P22_042.PDF
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Chapter 22 - 22.42
42. (a) Charge
Q
1
=+80
×
10
−
9
Cisonthe
y
axis at
y
=0
.
003 m, and charge
Q
2
=+80
×
10
−
9
Cison
10
−
9
C) is
due to the vector sum of the repulsive forces from
Q
1
and
Q
2
. In symbols,
F
31
+
F
32
=
F
3net
,
where
−
0
.
003 m. The force on particle 3 (which has a charge of
q
=+18
×
=
k
q
3
|
q
1
|
r
31
=
k
q
3
q
2
r
32
|
F
31
|
and
|
F
32
|
.
Using the Pythagorean theorem, we have
r
31
=
r
32
=0
.
005 m. In magnitude-angle notation
(particularly convenient if one uses a vector capable calculator in polar mode), the indicated vector
addition becomes
(0
.
518
−
37
◦
)+(0
.
518
37
◦
)=(0
.
829
0
◦
)
.
Therefore, the net force is 0
.
829 N in the +
x
direction.
(b) Switching the sign of
Q
2
amounts to reversing the direction of its force on
q
.Consequently,wehave
(0
.
518
−
37
◦
)+(0
.
518
−
143
◦
)=(0
.
621
−
90
◦
)
.
Therefore, the net force is 0
.
621 N in the
−
y
direction.
the
y
axis at
y
=
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