This preview shows page 1. Sign up to view the full content.
75. We take the derivative with respect to
x
of both sides of Eq. 3411:
∂
∂x
µ
∂E
∂x
¶
=
∂
2
E
∂x
2
=
∂
∂x
µ
−
∂B
∂t
¶
=
−
∂
2
B
∂x∂t
.
Now we diFerentiate both sides of Eq. 3418 with respect to
t
:
∂
∂t
µ
−
∂B
∂x
¶
=
−
∂
2
B
∂x∂t
=
∂
∂t
µ
ε
0
µ
0
∂E
∂t
¶
=
ε
0
µ
0
∂
2
E
∂t
2
.
Substituting
∂
2
E/∂x
2
=
−
∂
2
B/∂x∂t
from the ±rst equation above into the second one, we get
ε
0
µ
0
∂
2
E
∂t
2
=
∂
2
E
∂x
2
,
or
∂
2
E
∂t
2
=
1
ε
0
µ
0
∂
2
E
∂x
2
=
c
2
∂
2
E
∂x
2
.
Similarly, we diFerentiate both sides of Eq. 3411 with respect to
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

Click to edit the document details