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46. (a) Using Eq. 2711 and Eq. 2542, we obtain
¯
¯
¯
~
J
A
¯
¯
¯
=
¯
¯
¯
~
E
A
¯
¯
¯
ρ
=

∆
V
A

ρL
=
40
×
10
−
6
V
(100 Ω
·
m)(20 m)
=2
.
0
×
10
−
8
A
/
m
2
.
(b) Similarly, in region
B
we fnd
¯
¯
¯
~
J
B
¯
¯
¯
=

∆
V
B

ρL
=
60
×
10
−
6
V
(100 Ω
·
m)(20 m)
=3
.
0
×
10
−
8
A
/
m
2
.
(c) With
w
=1
.
0mand
d
A
=3
.
8 m (so that the crosssection area is
d
A
w
) we have (using Eq. 275)
i
A
=
¯
¯
¯
~
J
A
¯
¯
¯
d
A
w
=
(
2
.
0
×
10
−
8
A
/
m
2
)
(1
.
0 m)(3
.
8m)=7
.
6
×
10
−
8
A
.
(d) Assuming
i
A
=
i
B
we obtain
d
B
=
i
B
¯
¯
¯
~
J
B
¯
¯
¯
w
=
7
.
6
×
10
−
8
A
(3
.
0
×
10
−
8
A
/
m
2
)(1
.
0m)
=2
.
5m
.
(e) We do not show the graphandfgure here, but describe it brieﬂy. To be meaningFul (as a Function
oF
x
) we would plot
V
(
x
) measured relative to
V
(0) (the voltage at, say, the leFt edge oF the fgure,
which we are e±ectively setting equal to 0). ²rom the problem statement, we note that
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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

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