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22
2
.
B
B
c
tx
∂∂
=
∂
∂
(b) From
E
E f kx
t
m
=±
()
,
ω
∂
∂
=
∂±
∂
=
2
2
2
2
2
2
2
E
t
E
fk
x
t
t
E
df
du
mm
uk
x t
and
c
E
x
cE
x
t
t
cEk
du
2
2
2
2
2
2
2
2
∂
∂
=
∂
=
Since
=
ck
the righthand sides of these two equations are equal. Therefore,
2
.
EE
c
=
Changing
E
to
B
and repeating the derivation above shows that
BB
m
=
±
satisfies
2
.
B
B
c
=
∂
∂
103. (a) From Eq. 331,
∂
∂
=
∂
∂
−=
−
−
2
2
2
2
2
E
tt
Ek
x
t
x
t
sin(
)
sin(
),
ωω
and
c
E
x
c
x
x
t
k
ck
x
t
E
k
x
t
2
2
2
2
2
2
2
∂
∂
=
∂
∂
−
−
=
−
−
)
)
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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