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

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

, Stuvia.com - The Marketplace to Buy and Sell your Study Material




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


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




• Page 8, Prob. 1.26: Last Line Should Read
From Prob. 1.18: ∇ × Va = −6xz Xˆ + 2z Yˆ + 3z2 Zˆ⇒
∇ · (∇ × V A )= ∂ (−∂X6xz)+ ∂ (2 Z )∂+Y ∂
(3 Z2 )∂=Z −6z + 6z = 0. C

• Page 8, Prob. 1.27, In The Determinant ∇× ( ∇F ), 3rd Row, 2nd Column:
For Change Y3 To Y2 .

• Page 8, Prob. 1.29, Line 2: The Number In The Box Should Be -12 (Insert
Minus Sign).
3 3
• Page 9, Prob. 1.31, Line 2: Change 2x To 2z ; 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 Zˆa Fter ).
• 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 + 4πe−R = 4π −E−R + E−0 + 4πe−R = 4π. C
0


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


1

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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 Θd r Dφθˆ.
• 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 +(3 Y + Z) Dz = Y(2 − 2y)2 Dy − (3y +2 − 2y)2 Dy;
∫ ∫0 3
Y4 4y Y2 0
14
V · Dl =2 (2y3 − 4y2 + Y − 2) Dy = 2 − 3 + − 2y 1 = .
2 2 3
1


(3) X = Y = 0; Dx = Dy = 0; Z : 2 → 0. V · Dl = (3y + Z) Dz = Z Dz.
∫ ∫0 0
Z2 = −2.
V · Dl = Z Dz
2
= 2 2



Total: H V · Dl =0 + 14
. −2=
3
3
Mea Nwhile , Stokes’ Thereom Sa H V Dl ∫ ( V) Da. Here Da =
· ∇× ·
Ys
Dy Dz Xˆ , So∂ A Ll We Need Is =
(∇ ×V)X = (3 ∂Yy + Z) − ∂ (Yz 2 )=3 − 2yz. Therefore
∂Z
∫ ∫ ∫ ∫ ,
(∇×V) · Da = (3 − 2yz) Dy Dz = 1 2−2y(3 − 2yz) Dz Dy
N∫
∫ ∫1
0 0
= 1 3(2 − 2y) − 2y 1 (2 − 2y)2 Dy (−4y3 + 8y2 − 10y +6) Dy
=
0 2 1 8 0

= −Y 4 + 83 Y 3 — 5 y2 + 6y 0= —1 + 3
8
− 5 + 6 = 3 .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 Θdθ, Θ : Π
→Θ ≡
2 Sin Θ 2
Sin Θ 2 0
−1 1
Tan ( 2 ).

V L = Cos2 ( ) ( Cos Sin )( )= Cos2 Θ Cos Θ Cos Θ Sin Θ
− Dθ
·D R Θ Dr — R Θ Θ R Sin 2 Dθ − Sin 2 Θ
Sin
Dθ Θ Θ
Cos3 Θ Cos Θ Cos Θ Cos2 Θ +Sin 2 Θ Cos Θ
= + Dθ = − =
— Sin Θ Sin Θ Dθ − 3 Dθ.
Sin 3 Sin 2 Sin Θ
Θ Θ
Therefore
∫ ∫ Θ0 Θ0 5 1
1 1 1
V · Dl = − Cos Dθ = − = − = 2.
=
Θ
Sin 3 2 Sin2 Θ 2 · (1/5) 2 · (1) 2 2
Π/2
Θ Π/2



2

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√
(4) Θ = Θ0 , Φ = 2π ; R : 5 → 0.V · Dl = R Cos2 Θ (Dr)= 45r Dr.
0
∫ ∫0 4 R2 4 5
4
V · Dl = R Dr =− · = −2.
5 = 5 2 √
5 5 2
√
5


Total:
I 3π
V · Dl =0 +2 − 2= .
+ 2

• Page 21, Probl 1.61(E), Line 2: Change = Zzˆ To +Z Zˆ.
• Page 25, Prob. 2.12: Last Line Should Read
Q
Since Q = 4 Πr3 ρ, E = 1 R (As In Prob. 2.8).
Tot 3 4πα 0 R 3

• Page 26, Prob. 2.15: Last Expression In First Line Of (Ii) Should Be
Dφ, Not D Phi.
• Page 28, Prob. 2.21, At The End, Insert The Following Figure
V(R)




R
0.5 1 1.5 2 2.5 3


Q
In The Figure, R Is In Units Of R, And V (R) Is In Units Of .
4πα0 r

• Page 30, Prob. 2.28: Remove Right Angle Sign In The Figure.
• Page 42, Prob. 3.5: Subscript On V In Last Integralshould Be 3, Not 2.
• Page 45, Prob. 3.10: After The First Box, Add:

Q2 1 1 1
F = 4πξ − Xˆ − √2 Yˆ +
[Cos Θ Xˆ +Sin Θ Yˆ ] ,
0 2 2
(2a) (2b) (2 A + B2 )2
√ √
Where Cos Θ = A/ A2 + B2 , Sin Θ = B/ A2 + B2 .

a 1 b 1
− −
(a2 + b2 )3/2 a2 (a2 + b2 )3/2 b2

3