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Catégorie :Category: mViewer GX Creator Casio fx-CP400
Auteur Author: juanmunozp3
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21-2 Chapter 21

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: N A = 6.02 × 1023 atoms/mol. 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
⎛ 17.7 g ⎞
N A n = (6.02 × 1023 atoms/mol) ⎜ 22
⎟ = 5.41 × 10 atoms. The number of protons is
⎝ 197 g/mol ⎠
np = (79 protons/atom)(5.41×1022 atoms) = 4.27 ×1024 protons.
Q = (np )(1.60 × 10−19 C/proton) = 6.83 × 105 C.
(b) The number of electrons is ne = np = 4.27 × 1024.
EVALUATE: The total amount of positive charge in the ring is very large, but there is an equal amount of
negative charge.
21.5. IDENTIFY: Each ion carries charge as it enters the axon.
SET UP: The total charge Q is the number N of ions times the charge of each one, which is e. So Q = Ne,
where e = 1.60 × 10−19 C.
EXECUTE: The number N of ions is N = (5.6 × 1011ions/m)(1.5 × 10−2 m) = 8.4 × 109 ions. The total charge
Q carried by these ions is Q = Ne = (8.4 × 109 )(1.60 × 10−19 C) = 1.3 × 10−9 C = 1.3 nC.
EVALUATE: The amount of charge is small, but these charges are close enough together to exert large
forces on nearby charges.
21.6. IDENTIFY: Apply Coulomb’s law and calculate the net charge q on each sphere.
SET UP: The magnitude of the charge of an electron is e = 1.60 × 10−19 C.
1 q2
EXECUTE: F =
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4π ⑀0 r 2
. This gives

q = 4π ⑀0 Fr 2 = 4π ⑀ 0 (4.57 × 10−21 N)(0.200 m) 2 = 1.43 × 10−16 C. And therefore, the total
number of electrons required is n = q /e = (1.43 × 10−16 C)/(1.60 × 10−19 C/electron) = 890 electrons.
EVALUATE: Each sphere has 890 excess electrons and each sphere has a net negative charge. The two like
charges repel.
k q1q2
21.7. IDENTIFY: Apply F = and solve for r.
r2
SET UP: F = 650 N.
k q1q2 (8.99 × 109 N ⋅ m 2 /C2 )(1.0 C) 2
EXECUTE: r = = = 3.7 × 103 m = 3.7 km
F 650 N
EVALUATE: Charged objects typically have net charges much less than 1 C.
21.8. IDENTIFY: Use the mass of a sphere and the atomic mass of aluminum to find the number of aluminum
atoms in one sphere. Each atom has 13 electrons. Apply Coulomb’s law and calculate the magnitude of
charge q on each sphere.
SET UP: N A = 6.02 × 1023 atoms/mol. q = n′ee, where n′e is the number of electrons removed from one
sphere and added to the other.
EXECUTE: (a) The total number of electrons on each sphere equals the number of protons.
⎛ 0.0250 kg ⎞ 24
ne = np = (13)( N A ) ⎜ ⎟ = 7.25 × 10 electrons.
⎝ 0.026982 kg/mol ⎠

No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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Electric Charge and Electric Field 21-3

1 q2
(b) For a force of 1.00 × 104 N to act between the spheres, F = 1.00 × 104 N = . This gives
4π ⑀0 r 2
q = 4π ⑀ 0 (1.00 × 104 N)(0.800 m) 2 = 8.43 × 10−4 C. The number of electrons removed from one sphere
and added to the other is n′e = q /e = 5.27 × 1015 electrons.
(c) n′e /n e = 7.27 × 10−10.
EVALUATE: When ordinary objects receive a net charge the fractional change in the total number of
electrons in the object is very small.
21.9. IDENTIFY: Apply Coulomb’s law.
SET UP: Consider the force on one of the spheres.
(a) EXECUTE: q1 = q2 = q
1 q1q2 q2 F 0.220 N
F= = so q = r = 0.150 m = 7.42 × 10−7 C (on each)
4π ⑀0 r 2
4π ⑀0r 2 (1/4π ⑀0 ) 8.988 × 109 N ⋅ m 2 /C2
(b) q2 = 4q1
1 q1q2 4q12 F F
F= = so q1 = r = 12 r = 1 (7.42 × 10−7 C) = 3.71 × 10−7 C.
4π ⑀0 r 2
4π ⑀0r 2 4(1/4π ⑀0 ) (1/4π ⑀ 0 ) 2
And then q2 = 4q1 = 1.48 × 10−6 C.
EVALUATE: The force on one sphere is the same magnitude as the force on the other sphere, whether the
spheres have equal charges or not.
21.10. IDENTIFY: We first need to determine the number of charges in each hand. Then we can use Coulomb’s
law to find the force these charges would exert on each hand.
SET UP: One mole of Ca contains N A = 6.02 × 1023 atoms. Each proton has charge e = 1.60 × 10−19 C.

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The force each hand exerts on the other is F = k
r2
.
q2

EXECUTE: (a) The mass of one hand is (0.010)(75 kg) = 0.75 kg = 750 g . The number of moles of Ca is
750 g
n= = 18.7 mol. The number of atoms is
40.18 g/mol
N = nN A = (18.7 mol)(6.02 × 1023 atoms/mol) = 1.12 × 1025 atoms.
(b) Each Ca atom contains positive charge 20e. The total positive charge in each hand is
N e = (1.12 × 1025 )(20)(1.60 × 10−19 C) = 3.58 × 107 C. If 1.0% is unbalanced by negative charge, the net
positive charge of each hand is q = (0.010)(3.58 × 107 C) = 3.6 × 105 C.
(c) The repulsive force each hand exerts on the other would be
q2 (3.6 × 105 C) 2
F = k 2 = (8.99 × 109 N ⋅ m 2 /C2 ) = 4.0 × 1020 N. This is an immense force; our hands
r (1.7 m) 2
would fly off.
EVALUATE: Ordinary objects contain a very large amount of charge. But negative and positive charge is
present in almost equal amounts and the net charge of a charged object is always a very small fraction of
the total magnitude of charge that the object contains.
qq
21.11. IDENTIFY: Apply F = ma, with F = k 1 22 .
r
SET UP: a = 25.0 g = 245 m/s 2 . An electron has charge − e = −1.60 × 10−19 C.
EXECUTE: F = ma = (8.55 × 10−3 kg)(245 m/s 2 ) = 2.09 N. The spheres have equal charges q, so
q2 F 2.09 N
F =k 2
and q = r = (0.150 m) = 2.29 × 10−6 C.
r k 8.99 × 109 N ⋅ m 2 /C2

No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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21-4 Chapter 21

q 2.29 × 10−6 C
N= = = 1.43 × 1013 electrons. The charges on the spheres have the same sign so the
e 1.60 × 10−19 C
electrical force is repulsive and the spheres accelerate away from each other.
EVALUATE: As the spheres move apart the repulsive force they exert on each other decreases and their
acceleration decreases.
21.12. IDENTIFY: We need to determine the number of protons in each box and then use Coulomb’s law to
calculate the force each box would exert on the other.
SET UP: The mass of a proton is 1.67 × 10−27 kg and the charge of a proton is 1.60 × 10−19 C. The
q1 q2
distance from the earth to th...

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