0k 3 μ kR3 ⎛ s ⎞ s − α3 = − 0 ⎜ln ⎟ 9 3 ⎝ β⎠

(

)

we have

R3 − α 3 = 3R3 ln

R β

with

⎛ R⎞ R3⎜1 − 3 ln ⎟ = α 3 β⎠ ⎝ and

1 − 3 ln

R α3 = 3. β R

4-24

Electromagnetism

If α = β = R,

⎛ R ⎞ R3 1 − 3 ln⎜ ⎟ = 3 → 1 = 1. ⎝R⎠ R So

⎧ μ0k 3 3 ⎪ ⎪− 9 s − R zˆ s < R . A⃗ = ⎨ 3 s ⎪ μ0kR ln zˆ s > R − ⎪ ⎩ 3 R

(

)

Problem 4.16. A disk of radius R is carrying surface charge σ = kr , where k is a constant, and spinning at angular velocity ω. Find the magnetic dipole moment and the field it produces. Solution We have

K⃗ = συ ⃗ = σωrϕˆ = ωkr 2ϕˆ . So

dI = ωkr 2dr → I =

ωkr 3 . 3

Therefore,

m⃗ =

m⃗ =

∫

ωk I da ⃗ = 3

R

∫ 0

2πωk r 2πr dr zˆ = 3 3

R

∫

r 4 dr

0

2πωkR5 zˆ . 15

We have, in spherical coordinates,

⃗ = Bdip

μ0m 2 cos θ rˆ + sin θ θˆ , 4πr 3

(

)

which can be expressed in cylindrical coordinates by considering r = z s z s z s cos θ = r , sin θ = r , θˆ = r sˆ − r zˆ , and rˆ = r sˆ + r zˆ . Therefore,

μ0m ⎡ ⎛ z ⎞⎛ s s ⎞⎟⎤ z ⎞⎟ ⎛⎜ s ⎞⎟⎜⎛ z ⎜ ⎟⎜ s ˆ ˆ 2 zˆ ⎥ s z + ˆ + − ⎢ 4πr 3 ⎣ ⎝ r ⎠⎝ r r ⎠⎦ r ⎠ ⎝ r ⎠⎝ r μm = 0 5 2zssˆ + 2z 2zˆ + szsˆ − s 2zˆ 4πr

⃗ = Bdip

(

)

4-25

s2 + z2 ,

Electromagnetism

= ⃗ = Bdip

μ0m ⎡ 3zssˆ + 2z 2 − s 2 zˆ⎤⎦ 4πr 5 ⎣

(

μ0m 1 4π s 2 + z 2

(

)

⎡ 3zssˆ + 2z 2 − s 2 zˆ⎤ . ⎦

(

5/2 ⎣

)

)

Substitution of m yields,

⃗ = Bdip

μ0ωkR5 1 30 s2 + z2

(

⎡ 3zssˆ + 2z 2 − s 2 zˆ⎤ . ⎦

5/2 ⎣

)

(

)

Bibliography Griffiths D J 1999 Introduction to Electrodynamics 3rd edn (Englewood Cliffs, NJ: Prentice Hall) Griffiths D J 2013 Introduction to Electrodynamics 4th edn (New York: Pearson) Halliday D, Resnick R and Walker J 2010 Fundamentals of Physics 9th edn (New York: Wiley) Halliday D, Resnick R and Walker J 2013 Fundamentals of Physics 10th edn (New York: Wiley)

4-26

IOP Concise Physics

Electromagnetism Problems and solutions Carolina C Ilie and Zachariah S Schrecengost

Chapter 5 Electric fields in matter

Now we will address problems that deal with electric fields in matter, looking at problems involving dipole moments, media polarization, and electric displacement. Ideas developed in chapters 2 and 3 will be revisited and expanded upon in this chapter. Gauss’s law is reformulated for electric displacement and various ways to calculate the energy of a configuration. Some of the techniques practiced in chapter 3 will be applied now, including the Laplace equation and Legendre polynomials.

5.1 Theory 5.1.1 Induced dipole moment of an atom in an electric field Given an atom with polarizability α in an electric field E ⃗ , the induced dipole moment is

p ⃗ = αE ⃗ . 5.1.2 Torque on a dipole due to an electric field Given a dipole moment p ⃗ in an electric field E ⃗ , the torque on the dipole is

N⃗ = p ⃗ × E ⃗ . 5.1.3 Force on a dipole Given a dipole moment p ⃗ in an electric field E ⃗ , the force on the dipole is

(

)

F ⃗ = p ⃗ ⋅ ∇ E ⃗.

doi:10.1088/978-1-6817-4429-2ch5

5-1

ª Morgan & Claypool Publishers 2016

Electromagnetism

5.1.4 Energy of a dipole in an electric field Given a dipole moment p ⃗ in an electric field E ⃗ , the energy of the dipole is

U = −p ⃗ ⋅ E ⃗. 5.1.5 Surface bound charge due to polarization P ⃗ Given polarization P ⃗ and normal vector nˆ , the surface bound charge is

ˆ σ b = P ⃗ ⋅ n. 5.1.6 Volume bound charge due to polarization P ⃗ Given polarization P,⃗ the volume bound charge is

ρb = −∇ ⋅ P.⃗ 5.1.7 Potential due to polarization P ⃗ Given a volume V , the potential due to polarization P ⃗(r ⃗ ) is

