Instructor’s Solution Manual Introduction to Electrodynamics Fourth Edition

Instructor’s Solution Manual
Introduction to Electrodynamics
Fourth Edition

David J. Griffiths

2014

,Contents

1 Vector Analysis 4

2 Electrostatics 26

3 Potential 53

4 Electric Fields in Matter 92

5 Magnetostatics 110

6 Magnetic Fields in Matter 133

7 Electrodynamics 145

8 Conservation Laws 168

9 Electromagnetic Waves 185

10 Potentials and Fields 210

11 Radiation 231

12 Electrodynamics and Relativity 262




c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.

,Preface

Although I wrote these solutions, much of the typesetting was done by Jonah Gollub, Christopher Lee, and
James Terwilliger (any mistakes are, of course, entirely their fault). Chris also did many of the figures, and I
would like to thank him particularly for all his help. If you find errors, please let me know ().


David Griffiths




c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is
protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.

, CHAPTER 1. VECTOR ANALYSIS 3



Chapter 1
Chapter 1
Chapter 1
VectorChapter
Analysis
1
Vector Analysis
Vector Analysis
Vector Analysis
roblem 1.1
✒
✣



}} }
Problem 1.1
) From the diagram,
Problem |B1.1
+ C| cos θ3 = |B| cos θ1 + |C| cos θ2 . Multiply by |A|. ✒
✣
Problem 1.1
|A||B + C| cos θ(a) = |A||B|
thecos θ1 + |A||C| cos θ2θ.3 = |B| cos θ1 + |C| cos θ2 . ✒
✣




C
3 From diagram, |B + C| cos |C| sin θ2
So:(a)A·(B
From the diagram, + C| cos 3 = |B| cos ✓1θ ++|C| cos ✓2cos
. Multiply by |A|.




C
+
|B ✓=
+ C) = A·B
|A||B ++ A·C.
cos (Dot
3 + C| cos θ3 = is
product
|A||B| cos 1 distributive)
cos|A||C| θ2θ.2 .




C
C| θ
(a) From the diagram, |B |B| θ1 + |C| cos |C| sin θ2




B
|A||B + C| cos =A·(B θ3cos + |A||C| cos 2 . cosproduct




C
✓3 + |A||B| ✓=1 A·B ✓(Dot




+
So:
|A||B cos+ C)= |A||B| cos+θ1A·C.
+ |A||C| θ2 . is distributive)




C
C| |C| sin θ2




B
θ2
So: A·(B|B ++C| C) = A·B + A·C. (Dot product 2 . distributive)
is




C
Similarly: sin = |B| sin + |C| sin Mulitply by |A| n̂.




+
θ θ θ
+ |C| sin θ2 . Mulitply by |A| n̂.B 3 ✯
So: A·(B3 + C) = A·B 1+ A·C. (Dot product is distributive)




