terry (ect328) – homework 09 – Turner – (59130)
1
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001
10.0 points
Three identical point charges, each of mass
150 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 cm
150 g
+
q
10 cm
150 g
+
q
45
◦
150 g
+
q
If the lengths of the left and right strings
are each 10 cm, and each forms an angle of
45
◦
with the vertical, determine the value of
q
.
Correct answer: 0
.
80885
μ
C.
Explanation:
Let :
θ
= 45
◦
,
m
= 150 g = 0
.
15 kg
,
L
= 10 cm = 0
.
1 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
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
.
15 kg) (9
.
8 m
/
s
2
) tan 45
◦
= 1
.
47 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)
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terry (ect328) – homework 09 – Turner – (59130)
2
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
.
1 m) sin 45
◦
×
radicalBigg
(0
.
15 kg) (9
.
8 m
/
s
2
) tan 45
◦
5 (8
.
98755
×
10
9
N
·
m
2
/
C
2
)
= 8
.
0885
×
10
−
7
C =
0
.
80885
μ
C
.
002
10.0 points
An electron moves at 3
.
5
×
10
6
m
/
s into a
uniform electric field of magnitude 1194 N
/
C.
The
charge
on
an
electron is 1
.
60218
×
10
−
19
C and the mass
of an electron is 9
.
10939
×
10
−
31
kg
.
The field is parallel to the electron’s velocity
and acts to decelerate the electron.
How far does the electron travel before it is
brought to rest?
Correct answer: 2
.
91662 cm.
Explanation:
Let :
v
= 3
.
5
×
10
6
m
/
s
,
q
e
= 1
.
60218
×
10
−
19
C
,
m
= 9
.
10939
×
10
−
31
kg
,
and
E
= 1194 N
/
C
.
The kinetic energy
K
=
1
2
m v
2
is depleted by the amount of work done by
the electric force
F
=
q
e
E
on the particle:
W
=
integraldisplay
F dx
=
F x
=
q
e
E x
since the force is constant. When the electron
comes to rest, all its kinetic energy has been
converted, so
1
2
m v
2
=
q
e
E x .
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 Fall '09
 Turner
 Charge, Electric charge, Terry

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