V (r ⃗ ) =

1 4πε0

∫V r

ˆ ⋅ P (⃗ r ′⃗ ) dτ ′. r2

5.1.8. Electric displacement Given polarization P ⃗ and electric field E ⃗ , the electric displacement is given by

D⃗ = ε0E ⃗ + P.⃗ 5.1.9 Gauss’s law for electric displacement Considering electric displacement D⃗ and free charge density ρf , Gauss’s law can be written in differential form as

∇ ⋅ D⃗ = ρf and in integral form as

∮S D⃗ ⋅ da ⃗ = qf

enc

,

where qfenc is the total free charge enclosed in the volume. 5.1.10 Linear dielectrics Given a medium with electric susceptibility χe , the polarization is given by

P ⃗ = ε0 χe E ⃗ ,

5-2

Electromagnetism

where E ⃗ is the total electric field. The electric displacement is now

(

)

(

)

D⃗ = ε0E ⃗ + P ⃗ = ε0 + ε0 χe E ⃗ = ε0 1 + χe E ⃗ = ε0εrE ⃗ = εE ⃗ , where ε is the permittivity of the material and εr is the relative permittivity of the material. Also, the boundary conditions are now ⊥ ⊥ εaboveEabove − εbelowE below = σf

And

εabove

∂V ∂Vabove − εbelow below = −σf ∂n ∂n

while we still maintain

Vabove = Vbelow. 5.1.11 Energy in a dielectric system Given electric field E ⃗ and electric displacement D⃗ , the energy in a dielectric system is

W=

ε0 2

∫ εrE 2dτ = 12 ∫ D⃗ ⋅ E ⃗dτ.

5.2 Problems and solutions Problem 5.1. Given p1⃗ and p2⃗ below, find where to place point charge q such that there is no net torque on p2⃗ . Assume the center of p1⃗ is the origin and express your answer in spherical coordinates.

Solution The field at p2⃗ is given by

⃗ = Edip

p1 ˆ p1 zˆ . θ=− 3 4πε0r 4πε0r 3

5-3

Electromagnetism

Since the field is straight down, we must place the point charge q below it to cancel the field (thus resulting in zero torque on p2⃗ ). We must place q at a distance d from the dipole so that p1 q + = 0. Edip + Eq = − 4πε0r 3 4πε0d 2 Solving for d 2 yields

d2 =

qr 3 . p1

Therefore we have

with

rq2 =

⎛ qr ⎞ qr 3 + r 2 = r 2⎜ + 1⎟ p1 ⎝ p1 ⎠

so

rq = r

qr + 1. p1

From

cos φ =

r rq

we have

⎛ ⎞ ⎜ ⎟ 1 ⎟. −1⎜ φ = cos ⎜ qr ⎟ ⎜ p +1⎟ ⎝ ⎠ 1

5-4

Electromagnetism

So the spherical coordinates of q are

⎛ ⎛ ⎞⎞ ⎜ ⎜ ⎟⎟ π 1 qr ⎟⎟ . (r , θ ) = ⎜⎜r + 1 , + cos−1⎜ ⎜ qr ⎟⎟ 2 p1 ⎜ ⎜ p + 1 ⎟⎟ ⎝ ⎠⎠ 1 ⎝

Problem 5.2. Consider a neutral atom, with polarizability α , located z above a disk of radius R carrying surface charge σ . Find the force of attraction between the atom and the plate.

Solution The field at z is given by

E⃗ =

E⃗ =

1 4πε0

∫

σ σ rˆ da = 2 4πε0 r

zσ ⎛ 1 ⎜ − 2ε0 ⎝ z

R

∫ 0

2πrz

(z

2

+ r2

3/2

)

zˆ dr

⎞ ⎟zˆ . R2 + z 2 ⎠ 1

This induces a dipole

p ⃗ = αE ⃗ =

zσα ⎛ 1 ⎜ − 2ε0 ⎝ z

⎞ ⎟zˆ . R2 + z 2 ⎠ 1

The electric field due to the dipole is given by ⃗ = p 2 cos θ rˆ + sin θ θˆ Edip 4πε0r 3

(

=

zσα ⎛ 1 ⎜ − 8πε02r 3 ⎝ z

)

⎞ ⎟ 2 cos θ rˆ + sin θ θˆ . R2 + z 2 ⎠ 1

5-5

(

)

Electromagnetism

The force on a piece of charge dq is given by

dF ⃗ = E ⃗ dq so

F⃗ =

∫ dF ⃗ = ∫ E ⃗dq = ∫ E ⃗σ dA = ∫ E ⃗σ 2π l dl.

Consider the following

with side view

In our expression fix alignment for E ⃗ , we have the term (2 cos θ rˆ + sin θ θˆ ). Due to symmetry, we only have a zˆ -component of the force. Therefore,

z rˆ → −cos γ zˆ = − zˆ r

5-6

Electromagnetism

and

l θˆ = −sin γ zˆ = − zˆ , r where

z 2 + l2 .

r= Also, we have θ = π − γ . So

cos θ = cos(π − γ ) = cos(π ) cos( −γ ) + sin(π ) sin( −γ ) = − cos γ = −

z r

and

sin θ = sin(π − γ ) = sin(π ) cos( −γ ) + cos(π ) sin( −γ ) = −sin( −γ ) = sin γ =

l . r

Therefore, we have

⎛ l ⎞⎛ l ⎞ ⎛ z ⎞⎛ z ⎞ 1 2 cos θ rˆ + sin θ θˆ = 2⎜ − ⎟⎜ − ⎟zˆ + ⎜ ⎟⎜ − ⎟zˆ = − 2 −2z 2 + l 2 zˆ . ⎝ r ⎠⎝ r ⎠ ⎝ r ⎠⎝ r ⎠ r

