PART 4
ELECTROMAGNETISM
CHAPTER 23
ELECTRIC CHARGE, FORCE, AND FIELD
ActivPhysics
can help with these problems: Activities 11.1–11.8
Section 23

2: Electric Charge
Problem
1.
Suppose the electron and proton charges differed by one part in one billion. Estimate the net charge you would carry.
Solution
Nearly all of the mass of an atom is in its nucleus, and about one half of the nuclear mass of the light elements in living
matter (H, O, N, and C) is protons. Thus, the number of protons in a 65 kg averagesized person is approximately
1
2
27
28
65
167
10
2
10
()
(
.
)
kg
kg
=
××
−
¼
, which is also the number of electrons, since an average person is electrically neutral.
If there were a charge imbalance of
qq
e
proton
electron
−=
−
10
9
, a person’s net charge would be about
±×
×
×
−
21
0 1
0
28
9
16
10
32
19
..
,
×=
±
−
C
C
or several coulombs (huge by ordinary standards).
Problem
2.
A typical lightning flash delivers about 25 C of negative charge from cloud to ground. How many electrons are
involved?
Solution
The number is
Qe
==
=×=
×
−
25
10
156
10
19
20
C
.
Problem
3.
Protons and neutrons are made from combinations of the two most common quarks, the
u
quark and the
d
quark. The
u
quark’s charge is
+
2
3
e
while the
d
quark carries
−
1
3
e
. How could three of these quarks combine to make (a) a proton
and (b) a neutron?
Solution
(a) The proton’s charge is
1
2
3
2
3
1
3
eeee
=+−
, corresponding to a combination of
uud
quarks; (b) for neutrons, 0
=
2
3
1
3
1
3
−−
corresponds to
udd
. (See Chapter 39, or Chapter 45 in the extended version of the text.)
Problem
4.
A 2g pingpong ball rubbed against a wool jacket acquires a net positive charge of
1C
µ
. Estimate the fraction of the
ball’s electrons that have been removed.
Solution
If half the ball’s mass is protons, their number (equal to the original number of electrons) is 1 g
=
m
p
. The number of electrons
removed is 1 C
=
e
, so the fraction removed is
.
.
.
1
1
10
10
1
104
62
4
19
11
C
g
g
=
=
e
gm
p
=
=×
−
−
−
−
(a hundred billionth).
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View Full DocumentCHAPTER 23
541
Section 233:
Coulomb’s Law
Problem
5.
If the charge imbalance of Problem 1 existed, what would be the approximate force between you and another person
10 m away? Treat the people as point charges, and compare the answer with your weight.
Solution
The magnitude of the Coulomb force between two point charges of 3.2 C (see solution to Problem 1), at a distance of 10 m,
is
kq r
22
9
2 2
2
8
91
0
3
2 1
0
9
2
21
0
==
=×
⋅
= ×
(/
)
(
.
)
.
NmC
C m
N
. This is approximately 1.45 million times the weight of an
averagesized 65 kg person.
Problem
6.
Find the ratio of the electrical force between a proton and an electron to the gravitational force between the two. Why
doesn’t it matter that you aren’t told the distance between them?
Solution
At all distances (for which the particles can be regarded as classical point charges), the Coulomb force is stronger than the
gravitational force by a factor of:
F
F
ke
r
r
Gm m
pe
elec
grav
=
F
H
G
I
K
J
F
H
G
I
K
J
2
2
2
=
×⋅
×
×
×
×
−
−−
−
)
(
.
)
10
/
)( .
)( .
)
..
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 Spring '09
 staff
 Charge, Electric Fields, Electric charge

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