kavo (ak22862) – HW02 – Janow – (12121)
1
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001
15.0 points
Forces between two objects which are in
versely proportional to the square of the dis
tance between the objects include which of
the following?
I) Gravitational force between two celestial
bodies
II) Electrostatic force between two electrons
III) Nuclear force between two neutrons.
1.
III only
2.
II and III only
3.
I only
4.
I, II, and III
5.
I and II only
correct
Explanation:
I) Gravitational force
∝
1
r
2
.
II) Electrostatic force
∝
1
r
2
.
III) The nuclear force is very strong com
pared to the gravitational force and electro
static force over a very short range; outside
of that range, however, it is negligible and
cannot be proportional to
1
r
2
002
15.0 points
Two identical small charged spheres hang in
equilibrium with equal masses as shown in
the figure. The length of the strings are equal
and the angle (shown in the figure) with the
vertical is identical.
The acceleration of gravity is 9
.
8 m
/
s
2
and
the
value
of
Coulomb’s
constant
is
8
.
98755
×
10
9
N m
2
/
C
2
.
0
.
08 m
6
◦
0
.
04 kg
0
.
04 kg
Find the magnitude of the charge on each
sphere.
Correct answer: 3
.
58086
×
10
−
8
C.
Explanation:
Let :
L
= 0
.
08 m
,
m
= 0
.
04 kg
,
and
θ
= 6
◦
.
L
a
θ
m
m
q
q
From the right triangle in the figure above,
we see that
sin
θ
=
a
L
.
Therefore
a
=
L
sin
θ
= (0
.
08 m) sin(6
◦
)
= 0
.
00836228 m
.
The separation of the spheres is
r
= 2
a
=
0
.
0167246 m
.
The forces acting on one of the
spheres are shown in the figure 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
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kavo (ak22862) – HW02 – Janow – (12121)
2
vertical directions must separately add up to
zero:
summationdisplay
F
x
=
T
sin
θ
−
F
e
= 0
summationdisplay
F
y
=
T
cos
θ
−
m g
= 0
.
From the second equation in the system
above, we see that
T
=
m g
cos
θ
, so
T
can be
eliminated from the first equation if we make
this substitution. This gives a value
F
e
=
m g
tan
θ
= (0
.
04 kg)
(
9
.
8 m
/
s
2
)
tan(6
◦
)
= 0
.
0412009 N
,
for the electric force.
From Coulomb’s law, the electric force be
tween the charges has magnitude

F
e

=
k
e

q

2
r
2
,
where

q

is the magnitude of the charge on
each sphere.
Note:
The term

q

2
arises here because the
charge is the same on both spheres.
This equation can be solved for

q

to give

q

=
radicalBigg

F
e

r
2
k
e
=
radicalBigg
(0
.
0412009 N) (0
.
0167246 m)
2
(8
.
98755
×
10
9
N m
2
/
C
2
)
=
3
.
58086
×
10
−
8
C
.
003
(part 1 of 2) 15.0 points
Three charges are arranged as shown in the
figure.
The
Coulomb
cosnstant
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
x
y
3
.
9 nC
4
.
1 m
−
0
.
6 nC
1
.
2 nC
−
1
.
9 m
Find the magnitude of the electrostatic
force on the charge at the origin.
Correct answer: 2
.
18596 nN.
Explanation:
Let :
(
x
0
, y
0
) = (0 m
,
0 m)
,
q
0
=
−
0
.
6 nC
,
(
x
1
, y
1
) = (4
.
1 m
,
0 m)
,
q
1
= 3
.
9 nC
,
(
x
2
, y
2
) = (0 m
,
−
1
.
9 m)
,
and
q
2
= 1
.
2 nC
.
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 Fall '08
 Opyrchal
 Physics, Charge, Force, Correct Answer, Electric charge

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