(

)

Our force becomes

F⃗ =

2πzσ 2α ⎛ 1 ⎜ − 8πε02 ⎝ z

zσ 2α ⎛ 1 ⎜ − F⃗ = 4ε02 ⎝ z

⎞ ⎟ R2 + z 2 ⎠ 1

( −2z + l )l zˆ dl ∫ (z + l ) 0 R

2

2

−

2

⎡ ⎞⎢ R2 ⎟⎢ R2 + z 2 ⎠⎢⎣ R2 + z 2 1

2

5/2

⎤ σ 2αR2 R2 + z 2 − z ⎥ z zˆ . ˆ = 2 3/2 ⎥ 4ε02 R2 + z ⎥⎦

(

(

(

)

(

R2 + d 2 − d

So at z = d , the force of attraction is

σ 2αR2 F=

(

4ε02 R2 + d

2

)

).

We can verify this using

F ⃗ = ( p ⃗ ⋅ ∇) E ⃗ Note

p⃗ ⋅ ∇ =

σα ⎛ ⎜1 − 2ε0 ⎝

5-7

)

⎞∂ ⎟ R2 + z 2 ⎠ ∂z z

)

Electromagnetism

and

σα ⎛ ⎜1 − 2ε0 ⎝

F ⃗ = ( p ⃗ ⋅ ∇) E ⃗ = zσ 2α ⎛ 1 ⎜ − F⃗ = − 4ε02 ⎝ z

⎞∂⎡ σ ⎛ ⎜1 − ⎟ ⎢ R2 + z 2 ⎠ ∂z ⎢⎣ 2ε0 ⎝ z

⎡ ⎞⎢ R2 ⎟⎢ R2 + z 2 ⎠⎢⎣ R2 + z 2

⎤ σ 2αR2 R2 + z 2 − z ⎥ zˆ = − zˆ , 2 3/2 ⎥ 4ε02 R2 + z ⎥⎦

(

1

(

⎞ ⎤ ⎟zˆ⎥ R2 + z 2 ⎠ ⎥⎦ z

)

(

)

)

which is equal in magnitude and opposite in direction to what was found above. Why is this? In the first method, we calculated the force on the plate from the atom. So a positive force is ‘attractive’. In the second method, we are finding the force on the atom from the plate. A negative force at the atom ‘attracts’ it to the plate. Problem 5.3. Consider p1⃗ and p2⃗ below. Find the force of p2⃗ on p1⃗ and verify using the energy of the configuration.

Solution The electric field due to a dipole is

⃗ (r , θ ) = Edip

p 2 cos θ rˆ + sin θ θˆ . 4πε0r 3

(

)

Here, θ = π , so the field at p1⃗ due to p2⃗ is

⃗ = Edip

−2p2 −2p2 yˆ . rˆ = 3 4πε0r 4πε0y 3

The force is given by

(

)

F ⃗ = p⃗ ⋅ ∇ E⃗ where p ⃗ = p1⃗ = −p1 yˆ . So

p ⃗ ⋅ ∇ = − p1

5-8

∂ . ∂y

Electromagnetism

Therefore,

(

)

F ⃗ = p ⃗ ⋅ ∇ E ⃗ = −p1

2p p ∂ −3 ∂ ⎛ −2p2 ⎞ y yˆ yˆ ⎟ = 1 2 ⎜ 4πε0 ∂y ∂y ⎝ 4πε0y 3 ⎠

( )

F⃗ = −

3p1 p2 yˆ . 2πε0y 4

The energy stored in this configuration is given by

⎡ ⎛ −2p2 ⎞⎤ ⎛ −2p2 ⎞ 2p p ⎥ ˆ ( ) y p = − − U = −p1⃗ ⋅ E ⃗ = −⎢( −p1 yˆ ) ⋅ ⎜ ⎟ ⎜ ⎟ = − 1 23 . 1 3 3 4πε0y ⎝ 4πε0y ⎠⎦ ⎝ 4πε0y ⎠ ⎣ From this, the force is given by

⎛ −2p1 p2 ⎞ 2p1 p2 ∂ −3 y yˆ F ⃗ = −∇U = −∇⎜ ⎟= 3 4πε0 ∂y ⎝ 4πε0y ⎠

( )

F⃗ = −

3p1 p2 yˆ 2πε0y 4

as expected. Problem 5.4. Consider the two dipoles depicted below. Find the angle γ that maximizes and minimizes the magnitude of the torque on p2⃗ due to p1⃗ .