B
θ θ2
|A||B + C| sin θ3 n̂Similarly:
Similarly: |B
If n̂ is|A||B
the unit vector
= |A||B||B
+|A||B
C| sin+
Similarly: ✓C|
|B +=
sin+θ1C|n̂sin
C||B|
sinsin
θ ✓
=
θ3 = |B|sin
+ |A||C|
+
|B| |C|
sin sin
θ +
sinθθ1n̂.
✓ .
|C| 2
Mulitply
sin θ
3 sin θ3 n̂ =1 |A||B| sin θ21 n̂ + |A||C| sin θ2 n̂.
pointing out
3
of the θpage,
1
it✓ follows
2 . by
Mulitply
that
|A| n̂.
by |A| n̂. θ1 θ3 ✯
θ2
B ✲ }
θ3 ✯ |B| sin θ1
|B| sin θ1}
A×(B If +
+ C| sin
C)the
n̂ is
|A||B
= unit
(A×B)
✓3 n̂
If =
n̂+is
C|the
|A||B|
is the+unit
If n̂ vector
3 n̂sin
sin θunit
(A×C).
✓1 n̂ +pointing
=vector
|A||B|
vector out
pointing (Cross
pointing
sin
of the
1 n̂ +
|A||C|
product
out
sin
|A||C|
out 2ofn̂.sin
is
of theit page,
page,
theθ2page,
n̂.
distributive)
follows it follows
it follows that! "#θ1B $ !! "#
that that |B| cos θ
θ1
|C| cos
}
"# θ $ !✲ "# $ ✲
$|B| sinAθ1
A
A×(B + C) = (A×B) + (A×C). (Cross product is distributive)! "# 1 $ ! |B|"#cos θ2$1 |C| A cos θ2
A×(B + C) = (A×B) + (A×C). (Cross product is distributive)
) For the A⇥(B + C)case,
general = (A⇥B) see G. + E.
(A⇥C).
Hay’s (Cross
Vector product
and is distributive)
Tensor Analysis, Chapter
|B| cos θ1 |C| cos θ2
1, Section 71,(dot product) and
(b)(b)
ForFor
the the general
general case,case,
see G.see E. G.
Hay’sE. Hay’s
Vector Vector
and Tensor and Analysis,
Tensor Chapter 1,Chapter
Analysis, Section
Section 7 (dot 7 (dot
product) and product) and
Section
(b) For8 the(cross product)
general case,
Section
Section see G.
8 (cross
8 (cross E. product)
Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and
product)
Section
roblem 1.2 Problem 8 (cross
Problem product)
1.2 1.2
Problem 1.2 CC ✻ C
TheThe
The triple cross-product triple cross-product
triple cross-product
is not is not isinassociative.
in general general
not associative.
in general For example,
associative.
For example, For example, ✻ ✻
The triple cross-product
suppose
suppose A = is
B not
and in
C general
is
= B and C istoperpendicularassociative.
perpendicular to A, For
as example,
in the
to diagram. diagram.
A, as in the diagram.
suppose A = B and C isAperpendicular A, as in the ✲
suppose A Then
Then (B×C) points
= Then and(B×C)
B (B×C) is perpendicular
C points out-of-the-page,
points to A,
out-of-the-page, andasA×(B×C)
in
and theA×(B×C)
diagram.
points down,points down, ✲AA==BB ✲ A = B
Then (B⇥C) hasout-of-the-page,
andpoints magnitude ABC. and
out-of-the-page, Butand
A×(B×C)
A⇥(B⇥C)
(A×B) = 0, sopointspoints
(A×B)×C down,
down, = 0 ̸= ❂
and has and has magnitude ABC. But (A×B) = 0, so=(A×B)×C =B×C
❂
0 ̸= ❂
and magnitude
has magnitude ABC.
A×(B×C).
A×(B×C). ABC. But But(A×B)
(A⇥B)==0,0,so so (A×B)×C
(A⇥B)⇥C = 00 6≠= ❄A×(B×C)
B×C ❄ A×(B×C)
A×(B×C).A⇥(B⇥C).
B×C ❄ A×(B×C)
Problem 1.3 z
roblem 1.3 1.3 Problem 1.3
Problem z✻✻ z✻
√
A = +1 x̂ + 1√ ŷ −p1 ẑ; A = 3; B√= 1 x̂ + 1 ŷ + 1 ẑ; p
√
A =x̂ +1
A = +1 + 1x̂ŷ+−1 ŷ1Aẑ;=1A+1
ẑ;=A =3;1 ŷB3; B=1 =1 x̂1Ax̂
+=+1 1ŷ3; +
B131√
ŷ+√ ẑ;ẑ;1BB = 1 ŷ3.
x̂ = 3.+ 1 ẑ; ✣B
A·B = +1 +x̂1+ − 1 =− 1= ẑ;
ABpcos pθ = = 3 cos +
θ ⇒ cos θ.
√ √ √ √ 11 ✣ θB ✣
✲By
A·B =A·B+1=++1 1− + 11A·B
=
θ = cos
11−1
==
= 1%+1
=
& AB
1 +cos
AB 1 −cos
θ 1=✓=◦=1 3=3AB3 3cos
cos
cos
θ ✓θ⇒)
= cos
cos 3 θ =cos
✓ 3= 3 ..θ ⇒ cos θ.
3 ≈ 70.5288
θ θ
% &
3
❲ ✲y ✲y
✓ =−1cos1 1 3 ⇡
% & 1 θ= cos◦−1 13 ≈ 70.5288◦
70.5288 ✰ A
θ = cos 3 ≈ 70.5288
x
❲ ❲
✰ A✰ A
Problem 1.4 x x
Problem 1.4 The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,
roblem 1.4 Problem 1.4
we might pick the base (A) and the left side (B):
The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,
The cross-product The
of any two vectorsofin
cross-product anythe
twoplane willingive
vectors a vector
the plane willperpendicular to the plane.
give a vector perpendicular to For example,
the plane. For example,
we might pick the base (A) and the left side (B):
might pick the we Pearson
c base
⃝2005 might pick
and the
(A)Education, base
theInc.,
left (A)
side
Upper and
(B):
Saddle the NJ.
River, leftAllside (B):
rights reserved. This material is
protected under all copyright laws as they currently exist. No portion of this material may be
A= 1 x̂reproduced,
+ 2 ŷ +in0 any
ẑ; B =or by
form 1 x̂any+means,
0 ŷ +without
3 ẑ. permission in writing from the publisher.
A = −1 x̂ + 2 ŷ⃝2005
+ 0 ẑ;Pearson
c
B = Education,
−1 x̂ + 0Inc.,
ŷ +Upper
3 ẑ. Saddle River, NJ. All rights reserved. This material is
protected under all copyright laws as they currently exist. No portion of this material may be
c 2012
reproduced, in any form Pearson
or by Education,
any means, Inc.,
without Upper Saddle
permission River,from
in writing NJ. the
All rights reserved. This material is
publisher.
005 Pearson Education, Inc., Upper Saddleprotected under
River, NJ.all
Allcopyright laws as they
rights reserved. Thiscurrently exist.
material is No portion of this material may be
tected under all copyright laws as theyreproduced, in anyNo
currently exist. form or by any
portion means,
of this without
material permission
may be in writing from the publisher.
roduced, in any form or by any means, without permission in writing from the publisher.