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

Solution Manual For Electrodynamics 5th
Edition By David J. Griffiths Latest Edition
2025/2026 A+

, errata instructor’s
solutions manual
introductio n to electrody na mi cs,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 following figure




• page 8, prob. 1.26: last line should read
from prob. 1.18: ∇ × v a = −6xz x̂ + 2z ŷ + 3z2 ˆz ⇒
∂ ∂ ∂
∇ · (∇ × v a )= (−6xz)+
∂x (2 z)+
∂y (3 z 2 )=
∂z −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).
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 ˆz 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−r0 + 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 .


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
∫
(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 0
y4 4y3 y2 14
v · dl =2 (2y3 − 4y2 + y − 2) dy = 2 − 3 + − 2y = .
2 2 1 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 ys H v dl = ∫( v) da. here da =
· ∇× ·
dy dz x̂ , so a ll we need is
(∇ ×v)x = ∂ (3y ∂y + z) −
∂
(yz∂z2 )=3 − 2yz. therefore
∫ ∫∫ ∫ 1 n∫ 2−2y ,
(∇×v) · da = (3 − 2yz) dy dz = 0 0 (3 − 2yz) dz dy
∫1 ∫1
0 3(2 − 2y) − 2y 2(2 − 2y) dy = 0 (−4y3 + 8y2 − 10y +6) dy
2
= 1
1 8
−y 4 + 3 y3 — 5 y2 + 6y 0 = — 1 + 3 − 5 + 6 = 3 .c
8 8
=

• 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 2
sin θ 2 0
tan −1 ( 1 2).
v l = cos2 ( ) ( cos sin )( )= cos2 θ cos θ cos θ sin θ
·d r θ dr — r θ θ r dθ − dθ
sin θ sin2 θ dθ − sin2 θ
cos3 θ cos θ cos θ cos2 θ +sin 2 θ cos θ
= + dθ = − dθ = − dθ.
— sin θ sin θ 2
3
sin θ sin θ sin 3 θ
therefore
∫ ∫θ0 θ0
cos θ 1 1 1 5 1
v · dl = − dθ = = − = − = 2.
sin 3 θ 2 sin 2 θ π/2 2 · (1/5) 2 · (1) 2 2
π/2


2

, √
(4) θ = θ0 , φ = 2π ; r : 5 → 0. v · dl = r cos2 θ (dr)= 45 r dr.
∫ ∫0 0
4 4 r2 4 5
v · dl = r dr = = − · = −2.
5 5 2 √ 5 2
√ 5
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
q
read
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 integral should be 3, not 2.
• page 45, prob. 3.10: after the first box, add:


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

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

3