Solution The electric field due to a dipole is p ⃗ (r , θ ) = 2 cos θ rˆ + sin θ θˆ . Edip 4πε0r 3

(

Here, θ = π2 , so the field at p2⃗ due to p1⃗ is

⃗ = Edip

p1 ˆ θ, 4πε0r 3

which points down. The torque is given by

⃗ . N⃗ = p2⃗ × Edip

5-9

)

Electromagnetism

⃗ and p2⃗ by We can express Edip

⃗ =− Edip

p1 zˆ 4πε0y 3

and

p2⃗ = p2 cos γ yˆ + p2 sin γ zˆ . So

xˆ ⃗ = N⃗ = p2⃗ × Edip

yˆ

zˆ

0 p2 cos γ 0

0

p2 sin γ p1 − 4πε0y 3

=

−p1 p2 cos γ xˆ . 4πε0y 3

We can see that ∣N⃗ ∣ is maximum when γ = 0 and γ = π

N⃗ = and minimum when γ =

π 2

and γ =

p1 p2 4πε0y 3

3π 2

N⃗ = 0. Note the effect of aligning the dipole parallel to the field, ∣p ⃗ × E ∣⃗ = 0, and aligning the dipole perpendicular to the field ∣ p ⃗ × E ∣⃗ = pE . Problem 5.5. A sphere of radius R carries polarization P ⃗ = kr rˆ , where k is a constant, from r = a to r = R . Find the electric field in all regions.

5-10

Electromagnetism

Solution The bound charges are given by

σ b(a ) = P ⃗ ⋅ nˆ =

k k rˆ ⋅ ( −rˆ ) = − a a

σ b(R ) = P ⃗ ⋅ nˆ =

k k rˆ ⋅ rˆ = R R

ρb = −∇ ⋅ P ⃗ = −

1 ∂ ⎛ 2k⎞ k ⎜r ⎟ = − 2 . 2 ⎝ ⎠ r r r ∂r

When r < a , qenc = 0, so E ⃗ = 0. When a ⩽ r ⩽ R , we have

∮S E ⃗ ⋅ da ⃗ =

qenc ε0

with

qenc

k = − 4πa 2 + 4π a

(

)

r

∫

−

a

k (r′)2 dr′ = −4πkr . 2 (r′)

So

∮S E ⃗ ⋅ da ⃗ = E 4πr 2 = − 4πεkr 0 and

E⃗ = −

k r⃗ k 1 rˆ = − . ε0 r ε0 r 2

When r > R , we have

qenc = −

k 4πa 2 + 4π a

(

)

R

∫ a

−

k k (r′)2 dr′ + 4πR2 = 0 2 R (r′)

(

)

so E ⃗ = 0. Problem 5.6. Consider a very long cylinder of radius R hollowed out to a radius a and carrying a uniform, radial polarization P ⃗ and charge density ρ = ks , where k is a constant. Find the electric field in all three regions (P ⃗ from a to R , ρ from 0 to R ).

5-11

Electromagnetism

Solution For s < a , all we have is charge, so

∮S E ⃗ ⋅ da ⃗ =

qenc , ε0

where s

qenc = 2π l

∫

ks′s′ds′ =

0

2π lks 3 . 3

So

∮S E ⃗ ⋅ da ⃗ = E 2πsl = 2π3lεks0

3

and

E⃗ =

s 2k sˆ . 3ε 0

For a ⩽ s ⩽ R , we have bound charge

σ b(a ) = P ⃗ ⋅ nˆ = Psˆ ⋅ ( −sˆ ) = −P ρb = −∇ ⋅ P ⃗ = −

1 ∂ P (sP ) = − . s s ∂s

So,

qenc

2π lks 3 = −P 2πal+ + 2π l 3

s

∫ a

⎛ ks 3 ⎞ P − s′ ds′ = 2π l⎜ − Pa − Ps + Pa⎟ ⎝ 3 ⎠ s′

⎛ ks 2 ⎞ qenc = 2π ls⎜ − P⎟. ⎝ 3 ⎠ Therefore,

⎛

∮S E ⃗ ⋅ da ⃗ = E 2πsl = 2πε0ls ⎜⎝ ks3 5-12

2

⎞ − P⎟ ⎠

Electromagnetism

So

E⃗ =

1 ks 2 − 3P sˆ . 3ε 0

(

)

For s > R , we have

σ b(R ) = P ⃗ ⋅ nˆ = Psˆ ⋅ sˆ = P. So,

qenc

2π lkR3 = −P 2πal + + 2π l 3

R

∫ a

P − s′ ds′ + P 2πRl s′

⎛ kR3 ⎞ = 2π l⎜ − Pa − PR + Pa + PR⎟ ⎝ 3 ⎠ qenc =

2π lkR3 . 3

Therefore, 3

∮S E ⃗ ⋅ da ⃗ = E 2πsl = 2π3lεkR 0 and

E⃗ =

kR3 sˆ . 3sε0

Problem 5.7. Consider a cylinder of radius R and length L , carrying polarization P ⃗ = Pzˆ . Find the potential d from the cylinder.

Solution The potential is

V (r ⃗ ) =

1 4πε0

∫V r r⋅ 2P dτ.

5-13

ˆ

⃗

Electromagnetism

We can see from

that

r=

(l + d )2 + s 2 .

If z = 0 is the left side of the cylinder, then l goes from L to 0. So

r=

(L − z + d )2 + s 2 .

