P27_029.PDF
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Chapter 27 - 27.29
29. (a) The current
i
is shown in Fig. 27-22 entering the truncated cone at the left end and leaving at the
right. This is our choice of positive
x
direction. We make the assumption that the current density
J
at each value of
x
may be found by taking the ratio
i/A
where
A
=
πr
2
is the cone’s cross-section
area at that particular value of
x
. The direction of
J
is identical to that shown in the figure for
i
(our
+
x
direction). Using Eq. 27-11, we then find an expression for the electric field at each value
of
x
, and next find the potential difference
V
by integrating the field along the
x
axis, in accordance
with the ideas of Chapter 25. Finally, the resistance of the cone is given by
R
=
V/i
.Thus,
J
=
πr
2
=
E
i
ρ
where we must deduce how
r
depends on
x
in order to proceed. We note that the radius increases
linearly with
x
,so(with
c
1
and
c
2
to be determined later) we may write
r
=
c
1
+
c
2
x.
Choosing the origin at the left end of the truncated cone, the coe.cient
c
1
is chosen so that
r
=
a
(when
x
= 0); therefore,
c
1
=
a
. Also, the coe.cient
c
2
must be chosen so that (at the right end of
the truncated cone) we have
r
=
b
(when
x
=
L
); therefore,
c
2
=(
b
−
a
)
/L
. Our expression, then,
becomes
r
=
a
+
b
−
a
x.
L
Substituting this into our previous statement and solving for the field, we find
E
=
iρ
π
a
+
b
−
a
L
x
−
2
.
Consequently, the potential difference between the faces of the cone is
L
iρ
π
L
a
+
b
−
a
x
−
2
V
=
−
Edx
=
−
dx
L
0
0
x
−
1
iρ
π
L
a
+
b
a
L
=
iρ
π
L
1
a
−
1
b
−
=
b
−
a
L
b
−
a
0
=
iρ
π
L
b
−
a
=
iρL
πab
.
b
−
a
ab
The resistance is therefore
R
=
V
i
=
ρL
πab
.
(b) If
b
=
a
,then
R
=
ρL/πa
2
=
ρL/A
,where
A
=
πa
2
is the cross-sectional area of the cylinder.
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