homework 01 – FAUSAK, TAYLOR – Due: Jan 21 2008, 4:00 am
1
Question 1, chap 22, sect 1.
part 1 of 1
10 points
You have 3
.
2 kg of water. One mole of water
has a mass of 18 g
/
mol and each molecule
of water contains 10 electrons since water is
H
2
O.
What is the total electron charge contained
in this volume of water?
Correct answer:
−
1
.
71296
×
10
8
C (tolerance
±
1 %).
Explanation:
Let :
N
A
= 6
.
02214
×
10
23
molec
/
mol
,
q
e
=
−
1
.
6
×
10
−
19
C
/
electron
,
M
= 18 g
/
mol = 0
.
018 kg
/
mol
,
m
= 3
.
2 kg
,
and
Z
= 10 electrons
/
molec
.
The mass is proportional to the number of
molecules, so for
m
grams in
n
molecules and
M
grams in
N
A
molecules,
m
M
=
n
N
A
n
=
m
M
N
A
.
Since 10 electrons are in each molecule of
water, then the total number of electrons
n
e
in the coin is
n
e
=
Z n
=
Z
m
M
N
A
and the total charge
q
for the
n
e
electrons is
q
=
n
e
q
e
=
Z
m N
A
q
e
M
= (10 electrons
/
molec)
3
.
2 kg
0
.
018 kg
/
mol
×
(6
.
02214
×
10
23
molec
/
mol)
×
(
−
1
.
6
×
10
−
19
C
/
electron)
=
−
1
.
71296
×
10
8
C
.
Question 2, chap 22, sect 1.
part 1 of 1
10 points
What happens to the mass of an object
when it acquires a positive net charge by the
transfer of electrons?
1.
Increases
2.
Decreases
correct
3.
Doesn’t change
4.
Cannot be determined
Explanation:
When an object acquires a positive charge
it loses electrons, so its mass decreases.
Question 3, chap 22, sect 2.
part 1 of 1
10 points
A particle of mass 85 g and charge 26
μ
C
is released from rest when it is 44 cm from a
second particle of charge
−
10
μ
C.
Determine the magnitude of the initial ac
celeration of the 85 g particle.
Correct answer: 142 m
/
s
2
(tolerance
±
1 %).
Explanation:
Let :
m
= 85 g = 0
.
085 kg
,
q
= 26
μ
C = 2
.
6
×
10
−
5
C
,
r
= 44 cm = 0
.
44 m
,
Q
=
−
10
μ
C =
−
1
×
10
−
5
C
,
and
k
e
= 8
.
9875
×
10
9
N
·
m
2
/
C
2
.
The force exerted on the particle is
F
=
k
e

q
 
Q

r
2
=
m

a


a

=
k
e

q
 
Q

m r
2
= (8
.
9875
×
10
9
N
·
m
2
/
C
2
)
×
vextendsingle
vextendsingle
2
.
6
×
10
−
5
C
vextendsingle
vextendsingle
vextendsingle
vextendsingle
−
1
×
10
−
5
C
vextendsingle
vextendsingle
(0
.
085 kg) (0
.
44 m)
2
=
142 m
/
s
2
.
Question 4, chap 22, sect 2.
part 1 of 2
10 points
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homework 01 – FAUSAK, TAYLOR – Due: Jan 21 2008, 4:00 am
2
What can we conclude from the attrac
tive nature of the force between a positively
charged rod and an object?
+
+
+
F
F
?
1.
The object is an insulator.
2.
Cannot be determined
correct
3.
The object is a conductor.
Explanation:
The attractive nature of the force will be
the same whether the object is a conductor or
insulator.
Question 5, chap 22, sect 2.
part 2 of 2
10 points
What can we conclude from the attrac
tive nature of the force between a positively
charged rod and an object?
1.
The object does not have an appreciable
negative charge.
2.
The object has an appreciable negative
charge.
3.
The object has an appreciable positive
charge.
4.
The object does not have an appreciable
positive charge.
correct
5.
Cannot be determined.
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 Spring '07
 KBrowning
 Charge, Electric charge, Fnet

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