Also,

rˆ = cos θ zˆ =

l+d L−z+d zˆ = zˆ . r r

Since dτ = s ds dϕ dz , we have

1 V= 4πε0

L

R

2π

∫∫∫ 0

0

0

P (L − z + d )s dϕ ds dz . ⎡ (L − z + d )2 + s 2⎤3/2 ⎣ ⎦

Using u = L − z + d , we have du = −dz . Evaluating u at the endpoints yields

u(z = 0) = L + d And

u (z = L ) = d . So

P V= 2ε 0

L +d R

∫ ∫ d

0

P ds du = 3/2 2ε 0 u2 + s2 us

(

)

5-14

L +d

∫ d

⎛ ⎜1 − ⎝

⎞ ⎟du . R2 + u 2 ⎠ u

Electromagnetism

Using x = R2 + u 2 , we have dx = 2u du and

⎛ P ⎜ L +d V= u∣d − 2ε0 ⎜⎝ = V=

R 2 + (L + d ) 2

∫ R 2+ d 2

P ⎡ L+d−d− 2ε0 ⎣ P ⎡ L− 2ε0 ⎣

⎞ 1 −1 ⎟ x 2 dx⎟ 2 ⎠

R 2 + (L + d )2 +

R 2 + (L + d )2 +

R2 + d 2 ⎤⎦

R2 + d 2 ⎤⎦ .

Problem 5.8. Consider a sphere of radius R carrying polarization P ⃗(r ⃗ ) = kr nrˆ where n is an integer and k is a constant. Find the charge density required to cancel the polarization. Solution The bound charge is given by

1 ∂ 2 n r kr r 2 ∂r k (n + 2) n+1 k ∂ n +2 =− r = −k (n + 2)r n−1. =− 2 r r2 r ∂r

ρb = −∇ ⋅ P ⃗ = −

(

)

( )

Therefore, a charge density

ρ = k (n + 2)r n−1 will cancel the bound charge produced by P ⃗(r ⃗ ) = kr nrˆ . Problem 5.9. A spherical shell of radius R with surface charge density σ is surrounded up to radius a by an LIH dielectric material of susceptibility χe . Find the electric displacement and the electric field.

5-15

Electromagnetism

Solution Gauss’s law for electric displacement is given by

∮S D⃗ ⋅ da ⃗ = qf

enc

.

For r < R , we have

qfenc = 0. Therefore,

D = 0. For r > R , the left-hand side of Gauss’s law is given by

∮S D⃗ ⋅ da ⃗ = ∮S D da = D ∮S da = D4πr 2. Now, the enclosed free charge is

qfenc = σ 4πR2 . So

D 4πr 2 = σ 4πR2 and

D=

2 σR 2 ⃗ = σR r. ˆ D → r2 r2

Let us consider the electric field. For r < R , we have

E⃗ = 0 Since

D ⃗ = εE ⃗ , for R < r < a , the electric displacement is

D=

σR 2 , r2

so the electric field is

E⃗ =

D⃗ D⃗ D⃗ σR 2 rˆ . = = = ε ε0εr ε0 1 + χe ε0 1 + χe r 2

(

)

Finally, for r > a , the displacement is

D=

σR 2 r2

5-16

(

)

Electromagnetism

so

D⃗ σR 2 ˆ r. = ε0 ε0r 2

E⃗ =

Problem 5.10. For the previous problem, calculate the electric potential everywhere, relative to infinity. Solution The electric field in the three regions is given by

⎧ 0 r a , we have

Gauss’s law states

∮S D⃗ ⋅ da ⃗ = qf

enc

,

where the left-hand side is given by

∮S D⃗ ⋅ da ⃗ = ∮S D da = D ∮S da = D2πsl and the enclosed free charge is a

qfenc =

∫V ρ dτ = k ∫

2π 2

s ds

∫

0

l

dϕ

∫

0

dz =

0

2πka 3l . 3

So

D 2πsl =

2πka 3l 3

And

D= From this, E ⃗ =

D⃗ ε0

ka 3 ka 3 sˆ . → D⃗ = 3s 3s

for s > R ; if we knew P ⃗ , we could find the electric field as well.

5-18

Electromagnetism

For s < a ,

Now, the enclosed free charge is s

qfenc =

∫V ρ dτ = k ∫

2π

s′s′ds′

0

∫

l

dϕ

0

∫ 0

dz =

2πks 3l . 3

Therefore,

D 2πsl =

2πks 3l 3

and

D=

ks 2 ks 2 → D⃗ = sˆ . 3 3

The electric field can be easily obtained,

E⃗ =

D⃗ ks 2 = sˆ . ε0 3ε 0

Problem 5.12. A sphere of radius R carries a polarization P ⃗ = kr 2rˆ, k constant: (a) Calculate the bound charges σb and ρ b. (b) Find the electric field inside and outside the sphere by using the bound charges and Gauss’s law for E.⃗ (c) Calculate the field by using Gauss’s law for D.⃗ Solutions (a) Calculate the bound charges σb and ρ b. The bound charge density is given by

⎡1 ∂ 2 2 ⎤ ρb = −∇ ⋅ P ⃗ = −⎢ 2 r kr ⎥ = −4kr ⎣ r ∂r ⎦

(

)

and the surface charge density is given by

σ b = P ⃗ ⋅  nˆ = kr 2rˆ ⋅ nˆ r=R = kR2 = const.

