Solution Manual For Introduction to Electrodynamics, 5th Edition by David J. Griffiths. All Chapters 1-12

Solutions Manual
Introduction To Electrodynamics
Author: David J. Grif

5th Edition

, Table Of Content

Chapter 1 Vector Analysis

Chapter 2 Electrostatics

Chapter 3 Special Techniques

Chapter 4 Electrostatic Fields in Matter

Chapter 5 Magnetostatics

Chapter 6 Magnetostatic Fields in Matter

Chapter 7 Electrod ynamics

Chapter 8 Conservation Laws

Chapter 9 Electromagnetic Waves

Chapter 10 Potentials and Fields

Chapter 11 Radiation

Chapter 12 Electrodynamics and Relativity

, Errata
Instructor’s Solutions Manual
Introduction To Electrodynamics, 5th Ed
Author: David Griffiths
Date: September 1, 2004


• Page 4, Prob. 1.15 (B): Last Expression Should Read Y + 2z + 3x.
• Page 4, Prob.1.16: At The Beginning, Insert The Following Figure




• Page 8, Prob. 1.26: Last Line Should Read
From Prob. 1.18: ∇ × Va = −6xz X̂ + 2z Ŷ + 3z2 ẑ ⇒
∇ · (∇ × Va) = ∂x∂ (−6xz) + ∂y (2z)
∂
+ ∂
∂z (3z2) = −6z + 6z = 0. C
• Page 8, Prob. 1.27, In The Determinant For
∇× (∇F ), 3rd Row, 2nd
Column: Change Y3 To Y2.

• Page 8, Prob. 1.29, Line 2: The Number In The Box Should Be -12
(Insert Minus Sign).

• Page 9, Prob. 1.31, Line 2: Change 2x3 To 2z3; First Line Of Part (C):
Insert Comma Between Dx And Dz.
• Page 12, Probl 1.39, Line 5: Remove Comma After Cos Θ.
• Page 13, Prob. 1.42(C), Last Line: Insert ẑ After ).
• Page 14, Prob. 1.46(B): Change R· To A.

• Page 14, Prob. 1.48, Second Line Of J: Change The Upper Limit On The R
Integral From ∞ To R. Fix The Last Line To Read:
R
= 4π −E−R 0
+ 4πe−R = 4π −E−R + E−0 + 4πe−R = 4π. C

• Page 15, Prob. 1.49(A), Line 3: In The Box, Change X2 To X3.



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, • Page 15, Prob. 1.49(B), Last Integration “Constant” Should Be L(X, Z), Not
L(X, Y).

• Page 17, Prob. 1.53, First Expression In (4): Insert Θ, So Da = R Sin Θ Dr Dφ Θˆ.
• Page 17, Prob. 1.55: Solution Should Read As Follows:
Problem 1.55
∫
(1) X = Z = 0; Dx = Dz = 0; Y : 0 → 1. V · Dl = (Yz2) Dy = 0; V · Dl = 0.
(2) X = 0; Z = 2 − 2y; Dz = −2 Dy; Y : 1 → 0.
V · Dl = (Yz2) Dy + (3y + Z) Dz = Y(2 − 2y)2 Dy − (3y + 2 − 2y)2 Dy;

∫ ∫0 0
Y4 4y3 Y2 14
v · dl = 2 (2y3 − 4y2 + y − 2) dy = 2 − + − 2y = .
2 3 2 1 3
1


(3) X = Y = 0; Dx = Dy = 0; Z : 2 → 0. V · Dl = (3y + Z) Dz = Z Dz.

∫ ∫0 2 0
z
V · Dl = Z Dz = = −2.
2 2
2


H
Total: V · Dl = 0 + 314 − 2 =
Meanwhile, Stokes’ Thereom Says H V · Dl = ∫ (∇×V) · Da. Here Da =
Dy Dz X̂ , So All We Need Is
(∇×V)X =∂y∂ (3y + Z) −∂z∂ (Yz2) = 3 − 2yz. Therefore
∫ ∫∫ ∫ 1 n∫ 2−2y ,
(∇×v) · da = ∫ (3 − 2yz) dy dz = 0 0 (3 − 2yz)
∫1 dz dy
1
= 0
3(2 − 2y) − 2y 1
2
(2 − 2y) 2 dy =
0
(−4y 3
+ 8y2 − 10y + 6) dy
1
= −y 4 + 83 y3 — 5y 2 + 6y 0 = — 1 + 83 − 5 + 6 = 38 . C
• Page 18, Prob. 1.56: Change (3) And (4) To Read As Follows:
(3) Φ = Π ; R Sin Θ = Y = 1, So R = 1 , Dr = −1 Cos Θ Dθ, Θ : Π
→Θ ≡
2 Sin Θ Sin2 Θ 2 0
Tan−1( 1 2 ).

Cos
Cos2 Cos Θ Sin Θ
V · DL = Cos Θ (Dr) — Cos Sin Θ)(R ) = Dθ − Sin2 Θ Dθ —
Θ Θ
R 2
(R Θ Dθ Sin Θ
Sin2
Θ
Cos3 Θ Cos Θ Cos Θ Cos2 Θ + Sin2 Θ Cos Θ
= — + Dθ= − Dθ = − Dθ.
Sin3 Sin Sin Θ Sin2 Θ Sin3 Θ
Θ Θ
Therefore
∫ ∫Θ0 1 Θ0
1 1 5 1
v · dl = − = − = − = 2.
Cos 2 · (1/5) 2 · (1) 2 2
dθ = 2 sin θ
2 π/2
π/2 sin
Θ 3θ


2