Solutions Manual – Introduction to Electrodynamics, 5th Edition by Griffiths | All 12 Chapters Covered

SOLUTIONS

,Contents

1 Vector Analysis 4

2 Electrostatics 26

3 Potential 53

4 Electric Fields in Matter 92

5 Magnetostatics 110

6 Magnetic Fields in Matter 133

7 Electrodynamics 145

8 Conservation Laws 168

9 Electromagnetic Waves 185

10 Potentials and Fields 210

11 Radiation 231

12 Electrodynamics and Relativity 262

,Chapter 1

Vector Analysis

Problem 1.1
"
$
(a) From the diagram, |B + C| cos ✓3 = |B| cos ✓1 + |C| cos ✓2 . Multiply by |A|.
|A||B + C| cos ✓3 = |A||B| cos ✓1 + |A||C| cos ✓2 .
}

C
|C| sin θ2




C
+
So: A·(B + C) = A·B + A·C. (Dot product is distributive)




B
θ2
Similarly: |B + C| sin ✓3 = |B| sin ✓1 + |C| sin ✓2 . Mulitply by |A| n̂. θ3 #
|A||B + C| sin ✓3 n̂ = |A||B| sin ✓1 n̂ + |A||C| sin ✓2 n̂.
If n̂ is the unit vector pointing out of the page, it follows that !
B
θ1
"# $! "# $
}! |B| sin θ1

A
|B| cos θ1 |C| cos θ2
A⇥(B + C) = (A⇥B) + (A⇥C). (Cross product is distributive)
(b) For the general case, see G. E. Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and
Section 8 (cross product)
Problem 1.2 C
, %
The triple cross-product is not in general associative. For example,
.
suppose A = B and C is perpendicular to A, as in the diagram. !A=B
Then (B⇥C) points out-of-the-page, and A⇥(B⇥C) points down,
and has magnitude ABC. But (A⇥B) = 0, so (A⇥B)⇥C = 0 6= &
A⇥(B⇥C).
B×C '
A×(B×C)

Problem 1.3 z%
p p
A = +1 x̂ + 1 ŷ 1 ẑ; A = 3; B = 1 x̂ + 1 ŷ + 1 ẑ; B = 3.
p p !B
A·B = +1 + 1 1 = 1 = AB cos ✓ = 3 3 cos ✓ ) cos ✓ = 31 . . θ
!y
1 1
✓ = cos 3 ⇡ 70.5288 "
( A
x
Problem 1.4
The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,
we might pick the base (A) and the left side (B):
A= 1 x̂ + 2 ŷ + 0 ẑ; B = 1 x̂ + 0 ŷ + 3 ẑ.

, x̂ ŷ ẑ
1 2 0 = 6 x̂ + 3 ŷ + 2 ẑ.
A⇥B =
1 0 3
This has the right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by its
length:
p A⇥B 6 3 2
|A⇥B| = 36 + 9 + 4 = 7. n̂ = |A⇥B| = 7 x̂ + 7 ŷ + 7 ẑ .
Problem 1.5
x̂ ŷ ẑ
A⇥(B⇥C) = Ax Ay Az
(By Cz Bz Cy ) (Bz Cx Bx Cz ) (Bx Cy By Cx )
= x̂[Ay (Bx Cy By Cx ) Az (Bz Cx Bx Cz )] + ŷ() + ẑ()
(I’ll just check the x-component; the others go the same way)
= x̂(Ay Bx Cy Ay By Cx Az Bz Cx + Az Bx Cz ) + ŷ() + ẑ().
B(A·C) C(A·B) = [Bx (Ax Cx + Ay Cy + Az Cz ) Cx (Ax Bx + Ay By + Az Bz )] x̂ + () ŷ + () ẑ
= x̂(Ay Bx Cy + Az Bx Cz Ay By Cx Az Bz Cx ) + ŷ() + ẑ(). They agree.
Problem 1.6
A⇥(B⇥C)+B⇥(C⇥A)+C⇥(A⇥B) = B(A·C) C(A·B)+C(A·B) A(C·B)+A(B·C) B(C·A) = 0.
So: A⇥(B⇥C) (A⇥B)⇥C = B⇥(C⇥A) = A(B·C) C(A·B).
If this is zero, then either A is parallel to C (including the case in which they point in opposite directions, or
one is zero), or else B·C = B·A = 0, in which case B is perpendicular to A and C (including the case B = 0.)
Conclusion: A⇥(B⇥C) = (A⇥B)⇥C () either A is parallel to C, or B is perpendicular to A and C.
Problem 1.7
r = (4 x̂ + 6 ŷ + 8 ẑ) (2 x̂ + 8 ŷ + 7 ẑ) = 2 x̂ 2 ŷ + ẑ
p
r = 4+4+1= 3

= r = 2 2
+ 13 ẑ
3 x̂ 3 ŷ
r̂ r
Problem 1.8
(a) Āy B̄y + Āz B̄z = (cos Ay + sin Az )(cos By + sin Bz ) + ( sin Ay + cos Az )( sin By + cos Bz )
= cos2 Ay By + sin cos (Ay Bz + Az By ) + sin2 Az Bz + sin2 Ay By sin cos (Ay Bz + Az By ) +
2
cos Az Bz
= (cos2 + sin2 )Ay By + (sin2 + cos2 )Az Bz = Ay By + Az Bz . X
(b) (Ax )2 + (Ay )2 + (Az )2 = Σ3i=1 Ai Ai = Σ3i=1 Σ3j=1 Rij Aj Σ3k=1 Rik Ak = Σj,k (Σi Rij Rik ) Aj Ak .
⇢
2 2 2 3 1 if j = k
This equals Ax + Ay + Az provided Σi=1 Rij Rik =
0 if j 6= k

Moreover, if R is to preserve lengths for all vectors A, then this condition is not only sufficient but also
necessary. For suppose A = (1, 0, 0). Then Σj,k (Σi Rij Rik ) Aj Ak = Σi Ri1 Ri1 , and this must equal 1 (since we
2 2 2
want Ax +Ay +Az = 1). Likewise, Σ3i=1 Ri2 Ri2 = Σ3i=1 Ri3 Ri3 = 1. To check the case j 6= k, choose A = (1, 1, 0).
Then we want 2 = Σj,k (Σi Rij Rik ) Aj Ak = Σi Ri1 Ri1 + Σi Ri2 Ri2 + Σi Ri1 Ri2 + Σi Ri2 Ri1 . But we already
know that the first two sums are both 1; the third and fourth are equal, so Σi Ri1 Ri2 = Σi Ri2 Ri1 = 0, and so
on for other unequal combinations of j, k. X In matrix notation: R̃R = 1, where R̃ is the transpose of R.