Solucionario Serway Sexta Edic.
10.0 grams electrons 24 23 atoms 47.0 = 2.62 × 10 N= 6.02 × 10 atom mol 107.87 grams mol # electrons added = Q 1.00 × 10 −3 C = = 6.25 × 1015 e 1.60 × 10 -19 C electron
23.1
(a)
(b)
or
2.38 electrons for every 10 9 already present
23.2
(a)
8.99 × 10 9 N ⋅ m 2/ C 2 1.60 × 10 −19 C k q1q Fe = e 2 2 = r (3.80 × 10 − 10 m)2
()(
)
2
= 1.59 × 10 − 9 N
(repulsion)
(b)
6.67 × 10 −11 N ⋅ m 2 kg 2 (1.67 × 10 − 27 kg)2 G m1m2 Fg = = = 1.29 × 10 − 45 N r2 (3.80 × 10 −10 m)2 The electric force is larger by 1.24 × 10 36 times
(
)
(c)
If ke q = m
q1q2 mm = G 12 2 2 r r G = ke
with q1 = q2 = q and m1 = m2 = m, then
6.67 × 10 −11 N ⋅ m 2 / kg 2 = 8.61 × 10 −11 C / kg 8.99 × 10 9 N ⋅ m 2 /C 2
23.3
If each person has a mass of ≈ 70 kg and is (almost) composed of water, then each person contains 70, 000 grams protons 23 molecules 10 N≈ ≈ 2.3 × 10 28 protons 6.02 × 10 molecule mol 18 grams mol With an excess of 1% electrons over protons, each person has a charge q = (0.01)(1.6 × 10 −19 C)(2.3 × 10 28 ) = 3.7 × 107 C So F = ke q1q2 (3.7 × 107 )2 = (9 ×10 9 ) N = 4 × 10 25 N ~ 1026 N r2 0.6 2
This force is almost enough to lift a "weight" equal to that of the Earth: Mg = (6 × 10 24 kg)(9.8 m s 2 ) = 6 × 10 25 N ~ 1026 N
© 2000 by Harcourt, Inc. All rights reserved.
2
Chapter 23 Solutions
23.4
We find the equal-magnitude charges on both spheres: F = ke q1q2 q2 = ke 2 2 r r so q=r F 1.00 × 10 4 N = (1.00 m ) = 1.05 × 10 −3 C ke8.99 × 10 9 N ⋅ m 2/ C 2
The number of electron transferred is then N xfer = 1.05 × 10 −3 C
(
) (1.60 × 10 (
−19
C / e − = 6.59 × 1015 electrons
)
The whole number of electrons in each sphere is 10.0 g 23 − 24 − Ntot = 6.02 × 10 atoms / mol 47 e / atom = 2.62 × 10 e 107.87 g / mol
)(
)
The fraction transferred is then f= N xfer Ntot 6.59 × 1015 = =2.51 × 10–9 2.62 × 1024 = 2.51 charges in every billion
23.5
8.99 × 10 9 N ⋅ m 2 C 2 1.60 × 10 −19 C qq F = ke 1 2 2 = 2 r 2(6.37 × 106 m)
(
[
)(
]
) (6.02 × 10 )
2
23 2
= 514 kN
*23.6
(a)
The force is one of attraction. The distance r in Coulomb's law is the distance between centers. The magnitude of the force is F=
−9 −9 ke q1q2 N ⋅ m 2 12.0 × 10 C18.0 × 10 C = 8.99 × 10 9 = 2.16 × 10 − 5 N r2 C2 (0.300 m)2
(
)(
)
(b)
The net charge of − 6.00 × 10 −9 C will be equally split between the two spheres, or − 3.00 × 10 −9 C on each. The force is one of repulsion, and its magnitude is
2 3.00 × 10 −9 C 3.00 × 10 −9 C ke q1q2 9 N⋅m F= = 8.99 × 10 = r2 C2 (0.300 m)2
(
)(
)
8.99 × 10 −7 N
Chapter 23Solutions q1q2 (8.99 × 10 9 N ⋅ m 2/ C 2 )(7.00 × 10 −6 C)(2.00 × 10 −6 C) = = 0.503 N r2 (0.500 m)2 q1q2 (8.99 × 10 9 N ⋅ m 2 / C 2 )(7.00 × 10 −6 C)(4.00 × 10 −6 C) = = 1.01 N (0.500 m)2 r2
3
23.7
F1 = ke
F2 = k e
Fx = (0.503 + 1.01) cos 60.0° = 0.755 N Fy = (0.503 − 1.01) sin 60.0° = − 0.436 N F = (0.755 N)i − (0.436 N)j = 0.872 N at an angle of 330°
Goal Solution Three pointcharges are located at the corners of an equilateral triangle as shown in Figure P23.7. Calculate the net electric force on the 7.00− µ C charge. G: Gather Information: The 7.00− µ C charge experiences a repulsive force F1 due to the 2.00− µ C charge, and an attractive force F 2 due to the −4.00− µ C charge, where F2 = 2F1. If we sketch these force vectors, we find that the resultant appears to beabout the same magnitude as F2 and is directed to the right about 30.0° below the horizontal. Organize : We can find the net electric force by adding the two separate forces acting on the 7.00− µ C charge. These individual forces can be found by applying Coulomb’s law to each pair of charges. Analyze: The force on the 7.00− µ C charge by the 2.00− µ C charge is
9
O:
A:
F1
(8.99 × 10 =...
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