Solutions Manual – Introduction to Electrodynamics, 3rd Edition by Griffiths | All 12 Chapters Covered

INSTRUCTOR'S
SOLUTIONS
MANUAL


INTRODUCTION to
ELECTRODYNAMICS
Third Edition




David J. Griffiths

, Errata
Instructor’s Solutions Manual
Introduction to Electrodynamics, 3rd 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) = ∂ (−6xz)
∂x + ∂y(2z)
∂ + ∂z(3z2)
∂ = −6z + 6z = 0. X
• 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. X

• Page 15, Prob. 1.49(a), line 3: in the box, change x2 to x3.



1

,• 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
R
(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;

Z Z0 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.

Z Z0 0
z2
v · dl = z dz = = −2.
2 2
2

H
Total: v · dl = 0 + 14
3
−2= 3
.
H R
Meanwhile, Stokes’ thereom says v · dl = ( ∇× v) ·da. Here da =
dy dz x̂ , so all we need is
(∇×v)x = ∂y∂ (3y + z) − ∂ (yz2) = 3 − 2yz. Therefore
∂z
R RR R 1 nR 2−2y o
(∇×v) · da = (3 − 2yz) dy dz = 0 0 (3 − 2yz) dz dy
R1 R1
= 0
3(2 − 2y) − 2y 12 (2 − 2y)2 dy = 0 (−4y3 + 8y2 − 10y + 6) dy
1
= −y 4 + 83 y3 — 5y2 + 6y 0
= —1 + 8
3 − 5 + 6 = 83 . X

• 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( 12 ).
v l = cos2 ( ) ( cos sin )( ) = cos2 θ cos θ cos θ sin θ
·d r θ dr — r θ θ r dθ − dθ − dθ
sin θ sin2 θ sin2 θ
cos3 θ cos θ cos θ cos2 θ + sin2 θ cos θ
= — + dθ = − dθ = − dθ.
sin3 θ sin θ sin θ sin2 θ sin3 θ
Therefore
Z Zθ0 1 θ0 1
cos θ 1 5 1
v · dl = − dθ = = − = − = 2.
sin3 θ 2 sin2 θ π/2 2 · (1/5) 2 · (1) 2 2
π/2


2

, √
(4) θ = θ0, φ = π ; r : 5 → 0. v · dl = r cos2 θ (dr) = 4 r dr.
2 5

Z Z0 0
4 4 r2 4 5
v · dl = r dr = =− · = −2.
5 52 √
5
5 2
√
5


Total:
I
3π 3π
v · dl = 0 + +2−2= 2 .
2
• Page 21, Probl 1.61(e), line 2: change = z ˆz to +z ẑ .
• Page 25, Prob. 2.12: last line should read
Since Q = 4 πR3ρ, E = 1 Q r (as in Prob. 2.8).
tot 3 4πα0 R3

• 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



In the figure, r is in units of R, and V (r) is in units of 4παq R .
0

• Page 30, Prob. 2.28: remove right angle sign in the figure.
• Page 42, Prob. 3.5: subscript on V in last integral should be 3, not 2.
• Page 45, Prob. 3.10: after the first box, add:


q2 1 1 1
F= 2
−
x̂ − 2
ŷ + √ 2 [cos θ x̂ + sin θ ŷ ] ,
4πξ 0(2a) (2b) (2 a + b2)2
p p
where cos θ = a/ a2 + b2, sin θ = b/ a2 + b2.

q2 a 1 b 1
F= − x̂ + − ŷ .
16πξ0 (a2 + b2)3/2 a2 (a2 + b2)3/2 b2

3