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PART 4
ELECTROMAGNETISM
CHAPTER 23
ELECTRIC CHARGE, FORCE, AND FIELD
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
q
q
e
proton
electron

=

10
9
, a person’s net charge would be about
±
×
×
×

2
10
10
28
9
16
10
3 2
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
Q e
=
=
=
×
=
×

25
16
10
156
10
19
20
C
C
.
.
.
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
2
2
9
2
2
2
8
9
10
3 2
10
9 22
10
=
=
=
×
⋅
=
×
(
/
)( .
)
.
N m C
C
m
N. This is approximately 1.45 million times the weight of an
averagesized 65 kg person.
Problem
8.
How far apart should an electron and proton be so the force of Earth’s gravity on the electron is equal to the electric force
arising from the proton? Your answer shows why gravity is unimportant on the molecular scale!
Solution
The electric force between a proton and an electron has magnitude
ke r
2
2
=
, while the weight of an electron is
m g
e
. These are
equal when
r
ke m g
e
=
=
×
⋅
×
×
=


2
9
2
2
19
2
31
2
9
10
16
10
911
10
9 8
5 08
=
(
/
)( .
)
( .
)( .
/
)
.
N m C
C
kg
m s
m
(almost fifty billion atomic diameters).
Problem
9.
Two charges, one twice as large as the other, are located 15 cm apart and experience a repulsive force of 95 N. What is
the magnitude of the larger charge?
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541
Solution
The product of the charges is
q q
r F
k
1
2
2
2
9
2
2
10
2
015
95
9
10
2 38
10
=
=
×
⋅
=
×

Coulomb
m
N
N m C
C
=
=
( .
) (
) (
/
)
.
. If one charge
is twice the other,
q
q
1
2
2
=
, then
1
2
1
2
10
2 38
10
q
=
×

.
C and
q
1
218
= ±
.
.
C
μ
Problem
11. A proton is on the
x
axis at
x
=
16
. nm. An electron is on the
y
axis at
y
=
0 85
.
nm. Find the net force the two exert on a
helium nucleus (charge
+
2
e
) at the origin.
Solution
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 Spring '10
 SMITH
 Math

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