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Kapoor (mk9499) – oldmidterm 01 – Turner – (60230)
1
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beFore answering.
001
10.0 points
Two spheres, Fastened to “pucks”, are rid
ing on a Frictionless airtrack. Sphere “1” is
charged with 1 nC, and sphere “2” is charged
with 3 nC. Both objects have the same mass.
1 nC is equal to 1
×
10

9
C.
As they repel,
1.
sphere “2” accelerates 3 times as Fast as
sphere “1”.
2.
sphere “1” accelerates 9 times as Fast as
sphere “2”.
3.
sphere “2” accelerates 9 times as Fast as
sphere “1”.
4.
they have the same magnitude oF acceler
ation.
correct
5.
they do not accelerate at all, but rather
separate at constant velocity.
6.
sphere “1” accelerates 3 times as Fast as
sphere “2”.
Explanation:
The Force oF repulsion exerted on each mass
is determined by
F
=
1
4
π±
0
Q
1
Q
2
r
2
=
ma
where
r
is the distance between the centers oF
the two spheres.
±

F
12
±
=
±

F
21
±
Since both spheres have the same mass and
are subject to the same Force, they have the
same acceleration.
002
10.0 points
Three equal charges oF 3
μ
C are in the
x

y
plane. One is placed at the origin, another is
placed at (0.0, 99 cm), and the last is placed
at (85 cm, 0.0). The Coulomb constant is
9
×
10
9
Nm
2
/
C
2
.
Calculate the magnitude oF the Force on the
charge at the origin.
Correct answer: 0
.
13928 N.
Explanation:
Let :
q
=3
μ
C = 3
×
10

6
C
,
(
x
1
,y
1
) = (0
,
99 cm) = (0
,
0
.
99 m)
,
(
x
2
2
) = (85 cm
,
0) = (0
.
85 m
,
0)
and
(
x
0
0
) = (0
,
0)
.
The electric felds are
E
y
=
k
e
q
x
2
1
+
y
2
1
=
(
9
×
10
9
2
/
C
2
)(
3
×
10

6
C
)
0 + (0
.
99 m)
2
= 27548
.
2N
/
C
,
and
E
x
=
k
e
q
x
2
2
+
y
2
2
=
(
9
×
10
9
2
/
C
2
3
×
10

6
C
)
(0
.
85 m)
2
+0
= 37370
.
/
C
,
Thus
E
=
±
E
2
x
+
E
2
y
and
F
=
qE
=
q
±
E
2
x
+
E
2
y
=
(
3
×
10

6
C
)
·
±
(27548
.
/
C)
2
+ (37370
.
/
C)
2
=
0
.
13928 N
.
003
10.0 points
Two identical small charged spheres hang in
equilibrium with equal masses as shown in
the fgure. The length oF the strings are equal
and the angle (shown in the fgure) with the
vertical is identical.
The acceleration oF gravity is 9
.
8m
/
s
2
and the value oF Coulomb’s constant is
8
.
98755
×
10
9
2
/
C
2
.
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View Full Document Kapoor (mk9499) – oldmidterm 01 – Turner – (60230)
2
0
.
15 m
5
◦
0
.
03 kg
0
.
03 kg
Find the magnitude of the charge on each
sphere.
Correct answer: 4
.
4233
×
10

8
C.
Explanation:
Let :
L
=0
.
15 m
,
m
.
03 kg
,
and
θ
=5
◦
.
L
a
θ
m
m
q
q
From the right triangle in the ±gure above,
we see that
sin
θ
=
a
L
.
Therefore
a
=
L
sin
θ
= (0
.
15 m) sin(5
◦
)
.
0130734 m
.
The separation of the spheres is
r
=2
a
=
0
.
0261467 m
.
The forces acting on one of the
spheres are shown in the ±gure below.
θ
θ
m
g
F
T
e
T
sin
θ
T
cos
θ
Because the sphere is in equilibrium, the
resultant of the forces in the horizontal and
vertical directions must separately add up to
zero:
±
F
x
=
T
sin
θ

F
e
±
F
y
=
T
cos
θ

mg
.
From the second equation in the system
above, we see that
T
=
cos
θ
, so
T
can be
eliminated from the ±rst equation if we make
this substitution. This gives a value
F
e
=
tan
θ
= (0
.
03 kg)
(
9
.
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This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Turner

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