5-19

Electromagnetism

(b) Find the electric field inside and outside the sphere by using the bound charges and Gauss’s law for E ⃗ . What is the total bound charge (what do you expect it to be)? R

qb = σ b(area) +

∫V ρb dτ = kR 4πR 2

= 4πkR 4 − 4π 4k

r4 4

2

+

2π

∫ ( −4kr)r

2

0

dr

∫

π

dϕ

∫

0

sin θ dθ

0

R

= 4πkR 4 − 4πkR 4 = 0. 0

For r < R , we have Gauss’s law

∮S E ⃗ ⋅ da ⃗ =

qenc ε0

with the left-hand side given by

∮S E ⃗ ⋅ da ⃗ = E 4πr 2 and the enclosed charge r

qenc = ∫ ρb dτ = V

2π

∫(

π

)

−4kr′ r′2dr′ ∫ dϕ ∫ sin θ dθ

0

0

0

4 r

= −4π 4k

r′ 4

= −4πkr 4. 0

Therefore,

E 4πr 2 =

−4πkr 4 ε0

with

E=

kr 2 kr 2 ˆ → E⃗ = r. ε0 ε0

For r > R , we found qenc = 0 so E ⃗ = 0. (c) Calculate the field by using Gauss’s law for D⃗ . To find the electric displacement, we consider Gauss’s law for dielectrics

∮S D⃗ ⋅ da ⃗ = qf 5-20

enc

.

Electromagnetism

However, we have no free charge, qfenc = 0, and D⃗ = 0. Therefore, from

D⃗ = ε0E ⃗ + P ⃗ we have

E⃗ = −

P⃗ . ε0

So

E⃗ = −

kr 2rˆ (r < R ) ε0

and

E ⃗ = 0 (r > R ) as expected. Problem 5.13. A spherical conductor of radius R carries a surface charge density σ . The sphere is surrounded by alinear homogenous dielectric of susceptibility χe . Calculate the energy of this configuration. Solution The energy is given by

W=

ε0 2

∫ εrE 2dτ = 12 ∫ D⃗ ⋅ E ⃗dτ.

It is very easy to obtain D⃗ and E ⃗ . For r < R , we have Gauss’s law

qenc ε0

∮S E ⃗ ⋅ da ⃗ = with qenc = 0 so E ⃗ = 0. Also,

∮S D⃗ ⋅ da ⃗ = qf

enc

=0

So

D⃗ = 0. For r > R , the total enclosed charge is simply

qfenc = σ 4πR2 and

∮S D⃗ ⋅ da ⃗ = D4πr 2. 5-21

Electromagnetism

So

D 4πr 2 = σ 4πR2 With

D=

σR 2 . r2

For R < r < a , the polarization is given by

P ⃗ = ε0 χe E ⃗ so the electric displacement is

D⃗ = ε0E ⃗ + P ⃗ = εE ⃗ . Solving for the electric field, we have

E⃗ =

D⃗ σR 2 σR 2 . = = ε εr 2 ε0 1 + χe r 2

(

)

For r > a , the field is just

E=

σR 2 . ε0r 2

Therefore, our electric displacements are

⎧ 0 rR ⎩ r and our electric fields are

⎧ 0 r 0 indicates a minimum. Dropping all but the sign and the sine, we have

⎡ ⎛ ⎛ π⎞ ⎛ π ⎞⎤ π⎞ β = −sin⎜nπ − ⎟ = −⎢sin(nπ ) cos⎜ − ⎟ + cos(nπ ) sin⎜ − ⎟⎥ = cos(nπ ). ⎝ ⎝ 2⎠ ⎝ 2 ⎠⎦ ⎣ 2⎠ We have β is positive for n = 2, 4, 6, … so the torque is minimized for

⎛ 2n − 1 ⎞ ⎟π , γ=⎜ ⎝ 4 ⎠ when n = 2, 4, 6, … or for any positive integer n ,

⎛ 2(2n ) − 1 ⎞ ⎛ π 1⎞ ⎟π = ⎜n − ⎟π = nπ − . γ=⎜ ⎝ ⎠ ⎝ ⎠ 4 4 4 Since each dipole moment is at an angle γ , the angle between them is π 2γ = 2nπ − 2 Note the multiple of 2nπ is just the addition of another complete circle. The result that minimizes the torque is a γ that causes the dipoles to be perpendicular to each other. Problem 6.4. Consider the rotating cylindrical shell in problem 4.6, where the z-axis starts at the left side of the cylinder. Suppose we place a dipole m⃗ = mzˆ at a distance d from the right-hand side of the cylinder, as depicted below. Find the force on the dipole.

Solution From problem 4.6, the field is given by

B⃗ =

μ0σωR ⎡⎢ d+L – 2 ⎢⎣ R2 + (d + L )2

⎤ ⎥zˆ , R2 + d 2 ⎥⎦ d

where d was the distance from our point to the right-hand side of the cylinder. We can rewrite this considering d = z − L . So

6-6

Electromagnetism

B⃗ =

μ0σωR ⎡⎢ z – 2 ⎢⎣ R2 + z 2

⎤ ⎥zˆ . R2 + (z − L )2 ⎥⎦ z−L

The force is given by

(

F ⃗ = ∇ m⃗ ⋅ B ⃗

)

with

m⃗ ⋅ B ⃗ =

μ0σωRm ⎡⎢ z – 2 ⎢⎣ R + z 2 2

⎤ ⎥. R2 + (z − L )2 ⎥⎦ z−L

So

F⃗ =

μ0σωRm ⎡⎢ z ∇ – ⎢⎣ R2 + z 2 2

⎤ ⎥ 2 2 ⎥ R + (z − L ) ⎦ z−L

and

⎡ μ0σωR3m ⎢ 1 F⃗ = ⎢ 2 2 ⎢⎣ R + z 2

(

⎤ ⎥ − zˆ . ⎡ R2 + (z − L )2 ⎤3/2 ⎥ ⎣ ⎦ ⎥⎦ 1

3/2

)

At a distance d from the right-hand side of the cylinder, z = d + L . Therefore

F⃗ =

⎡ μ0σωR3m ⎢ 1 1 − ⎢ 3/2 ⎡ ⎤ 2 R2 + d 2 ⎢⎣ ⎣ R2 + (d + L )2 ⎦

(

⎤ ⎥ zˆ . 3/2 ⎥ ⎥⎦

)

Problem 6.5. An infinitely long cylinder has a constant magnetization M⃗ parallel to the axis of the cylinder. Find the magnetic field due to M⃗ everywhere.

