Solution Manual for Introduction to Electrodynamics 3rd Edition by David J. Griffiths, All Chapters

INSTRUCTOR'S SOLUTIONS MANUAL
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For


Introduction to Electrodynamics 3rd Edition
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by David J. Griffiths
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All Chapters Included
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, Errata
Instructor’s Solutions Manual
Introduction to Electrodynamics, 3rd ed
Author: David Griffiths
Date: September 1, 2004
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• 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
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• Page 8, Prob. 1.26: last line should read
From Prob. 1.18: ∇ × va = −6xz x̂ + 2z ŷ + 3z 2 ẑ ⇒
∂ ∂ ∂ 2
∇ · (∇ × va ) = ∂x (−6xz) + ∂y (2z) + ∂z (3z ) = −6z + 6z = 0.


• Page 8, Prob. 1.27, in the determinant for ∇×(∇f ), 3rd row, 2nd column:
change y 3 to y 2 .
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• 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 2z 3 ; first line of part (c): insert
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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 ).
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• 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.


• 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 θ̂θ̂.
• Page 17, Prob. 1.55: Solution should read as follows:
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Problem 1.55

(1) x = z = 0; dx = dz = 0; y : 0 → 1. v · dl = (yz 2 ) dy = 0; v · dl = 0.
(2) x = 0; z = 2 − 2y; dz = −2 dy; y : 1 → 0.
v · dl = (yz 2 ) dy + (3y + z) dz = y(2 − 2y)2 dy − (3y + 2 − 2y)2 dy;

0 0
y4 4y 3 y2

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3 2

v · dl = 2 (2y − 4y + 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.
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0 0
z 2 

v · dl = z dz = = −2.
2 2
2
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14 8
Total: v · dl = 0 + 3 −2= 3.

Meanwhile, Stokes’ thereom says v · dl = (∇×v) · da. Here da =
dy dz x̂, so all we need is
∂ ∂
(∇×v)x = ∂y (3y + z) − ∂z (yz 2 ) = 3 − 2yz. Therefore
   1  2−2y
(∇×v) · da = (3 − 2yz) dy dz = 0 (3 − 2yz) dz dy
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0
1 1
1
= 0 3(2 − 2y) − 2y 2 (2 − 2y)2 dy = 0 (−4y 3 + 8y 2 − 10y + 6) dy
1
= −y 4 + 83 y 3 − 5y 2 + 6y 0 = −1 + 38 − 5 + 6 = 83 .

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• Page 18, Prob. 1.56: change (3) and (4) to read as follows:
(3) φ = π2 ; r sin θ = y = 1, so r = 1
sin θ , dr = −1
sin2 θ
cos θ dθ, θ : π
2 → θ0 ≡
tan−1 ( 12 ).

cos2 θ
 
cos θ cos θ sin θ
r cos2 θ (dr) − (r cos θ sin θ)(r dθ) =
 
v · dl = − 2 dθ − dθ
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sin θ sin θ sin2 θ
cos θ cos2 θ + sin2 θ
 3   
cos θ cos θ cos θ
= − 3 + dθ = − 2 dθ = − 3 dθ.
sin θ sin θ sin θ sin θ sin θ
Therefore
θ0 θ
cos θ 1  0 1 1 5 1

v · dl = − dθ = = − = − = 2.
sin3 θ 2 sin2 θ π/2 2 · (1/5) 2 · (1) 2 2
π/2


2

, π
√
5 → 0. v · dl = r cos2 θ (dr) = 54 r dr.
 
(4) θ = θ0 , φ = 2; r:

0 0
4 4 r2  4 5

v · dl = r dr = = − · = −2.
5√ 5 2 5  √ 5 2
5
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Total:

3π

3π
v · dl = 0 + +2−2= 2 .
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• Page 21, Probl 1.61(e), line 2: change = z ẑ to +z ẑ.
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• Page 25, Prob. 2.12: last line should read
Q
Since Qtot = 34 πR3 ρ, E = 1
4π 0 R3 r (as in Prob. 2.8).
• Page 26, Prob. 2.15: last expression in first line of (ii) should be dφ, not
R
d phi.
• Page 28, Prob. 2.21, at the end, insert the following figure
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V(r)
1.6

1.4

1.2

1

0.8

0.6

0.4
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0.2

0.5 1 1.5 2 2.5 3
r


q
In the figure, r is in units of R, and V (r) is in units of 4π 0R
.
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• 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:
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q2
 
1 1 1
F= − x̂ − ŷ + √ [cos θ x̂ + sin θ ŷ] ,
4π0 (2a)2 (2b)2 (2 a2 + b2 )2
 
where cos θ = a/ a2 + b2 , sin θ = b/ a2 + b2 .

q2
    
a 1 b 1
F= − 2 x̂ + − 2 ŷ .
16π0 (a2 + b2 )3/2 a (a2 + b2 )3/2 b


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