14.3 Solutions
665
14.3.2 Maxwell’s Equations and Electromagnetic Waves,
Poynting Vector
14.33 E z = 100 cos(6 × 10
8 t + 4x) (by problem)
(1)
E z = A cos(ωt + kx) (standard equation)
(2)
Comparison of (1) and (2) shows that
ω = 6 × 10
8 and k = 4
v =
ω
k
=
6 × 10 8
4
= 1.5 × 10
8 m/s
Dielectric constant, K =
c
v
=
3 × 10 8
1.5 × 10 8 = 2.0
14.34
E · ds = q/ε 0 (Gauss’ law)
E(2π rl) = q/ε 0
∴ E =
λ ˆ
e r
2πr ε 0
(1)
B · ds = μ 0 I (Ampere’s law)
2π rB = μ 0 I
∴ B =
μ 0 I
2π r
ˆ
e φ
(2)
E
r = γ (E r − v B φ ) (Lorentz transformation)
(3)
= γ
λ
2πr ε 0
−
v μ 0 I
2πr
=
γ
2π r
λ
ε 0
− vμ 0 I
Thus E
r = 0 if vμ 0 I =
λ
ε 0
or v =
λ
I μ 0 ε 0
=
λc 2
I
14.35 Maxwell’s equations in vacuum are
∇ · E = 0
( 1 )
∇ · B = 0
( 2 )
∇ × E = −
∂B
∂t
(3)
∇ × B = μ 0 ε 0
∂E
∂t
(4)