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14
CHAPTER 15
15.16
The required position is shown in the
sketch at the right. Note that this places
q
closer to the smaller charge, which will
allow the two forces to cancel. Requiring
that
gives
6
FF
=
3
()
2
2
nC
3.00 nC
00 m
ee
kq
k
x
=
6.00
0.6
x
+
q
2
2
2
0.600 m
xx
=+
, or
Solving for
x
gives the equilibrium position as
0.6
x
00 m
1.45 m beyond the
21
==
−
3.00 nC charge
−
±²³³´µ¶
·¸²³³´µ¶
±
²
³²±³³´¹
±
¸
±²
±
±
15.17
For the object to “float” it is necessary that the electrical force support the weight, or
( )
( )
6
3
2
24
10
C
610 N C
or
1.5
10
kg
9.8 m s
qE
qE
mg
m
g
−
−
×
=
=
×
15.18
(a) Taking to the right as
positive, the resultant
electric field at point
P
is
given by
132

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