Errata Instructors Solutions Manual Introduction to Electrodynamics, 3rd ed Author: David Griths 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 gure
Page
Chapter
12
Electrodynamics
and Relativity
Problem 12.1 Let u be the velocity of a particle in S, u its velocity in 5, and v the velocity of 5 with respect to S. Galileo's velocity addition rule says that u = u + v. For a free particle, u is constant (that
Chapter
11
Radiation
Problem 11.1 From Eq. 11.17, A =  ILo7r r sin[UJ(t POUJ 4 V.A
~

rlc)](cos(Jr
 sin B 8), so
1
= 
ILoPoUJ I
= 
[ ] I . UJr 2 sin B cos B . ILoPoUJ 47r cfw_ r 2" sm[UJ(t riG)]  c cos[UJ(t riG)] coBB r 2'sm () sm[UJ(t riG)] 
47r
Chapter 10
Potentials and Fields
Problem 10.1 2
2
oL
2
2
02V
J.LOfO
0
02V
2
0
1
fO
0 V + 7ft = 'VV 0 AVL Problem 10.2
Ot2 + ot (V .A) + J.LOfoOt2 = 'V V + ot (V . A) = V
p. ,(
02 A
= 'V AJ.LOfO Ot2
(
OV
V.A+J.LOfcfw_jt
)
=J.LOJ.,(
(foE2 + :0 B2) dr. A
Chapter 8
Conservation
Problem 8.1 Example 7.13.
Laws
E  8 1 A 211"100 8
B
~
Pol!.
211" 8 cp
s =~ (E x B) =~ 1 . 411"210082 Zj
>.I
p=
I
E
S. da=
b
I
a
b
S211"8d8_ =
b
211"100
I
a
b
I d8= _>.I In(b/a). 211"100 8
ButV= Problem 7.58.
/
a
E.dl=
A 211"10
Chapter 7
Electrodynamics
Problem 7.1 (a) Let Q be the charge on the inner shell. Then E = 4;EO in the spacebetweenthem, and (Va Vb)= ~f  Jb dr  41rEO Q Jb J dr  .sL(1  1 ) faE. L fa r~  41rEO a b'
1=
I .da = I
J
u
E. da =u Q
100
= u 47ro(Va100 (I/a
Chapter 6
Magnetostatic
Problem 6.1 N
Fields in Matter
= m2
X
AA B 1; B 1 = J.Lo1 [3(mi. r ) r  ml ] ; r "3 41rr
_
A
= y; ml = mlZ;
A
A
m2
= m2Y'
A
B
1
= 41rr 3z,
J.Lo ml
A
N =J.Lomlm2 (yxz ) ~ 47rr
A A
= 4 7rr ~x.
J.Lomlm2A
Hereml
=1ra2 1 , m2 = b2 I
Chapter 5
Magnetostatics
Problem 5.1 terms of a and d, use the pythagorean theorem:
Sincev x B points upward, and that is also the direction of the force, q must be positive. To find R, in
I
I
(R  d)2 + a2 = R2 =? R2 The cyclotron formula then giyes

2R
Chapter 4
Electrostatic Fields in Matter
Problem 4.1
E =V/x = 500/103 = 5x 105. Table 4.1: a/47r0 = 0.66x 1030, so a = 47r(8.85x 1012)(0.66x 1030) = 7.34X1041. p =aE =ed ~ d =aE/e = (7.34x 1041)(5 x 105)/(1.6 x 1019) =2.29 X 1016 m.
d/R
=(2.29 x 1
Chapter
3
Special Techniques
Problem . 3.1 The argument is exactly the same as in Sect. 3.1.4, except that since z < R, vz2 + R2  2zR = (R  z), q 1 q.
Insteadof (z  R). HenceVave _7ro2z R [(z+ R) (R z)]= _7roR ' =4 4
charges is Vcenten Vave Vcenter b
Chapter 2
Electrostatics
Problem (a)
I Zero.!
2.1
1 q~, where r is the distance from center to each numeral. F points toward the missing q. Explanation: by superposition, this is equivalent to (a), with an extra q at 6 o'clocksince the force of all twel