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

SOLUTION MANUAL FOR INTRODUCTION TO ELECTRODYNAMICS

5TH EDITION BY DAVID J GRIFFITHS

,Contents

1 Vector Ana lysis 4

2 Electrosta tics 26

3 Pote ntia l 53

4 Ele ctric Fie lds in Ma tter 92

5 Ma gne tosta tics 110

6 Ma gne tic Fie lds in Ma tter 133

7 Electrodyna m ics 145

8 Conserva tion La ws 168

9 Ele ctrom a gne tic Wa ves 185

10 Pote ntia ls a nd Fie lds 210

11 Radia tion 231

12 Ele ctrodyna m ics a nd Re la tivity 262

,Chapter 1

Vector Analysis

Proble m 1.1
(a) From the diagram, |B + C| cos θ3 = |B| cos θ1 + |C| cos θ2. Multiply by |A|.
|A||B + C| cos θ3 = |A||B| cos θ1 + |A||C| cos θ2.
So: A·(B + C) = A·B + A·C. (Dot product is distributiv e)
Similarly:| B + C | sin θ3 = |B |sin θ1 + C | sin
| θ2. Mulitply by |A |n̂.
A
| || B + C |sin θ3 n̂ = A B
| || | sin θ1 n̂ + | || C| sin θ2 n̂ .
A | sin θ

If n̂ is the unit vector pointing out of the page, it follows that ` ˛¸ x` ˛¸ x
|B| c os θ1 |C| c os θ2
A⇥(B + C) = (A⇥B) + (A⇥C). (Cross product is distributi v e)
(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)
Proble m 1.2
The triple cross-product is not in general associative. For example,
suppos e A = B and C is perpendi cul ar to A, as in the diagram.
= B
Then (B⇥C) points out-of-the-page, and A⇥(B⇥C) points down,
and has magnitu de ABC. But (A⇥B) = 0, so (A⇥B)⇥C = 0 /=
A⇥(B⇥C).

Proble m 1.3
√ √
A = +1 x̂ + 1 ŷ — 1 ẑ ; A =
3; B = 1 x̂ + 1 ŷ + 1 ẑ ; B = 3.
√ √
A·B = +1 + 1 — 1 = 1 = AB cos θ = 3 3 cos θ ⇒ cos θ = 13.
θ
y
1 1
3 70

x
Proble m 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
A⇥B = —1 2 0 = 6 x̂ + 3 ŷ + 2 ẑ .
—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:
√
|A⇥B| = 36 + 9 + 4 = 7. A⇥
n̂ = |A B
⇥B| = 7 7 7 .

Proble m 1.5
x̂ ŷ ẑ
A⇥(B⇥C) = Ax Ay Az
(By Cz — BzCy ) (BzCx — BxCz ) ( BxCy — By Cx )
= x̂[A y (B xCy — B y C x) — A z(B zC x — BxCz)] + ŷ() + ẑ ()
(I’ll just check the x-component; the others go the same way)
= x̂ (A y B xC y — A y B y C x — A zB zC x + A zB xC z) + ŷ ( ) + ẑ ().
B(A·C) — C(A·B) = [B x(A xCx + A y Cy + A z Cz ) — Cx(A xB x + A y B y + A z B z )] x̂ + () ŷ + () ẑ
= x̂ (A y B x C y + AzBxC z — Ay By Cx — AzBzC x) + ŷ ( ) + ẑ ( ) . They agree.
Proble m 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 perpendicul ar 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
= (4 x̂ + 6 ŷ + 8 ẑ ) — (2 x̂ + 8 ŷ + 7 ẑ ) = 2 —2
√
= 4 +4 + 1 = 3
2
3 — 23 3

Proble m 1.8
(a) A¯ y B̄ y + A¯ z B̄ z = (cos Ay + sin Az)(cos By + sin Bz) + (— sin Ay + cos Az)( —sin By + cos Bz)
= cos 2 Ay By + sin cos (Ay Bz + AzBy ) + sin2 AzBz + sin2 Ay By — sin cos (Ay Bz + AzBy ) +
2
cos AzBz
= (cos 2 + sin2 )Ay By + (sin2 + cos 2 )AzBz = Ay By + AzBz. X
(b) (A x) 2 + (A y ) 2 + (A z) 2 = C3 A i A i = C3 C 3 R ijA j C3 R i k A k = Cj,k (Ci R i j R i k ) A j A k .
i =1 i =1 j=1 k=1
⇢
1 if j = k
This equals A 2x + A 2y + A 2z provided C 3
i =1 R i jR i k = 0 if j /= k

Moreover, if R is to preserve lengths for all vectors A, then this condition is not only sufficient but also
necessary.
2 2
For2suppose A = (1, 0, 0). Then Cj,k (Ci R i jR i k ) A jA k = Ci R i 1R i 1, and this must equal 1 (since we
want A x + Ay + A z= 1). Likewis e, C3i =1R i2R i2 = C3i =1 R i3 R i3 = 1. To check the case j /= k, choose A = (1, 1, 0).
Then we want 2 = Cj,k (Ci R i jR i k ) A jA k = Ci R i 1R i 1 + Ci R i 2R i 2 + Ci R i 1R i 2 + Ci R i 2R i 1. But we already
know that the first two sums are both 1; the third and fourth are equal, so Ci R i 1R i 2 = Ci R i 2R i 1 = 0, and so
on for other unequal combinations of j, k. X In matrix notation: R̃ R = 1, where R̃ is the transpose of R.