kinsey (cbk323) – oldhomework 01 – Turner – (56725)
1
This
printout
should
have
10
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
Two uncharged metal balls,
X
and
Z
, stand
on insulating glass rods. A third ball, carrying
a negative charge, is brought near the ball
Z
as shown in the figure. A conducting wire is
then run between
X
and
Z
and then removed.
Finally the third ball is removed.
Z
X
−
conducting wire
When all this is finished
1.
ball
X
is positive and ball
Z
is negative.
2.
ball
X
is negative and ball
Z
is positive.
correct
3.
ball
X
is negative and ball
Z
is neutral.
4.
ball
X
is neutral and ball
Z
is negative.
5.
balls
X
and
Z
are both positive, but ball
X
carries more charge than ball
Z
.
6.
ball
X
is neutral and ball
Z
is positive.
7.
balls
X
and
Z
are both positive, but ball
Z
carries more charge than ball
X
.
8.
balls
X
and
Z
are both negative.
9.
ball
X
is positive and ball
Z
is neutral.
10.
balls
X
and
Z
are still uncharged.
Explanation:
When the conducting wire is run between
X
and
Z
, some negative charge flows from
Z
to
X
under the influence of the negative
charge of the third ball.
Therefore, after the wire is removed,
X
is
charged negative and
Z
is charged positive.
002
10.0 points
Three identical point charges hang from three
strings, as shown.
45
◦
45
◦
F
g
28.0 cm
28.0 cm
+
+
+
+q
+q
+q
0.10 kg
0.10 kg
0.10 kg
What is the value of
q
?
The Coulomb
constant is 8
.
98755
×
10
9
N
·
m
2
/
C
2
, and the
acceleration of gravity is 9
.
81 m
/
s
2
.
Correct answer: 1
.
85013
×
10
−
6
C.
Explanation:
Let :
m
= 0
.
10 kg
,
L
= 28
.
0 cm
,
θ
= 45
◦
,
and
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
r
= 2
L
sin
θ
= 2
L
sin 45
◦
= 2
L
√
2
2
=
L
√
2
F
T,x
=
F
T
sin
θ
F
T,y
=
F
T
cos
θ
Each sphere is in equilibrium horizontally
F
electric
−
F
T,x
= 0
F
electric
−
F
T
sin
θ
= 0
and vertically
F
T,y
−
F
g
= 0
F
T
cos
θ
−
F
g
= 0
F
T
=
F
g
cos
θ
.
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kinsey (cbk323) – oldhomework 01 – Turner – (56725)
2
From the horizontal equilibrium,
F
electric
=
parenleftbigg
F
g
cos
θ
parenrightbigg
sin
θ
F
electric
=
F
g
tan
θ
=
F
g
(tan 45
◦
) =
F
g
.
For either of the end charges,
F
electric
=
k
e
q
2
r
2
+
k
e
q
2
(
r
2
)
2
=
k
e
q
2
r
2
+ 4
k
e
q
2
r
2
= 5
k
e
q
2
r
2
5
k
e
q
2
r
2
=
m g .
Thus

q

=
radicalBigg
r
2
m g
5
k
e
=
radicalBigg
(
L
√
2)
2
m g
5
k
e
=
L
·
radicalbigg
2
m g
5
k
e
= (28 cm)
parenleftbigg
1 m
100 cm
parenrightbigg
×
radicalBigg
2(0
.
1 kg)(9
.
81 m
/
s
2
)
5(8
.
98755
×
10
9
N
·
m
2
/
C
2
)
=
1
.
85013
×
10
−
6
C
.
003
(part 1 of 2) 10.0 points
Two identical small metal spheres with
q
1
>
0
and

q
1

>

q
2

attract
each other with a force
of magnitude 65
.
2 mN, as shown in the figure
below.
F
21
F
12
2
.
45 m
q
1
q
2
r
1
= 20
μ
m
r
2
= 20
μ
m
Figure 1:
Before touching
The spheres are then brought together until
they are touching. At this point, the spheres
are in electrical contact so that the charges
can move from one sphere to the other until
both spheres have the same final charge,
q
.
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 Spring '08
 Turner
 Physics, Charge, Work, Electric charge, KE, Kinsey

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