Version 068/ABABA – midterm 01 – turner – (56705)
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have
19
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001
10.0 points
Three identical point charges, each of mass
120 g and charge +
q
, hang from three strings,
as in the figure.
9
.
8 m
/
s
2
10
.
4 cm
120 g
+
q
10
.
4 cm
120 g
+
q
37
◦
120 g
+
q
If the lengths of the left and right strings
are each 10
.
4 cm, and each forms an an
gle of 37
◦
with the vertical, determine the
value
of
q
.
The
acceleration
of
gravity
is 9
.
8 m
/
s
2
, and the Coulomb constant is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
1. 0.618551
2. 0.567833
3. 0.488958
4. 0.651312
5. 0.775448
6. 1.014
7. 0.555881
8. 0.934425
9. 0.711318
10. 0.517167
Correct answer: 0
.
555881
μ
C.
Explanation:
Let :
θ
= 37
◦
,
m
= 120 g = 0
.
12 kg
,
L
= 10
.
4 cm = 0
.
104 m
,
g
= 9
.
8 m
/
s
2
,
and
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
Consider the forces acting on the charge on
the right. There must be an electrostatic force
F
acting on this charge, keeping it balanced
against the force of gravity
m g
. The electro
static force is due to the other two charges
and is therefore horizontal.
In the
x
direction
F

T
sin
θ
= 0
.
In the
y
direction
T
cos
θ

m g
= 0
.
F
sin
θ
F
cos
θ
=
F
e
m g
F
=
m g
tan
θ
(1)
= (0
.
12 kg) (9
.
8 m
/
s
2
) tan 37
◦
= 0
.
88618 N
.
The distance between the right charge and
the middle charge is
L
sin
θ
, and the distance
to the left one is twice that. Since all charges
are of the same sign, both forces on the right
charge are repulsive (pointing to the right).
We can add the magnitudes
F
=
k
e
q q
(
L
sin
θ
)
2
+
k
e
q q
(2
L
sin
θ
)
2
F
=
5
k
e
q
2
4
L
2
sin
2
θ
(2)
q
2
=
4
F L
2
sin
2
θ
5
k
e
(3)
q
= 2
L
sin
θ
F
5
k
e
= 2 (0
.
104 m) sin 37
◦
×
0
.
88618 N
5 (8
.
98755
×
10
9
N
·
m
2
/
C
2
)
= 5
.
55881
×
10

7
C =
0
.
555881
μ
C
.
002
10.0 points
Three small spheres carry equal amounts of
electric charge. They are equally spaced and
lie along the same line.
+


What is the direction of the net electric
force on each charge due to the other charge?
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Version 068/ABABA – midterm 01 – turner – (56705)
2
1.
+


2.
+


3.
+


4.
+


5.
+


correct
6.
+


7.
+


8.
+


9.
+


10.
+


Explanation:
Since like charges repel and unlike charge
attract,
+


003
10.0 points
A charge of 8
.
4
μ
C is 20 cm above the center
of a square of side length 40 cm.
The permittivity of free space is 8
.
85
×
10

12
C
2
/
N
·
m
2
.
Find the flux through the square.
1. 154426.0
2. 73446.3
3. 116761.0
4. 158192.0
5. 139360.0
6. 82862.5
7. 90395.5
8. 165725.0
9. 124294.0
10. 94162.0
Correct answer: 1
.
58192
×
10
5
N
·
m
2
/
C.
Explanation:
Let :
q
= 8
.
4
μ
C = 8
.
4
×
10

6
C
,
= 20 cm = 0
.
2 m
,
s
= 40 cm = 0
.
4 m
,
and
0
= 8
.
85
×
10

12
C
2
/
N
·
m
2
.
The point charge is at the center of the cube.
Applying Gauss’ law, the flux through one
face of the square is
Φ
square
=
1
6
Φ
total
=
1
6
q
inside
0
=
1
6
8
.
4
×
10

6
C
8
.
85
×
10

12
C
2
/
N
·
m
2
=
1
.
58192
×
10
5
N
·
m
2
/
C
.
keywords:
004
10.0 points
A charge moves a distance of 2
.
3 cm in the
direction of a uniform electric field having
a magnitude of 239 N
/
C.
The electrical
potential energy of the charge decreases by
72
.
3544
×
10

19
J as it moves.
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 Spring '10
 Turner
 Charge, Electrostatics, Electric charge, Version 068/ABABA

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