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: ∇ × v a = −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 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 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 .


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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 ys h v dl = ∫ ( v) da. here da =
· ∇× ·
dy dz xˆ , so ∂a ll we need is ∂
(∇ ×v)x = (3 ∂yy + z) − (yz 2 )=3 − 2yz. therefore
∂z
∫ ∫ ∫ n ,
(∇×v) · da = (3 − 2yz) dy dz = ∫ 1 ∫ 2−2y(3 − 2yz) dz dy
0 0 ∫1
= ∫ 1 3(2 − 2y) − 2y 1 (2 − 2y)2 dy = (−4y3 + 8y2 − 10y +6) dy
0 2 1 8 0

= −y 4 + 83 y3
— 5 y2 + 6y 0= —1 + 3
8
− 5 + 6 = 3 .c
• page 18,π prob. 1.56: change (3) and (4) to read as follows:
(3) φ = ; r sin θ = y = 1, so r = 1 , dr = −1 cos θdθ, 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
− dθ
·d r θ dr — r θ θ r dθ sin θ
2 dθ − sin2 θ
sin θ
cos3 θ cos θ cos θ cos2 θ +sin 2 θ cos θ
= + dθ = − =
— sin θ sin θ 2 dθ − dθ.
sin 3 θ sin θ sin 3 θ
therefore
∫ ∫ θ0 θ0 5 1
1 1 1
v · dl = − cos θ dθ = − = − = 2.
=
sin3 θ 2 sin2 θ π/2 2 · (1/5) 2 · (1) 2 2
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 ŷ ] ,
0 2 2
(2a) (2b) (2 a + b2 )2
√ √
where cos θ = a/ a 2 + b 2 , sin θ = b/ a 2 + b 2 .

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

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