C
HAPTER
15
Quick Quizzes
1.
(d). Object A could possess a net charge whose sign is opposite that of the excess charge
on B. If object A is neutral, B would also attract it by creating an induced charge on the
surface of A. This situation is illustrated in Figure 15.5 of the textbook.
2.
(b). By Newton’s third law, the two objects will exert forces having equal magnitudes but
opposite directions on each other.
3.
(c). The electric field at point
P
is due to charges
other
than the test charge. Thus, it is
unchanged when the test charge is altered. However, the direction of the force this field
exerts on the test change is reversed when the sign of the test charge is changed.
4.
(b). The magnitude of the upward electrical force must equal the weight of the ball. That
is:
, so
qE
m g
=
( )
( )
32
4
6
5.
01
0 kg 9.80 m s
1.
21
0
NC
4.
0 C
mg
E
q
−
−
×
==
=×
×
5.
(a). If a test charge is at the center of the ring, the force exerted on the test charge by charge
on any small segment of the ring will be balanced by the force exerted by charge on the
diametrically opposite segment of the ring. The net force on the test charge, and hence the
electric field at this location, must then be zero.
6.
(c) and (d) The electron and the proton have equal magnitude charges of opposite signs.
The forces exerted on these particles by the electric field have equal magnitude and
opposite directions. The electron experiences an acceleration of greater magnitude than
does the proton because the electron’s mass is much smaller than that of the proton.
7.
A, B,
and
C
. The field is greatest at point
A
because this is where the field lines are closest
together. The absence of lines at point
C
indicates that the electric field there is zero.
8.
Statements (b) and (d) are true and follow from Gauss’s law. Statement (a) is not
necessarily true because Gauss’s law says that the net flux through any closed surface
equals the net charge inside the surface divided by
0
∈
. For example, a positive and a
negative charge could be inside the surface. Statement (c) is not necessarily true. Although
the net flux through the surface is zero, the electric field in that region may not be zero.
3
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View Full DocumentCHAPTER 15
Problem Solutions
15.1
Since the charges have opposite signs, the force is one of attraction .
Its magnitude is
( )( )
()
99
2
12
9
2
22
4.
51
0
C2
.
81
0 C
Nm
8.99
10
C
3.2 m
e
kqq
F
r
−−
××
⎛⎞
⋅
==
×
=
8
1.
11
0 N
−
×
⎜⎟
⎝⎠
15.2
(a) The force is
( )
( )
2
14
2
2
9
2
6
1.60
10
C
C
2.
0 m
e
kq
F
r
−
−
×
⋅
== ×
=
(
×
)
7
3.
71
0 N rep u lsion
−
×
(b) With
r
two times larger,
is four times larger and the force is
2
r
red u ced to one fourth of its previous value
.
15.3
( )
2
27
9
e
ke e
F
r
=
( )
( )
2
19
2
9
2
2
158 1.60
C
C
01
m
−
−
×
⋅
=×
=
×
( )
91 N rep u lsion
15.4
The electrical force is
2
2
e
e
ke
F
r
=
, and the gravitational force is
2
2
g
Gm
F
r
=
. Thus, if
ge
FF
=
,
the mass would have to be
( )
92
2
2
2
10 N
m
C
C
6.67
N
m
k
e
k
me
Gg
−
−
×⋅
==×
=
9
1.86
kg
−
×
This result is
times greater than the actual mass of a proton.
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 Spring '08
 Chilton
 Charge, Electric charge, NC

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