22
2
.
EE
B
B
x
xxx
t
x
t
∂∂
∂
∂ ∂
∂
⎛⎞
⎛
⎞
==
−=
−
⎜⎟
⎜
⎟
∂
∂
∂
∂
∂
⎝⎠
⎝
⎠
Now we differentiate both sides of Eq. 3318 with respect to
t
:
∂
∂
−
∂
∂
F
H
G
I
K
J
=−
∂
=
∂
∂
∂
∂
F
H
G
I
K
J
=
∂
∂
t
B
x
B
xt
t
E
t
E
t
2
00
2
2
εµ
.
Substituting
222
Ex
Bx
t
=
−
∂
from the first equation above into the second one, we
get
2
2
2
2
2
2
2
1
.
E
E
E
c
tx
t
x
x
∂
∂
∂
=⇒
=
=
∂
∂
∂
Similarly, we differentiate both sides of Eq. 3311 with respect to
t
2
,
EB
x
tt
∂
and differentiate both sides of Eq. 3318 with respect to
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Any
 Physics

Click to edit the document details