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986
CHAPTER 16
MULTIPLE INTEGRATION
(ET CHAPTER 15)
If you had arranged the axes differently, you could have computed the answer as
³
0
,
1
12
,
37
48
´
(depending on orientation).
66.
According to Coulomb’s Law, the force between two electric charges of magnitude
q
1
and
q
2
separated by
a distance
r
is
Cq
1
q
2
/
r
2
acting along the line through the two charges (
C
is a constant). Suppose that a circular disk of
radius
R
has a uniformly distributed charge of density
k
Coulombs per square centimeter (Figure 25). Let
F
be the net
force on a charged particle
P
of magnitude
Q
coulombs located at a distance
d
cm vertically above the center of the disk
(by symmetry,
F
acts in the vertical direction).
(a)
Let
R
be a small polar rectangle of size
1
r
×
1
θ
located at distance
r
. Show that
R
exerts a force on
P
whose
vertical component is
±
kCQd
(
r
2
+
d
2
)
3
/
2
¶
r
1
r
1
(b)
Explain why
F
is equal to the following double integral and evaluate:
F
=
Z
2
π
0
Z
R
0
rdrd
(
r
2
+
d
2
)
3
/
2
R
d
Charged plate
Δ
r
Δ
P
FIGURE 25
SOLUTION
(a)
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 Winter '08
 Rogawski

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