6-7

Electromagnetism

Solution The magnetization is given by

M⃗ = Mz. ˆ The bound volume current Jb⃗ is

Jb⃗ = ∇ × M⃗ = 0 since M⃗ = constant. The bound surface current Kb⃗ is Kb⃗ = M⃗ × nˆ = Mzˆ × sˆ = Mϕˆ . For s < R

B ⃗ = μ0M⃗ = μ0Mzˆ and for s > R

B ⃗ = 0. Outside, the field is zero (B ⃗ = 0 outside a solenoid). Problem 6.6. A long circular cylinder of radius R has a magnetization M⃗ = ksϕˆ , where k is a constant, s the distance from the axis of the cylinder, and ϕˆ the azimuthal unit vector. Find the magnetic field due to M⃗ for s < R and s > R . Solution Let us first find the bound volume current Jb⃗ and the bound surface current Kb⃗ . The bound volume current is given by 1 ∂ 1 ∂ 1 Jb⃗ = ∇ × M⃗ = sMϕ )zˆ = ks 2 zˆ = 2kszˆ = 2kzˆ = const, ( s ∂s s ∂s s and the bound surface current is given by

( )

(

Kb⃗ = M⃗ × nˆ = ks ϕˆ × sˆ

)

= −kRz. ˆ s=R

So the bound current flows up the cylinder, and returns down the surface. Let us check that the total current is zero. The total current due to the bound volume current is given by R

Itot,Jb =

∫

Jb⃗ ⋅ da ⃗ =

∫ Jb da = ∫

(2k )(2πs ds ) =

0

4πkR2 = 2πkR2 2

and the total current due to the bound surface current is

Itot,Kb =

∫ Kb dl = (−kR)2πR = −2πkR2.

Since they are equal and opposite, the total current is zero. Now we can find the magnetic field using Ampère’s law. For s < R ,

∮S

B ⃗ ⋅ dl⃗ = μ0Ienc.

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Electromagnetism

The left-hand side is given by

∮S

B ⃗ ⋅ dl⃗ = B 2πs

and enclosed current is given by s

Ienc =

∫

s

Jb da =

∫

0

2k 2πs′ds′ = 2kπs 2 .

0

Therefore,

B ⃗ = μ0ksϕˆ = μ0M.⃗ For s > R , our enclosed current is zero, so B ⃗ = 0. Problem 6.7. A long cylinder of radius R carries a magnetization M⃗ = ks 3ϕˆ , where k is a constant. Find the magnetic field due to M⃗ everywhere. Solution Let us start by finding the bound currents. The volume bound current is given by 1 ∂ 1 ∂ 1 ∂ 1 Jb⃗ = ∇ × M⃗ = sMϕ )zˆ = sks 3 zˆ = ks 4 zˆ = 4ks 3zˆ = 4ks 2zˆ ( s ∂s s ∂s s ∂s s and the surface bound current is

(

)

(

Kb⃗ = M⃗ × nˆ = ks 3 ϕˆ × sˆ

( )

)

= −kR3z. ˆ s=R

We can check that the total bound current is zero. From the bound volume current, we have R

Itot,Jb =

∫

Jb⃗ ⋅ da ⃗ =

∫ ( 4ks 2)(2πs ds) = 8π4ks 0

4 R

= 2πkR 4 0

and from the bound surface current, we have

Itot,Kb =

∫ Kb dl = (−kR3)2πR = −2πkR 4.

Therefore, the total bound current is zero, Ib = 0. Now we can find the field by using Ampère’s law. For s < R ,

∮S

B ⃗ ⋅ dl⃗ = μ0Ienc.

∮S

B ⃗ ⋅ dl⃗ = B 2πs

The left-hand side is

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Electromagnetism

and the enclosed current is

Ienc =

s

s

∫

∫

Jb da =

0

s 2

4ks′ 2πs′ds′ =

0

∫

8πk (s′)3ds′ = 2πks 4.

0

Therefore,

B 2πs = μ02πks 4 = μ0ks 3 And

B ⃗ = μ0ks 3ϕˆ = μ0M.⃗ For s > R , we have zero enclosed current. So,

B ⃗ = 0. Problem 6.8. A sphere of radius R carries magnetization M⃗ = krϕˆ . Find the magnetic field inside and outside. Solution Since there is no free current, H⃗ = 0. Inside, we have magnetization, so

1 H⃗ = Bin⃗ − M⃗ . μ0 So Bin⃗ is given by

Bin⃗ = μ0M⃗ = μ0krϕˆ . Outside, we have no magnetization either, so

⃗ = 0. Bout Problem 6.9. An infinitely long wire carries current I and is surrounded by material, out to radius R , with magnetization M⃗ = kϕˆ . Find the magnetic field for s < R and s > R. Solution For s < R, we have

∮S

H⃗ ⋅ dl⃗ = Ifenc

with

∮S

H⃗ ⋅ dl⃗ = H 2πs

and

Ifenc = I.

