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211
E
LECTRIC
C
HARGE AND
E
LECTRIC
F
IELD
21.1.
(a)
IDENTIFY
and
SET UP:
Use the charge of one electron
19
( 1.602 10
C)
−
−×
to find the number of electrons
required to produce the net charge.
EXECUTE:
The number of excess electrons needed to produce net charge
q
is
9
10
19
3.20 10 C
2.00 10 electrons.
1.602 10
C/electron
q
e
−
−
==
×
−−
×
(b) IDENTIFY
and
SET UP:
Use the atomic mass of lead to find the number of lead atoms in
3
8.00 10 kg
−
×
of
lead. From this and the total number of excess electrons, find the number of excess electrons per lead atom.
EXECUTE:
The atomic mass of lead is
3
207 10 kg/mol,
−
×
so the number of moles in
3
−
×
is
3
tot
3
0.03865 mol.
207 10 kg/mol
m
n
M
−
−
×
=
×
A
N
(Avogadro’s number) is the number of atoms in 1 mole, so the
number of lead atoms is
23
22
A
(0.03865 mol)(6.022 10 atoms/mol) = 2.328 10 atoms.
Nn
N
×
×
The number of
excess electrons per lead atom is
10
13
22
2.00 10 electrons
8.59 10 .
2.328 10 atoms
−
×
=×
×
EVALUATE:
Even this small net charge corresponds to a large number of excess electrons. But the number of
atoms in the sphere is much larger still, so the number of excess electrons per lead atom is very small.
21.2.
IDENTIFY:
The charge that flows is the rate of charge flow times the duration of the time interval.
SET UP:
The charge of one electron has magnitude
19
1.60 10
C.
e
−
EXECUTE:
The rate of charge flow is 20,000 C/s and
4
100 s
1.00 10 s.
t
μ
−
×
4
(20,000 C/s)(1.00 10 s)
2.00 C.
Q
−
=
The number of electrons is
19
e
19
1.25 10 .
1.60 10
C
Q
n
−
×
×
EVALUATE:
This is a very large amount of charge and a large number of electrons.
21.3.
IDENTIFY:
From your mass estimate the number of protons in your body. You have an equal number of electrons.
SET UP:
Assume a body mass of 70 kg. The charge of one electron is
19
1.60 10
C.
−
EXECUTE:
The mass is primarily protons and neutrons of
27
1.67 10
kg.
m
−
The total number of protons and
neutrons is
28
p and n
27
70 kg
4.2 10 .
1.67 10
kg
n
−
×
×
About onehalf are protons, so
28
p
e
2.1 10
nn
=
×=
. The number of
electrons is about
28
2.1 10 .
×
The total charge of these electrons is
19
28
9
( 1.60 10
C/electron)(2.10 10 electrons)
3.35 10 C.
Q
−
=−
×
×
×
EVALUATE:
This is a huge amount of negative charge. But your body contains an equal number of protons and
your net charge is zero. If you carry a net charge, the number of excess or missing electrons is a very small fraction
of the total number of electrons in your body.
21.4.
IDENTIFY:
Use the mass
m
of the ring and the atomic mass
M
of gold to calculate the number of gold atoms.
Each atom has 79 protons and an equal number of electrons.
SET UP:
23
A
6.02 10 atoms/mol
N
. A proton has charge +
e
.
EXECUTE:
The mass of gold is 17.7 g and the atomic weight of gold is 197
g mol. So the number of atoms
is
23
22
A
17.7 g
(6.02 10 atoms/mol)
5.41 10 atoms
197 g mol
⎛⎞
=
×
⎜⎟
⎝⎠
. The number of protons is
22
24
p
(79 protons/atom)(5.41 10 atoms)
4.27 10 protons
n
=
×
.
19
5
p
(
)(1.60 10
C/proton)
6.83 10 C
Qn
−
=
×
.
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 Spring '08
 Jabbour
 Materials Science And Engineering

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