33. (a) The current
i
is shown in Fig. 2629 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
=
π
r
2
is the cone’s crosssection area at that particular value of
x
. The direction of
G
J
is
identical to that shown in the figure for
i
(our +
x
direction). Using Eq. 2611, 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
i
r
E
=
=
π
2
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
c x
=
+
1
2
.
Choosing the origin at the left end of the truncated cone, the coefficient
c
1
is chosen so
that
r = a
(when
x
= 0); therefore,
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 Spring '10
 Rebello,NobelS
 Vector Space, Current, Electric Potential, 1 L, 10 m, 2 m

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