Version 096/ABCAA – midterm 01 – Turner – (59070)
1
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001
10.0 points
A
vertical
electric
field
of
magnitude
19100 N
/
C exists above Earth’s surface on
a day when a thunderstorm is brewing. A car
with a rectangular size of 6
.
02 m by 4
.
09 m is
traveling along a roadway sloping downward
at 9
.
7
◦
.
Determine the electric flux through the bot
tom of the car.
1. 573419.0
2. 514644.0
3. 1322320.0
4. 553942.0
5. 235821.0
6. 472045.0
7. 1064980.0
8. 463560.0
9. 567291.0
10. 206298.0
Correct answer: 4
.
6356
×
10
5
N
·
m
2
/
C.
Explanation:
Let :
a
= 6
.
02 m
,
b
= 4
.
09 m
,
E
= 19100 N
/
C
,
and
θ
= 9
.
7
◦
.
Flux is
Φ =
E A
cos
θ
=
E a b
cos
θ
= (19100 N
/
C) (6
.
02 m) (4
.
09 m) cos 9
.
7
◦
=
4
.
6356
×
10
5
N
·
m
2
/
C
002
10.0 points
Three identical point charges, each of mass
160 g and charge +
q
, hang from three strings,
as in the figure.
The acceleration of gravity is 9
.
8 m
/
s
2
,
and
the
Coulomb
constant
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
9
.
8 m
/
s
2
10
.
9 cm
160 g
+
q
10
.
9 cm
160 g
+
q
42
◦
160 g
+
q
If the lengths of the left and right strings
are each 10
.
9 cm, and each forms an angle of
42
◦
with the vertical, determine the value of
q
.
1. 0.786265
2. 0.918923
3. 0.622486
4. 0.817624
5. 0.44089
6. 1.14218
7. 0.528588
8. 0.676143
9. 0.738161
10. 0.320938
Correct answer: 0
.
817624
μ
C.
Explanation:
Let :
θ
= 42
◦
,
m
= 160 g = 0
.
16 kg
,
L
= 10
.
9 cm = 0
.
109 m
,
g
= 9
.
8 m
/
s
2
,
and
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
Newton’s 2nd law:
summationdisplay
F
=
m a .
Electrostatic force between point charges
q
1
and
q
2
separated by a distance
r
F
=
k
e
q
1
q
2
r
2
.
All three charges are in equilibrium, so for
each holds
summationdisplay
F
= 0
.
Consider the forces acting on the charge on
the right. There must be an electrostatic force
F
acting on this charge, keeping it balanced
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Version 096/ABCAA – midterm 01 – Turner – (59070)
2
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
.
These can be rewritten as
F
=
T
sin
θ
and
m g
=
T
cos
θ .
Dividing the former by the latter, we find
F
m g
= tan
θ ,
or
F
=
m g
tan
θ
(1)
= (0
.
16 kg) (9
.
8 m
/
s
2
) tan 42
◦
= 1
.
41183 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
or
F
=
5
k
e
q
2
4
L
2
sin
2
θ
.
(2)
We have already found
F
, and the other quan
tities are given, so we solve for the squared
charge
q
2
q
2
=
4
F L
2
sin
2
θ
5
k
e
(3)
or, after taking the square root (we know
q >
0) and substituting
F
from Eq. 1 into Eq.
3 and solving for
q
, we have
q
= 2
L
sin
θ
radicalbigg
m g
tan
θ
5
k
e
= 2 (0
.
109 m) sin 42
◦
×
radicalBigg
(0
.
16 kg) (9
.
8 m
/
s
2
) tan 42
◦
5 (8
.
98755
×
10
9
N
·
m
2
/
C
2
)
= 8
.
17624
×
10
−
7
C =
0
.
817624
μ
C
.
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 Spring '08
 Turner
 Electric charge, Qin, 42◦

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