48. (a) Current is the transport of charge; here it is being transported
“in bulk”
due to the
volume rate of flow of the powder. From Chapter 14, we recall that the volume rate of
flow is the product of the crosssectional area (of the stream) and the (average) stream
velocity. Thus,
i
=
ρ
Av
where
is the charge per unit volume. If the crosssection is that
of a circle, then
i
=
π
R
2
v
.
(b) Recalling that a Coulomb per second is an Ampere, we obtain
i
=×
=
×
−−
11 10
2 0
17 10
33
2
5
..
.
C/m
m
m/s
A.
c
h
b
g b
g
π 0.050
(c) The motion of charge is not in the same direction as the potential difference computed
in problem 57 of Chapter 25. It might be useful to think of (by analogy) Eq. 748; there,
the scalar (dot) product in
P
F v
=⋅
G
G
makes it clear that
P
= 0 if
G
G
F v
⊥
. This suggests that
a radial potential difference and an axial flow of charge will not together produce the
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 Spring '10
 Rebello,NobelS
 Charge, Current, Volt, Potential difference, Ampere, volume rate, radial potential difference

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