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Electromagnetism

Therefore,

H=

I ˆ ϕ. 2πs

Using

1 H⃗ = Bin⃗ − M⃗ μ0 the magnetic field inside is given by

⎛ I ⎞ Bin⃗ = μ0 H⃗ + M⃗ = μ0⎜ + k ⎟ϕˆ . ⎝ 2πs ⎠

(

For s > R , we still have H =

I ˆ ϕ, 2πs

)

but we do not have any magnetization. So

μ0I ˆ ϕ, 2πs

⃗ = μ0H⃗ = Bout

which is exactly what we would expect from a wire carrying current I . Problem 6.10. An infinitely long wire, of radius R carries magnetization M⃗ = ks 2zˆ . At s = R , there is a surface current K⃗ = K 0ϕˆ . Find the field for s < R and s > R. Solution For s < R , both M⃗ and K⃗ contribute to the field. We can see the contribution due to K⃗ is that of a solenoid,

BK⃗ = μ0K 0z. ˆ We also have H⃗ = 0, so

BM⃗ = μ0ks 2z. ˆ Therefore,

(

)

Bin⃗ = BK⃗ + BM⃗ = μ0 K 0 + ks 2 z. ˆ For s > R , we have zero magnetization, so

BM⃗ = 0. Also, there is zero field outside of a solenoid, so

BK⃗ = 0. Therefore,

⃗ = 0. Bout

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Electromagnetism

Problem 6.11. Find the H-field produced from a current density Jf⃗ = J0szˆ in two ways. Solution First, we will use

∇ × H⃗ = Jf⃗ . Note, we must have H⃗ = H (s )ϕˆ . So

1 ∂ [sH (s )]zˆ = J0 szˆ ∇ × H⃗ = s ∂s and

∂ [s H (s )] = J0 s 2 . ∂s From this, we have

1 2 J0 s + C . 3

H (s ) =

Since there is zero current at s = 0, we have H (0) = 0 → C = 0. Therefore,

J s2 H⃗ = 0 ϕˆ . 3 Now we will use

∮S

H⃗ ⋅ dl⃗ = Ifenc,

where s

Ifenc =

∫ J da = ∫

2πs′J0 s′ds′ =

0

2πJ0 s 3 3

and

∮S

H⃗ ⋅ dl⃗ = H 2πs .

Therefore,

J s2 H⃗ = 0 ϕˆ . 3 As expected from the first method. Note each equation we used is simply Stoke’s theorem applied to the other. Problem 6.12. This problem was inspired by a different problem presented in the Electrodynamics graduate course by Dr Charles Ebner at the Ohio State University in 2002. A sphere of radius R is uniformly polarized with a polarization P ⃗ . Within 6-12

Electromagnetism

⃗

such a sphere, one can show that D⃗ = 23 P ⃗ and E ⃗ = − 3Pε . By using the similarity of 0

the equations of electrostatics and magnetostatics, find B ⃗ and H⃗ within a uniformly magnetized sphere having magnetism M⃗ . Solution The equivalent equations for electrostatics and magnetostatics are the following

∇ × E⃗ = 0 ∇ ⋅ D⃗ = 0 D⃗ P⃗ E⃗ = − . ε0 ε0 For a system with no free current,

∇ × H⃗ = 0 ∇ ⋅ B⃗ = 0 B⃗ H⃗ = − M⃗ . μ0 Comparing the equations, we note that E ⃗ is equivalent to H⃗ , E ⃗ ⇔ H⃗ ; equivalent to

B ⃗ D⃗ , μ0 ε 0

⇔

B⃗ ; μ0

and

P⃗ ε0

is equivalent to M⃗ ,

Starting with

D⃗ =

2 ⃗ P 3

we can divide both sides by ε0

D⃗ 2 ⃗ = P. ε0 3ε0 Since

P⃗ ε0

⇔ M,⃗ we have

2 ⃗ 2 P ⇔ M⃗ . 3ε0 3 Using

D⃗ ε0

⇔

B⃗ , μ0

we have

B⃗ 2 = M⃗ . μ0 3 Therefore,

B⃗ =

2μ 0 M⃗ . 3

6-13

P⃗ ε0

⇔ M⃗ .

D⃗ ε0

is

Electromagnetism

Now we will look at the electric field,

H⃗ ⇔ E ⃗ . Using

P⃗ ε0

⇔ M,⃗

E⃗ = −

P⃗ M⃗ =− . 3ε0 3

Therefore,

M⃗ H⃗ = − . 3

Bibliography Griffiths D J 1999 Introduction to Electrodynamics 3rd edn (Englewood Cliffs, NJ: Prentice Hall) Griffiths D J 2013 Introduction to Electrodynamics 4th edn (New York: Pearson) Halliday D, Resnick R and Walker J 2010 Fundamentals of Physics 9th edn (New York: Wiley) Halliday D, Resnick R and Walker J 2013 Fundamentals of Physics 10th edn (New York: Wiley) Jackson J D 1998 Classical Electrodynamics 3rd edn (New York: Wiley) Purcell E M and Morin D J 2013 Electricity and Magnetism 3rd edn (Cambridge: Cambridge University Press)

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