Solucionario De Serway Vol 2 Sexta Edicion

Páginas: 344 (85908 palabras) Publicado: 28 de octubre de 2012
© 2000 by Harcourt, Inc. All rights reserved.
Chapter 23 Solutions
23.1 (a)
N = 10.0 grams
107.87 grams mol



 6.02 × 1023 atoms
mol



 47.0
electrons
atom



 = 2.62 × 1024
(b)
# electrons added = Q
e
= 1.00 × 10−3 C
1.60 × 10-19 C electron
= 6.25 × 1015
or 2.38 electrons for every 109 already present
23.2 (a)
Fe =
ke q1q2
r2 =
8.99 × 109 N ⋅ m2/C2( )1.60 × 10−19 C ( )2
(3.80 × 10− 10 m)2 = 1.59 × 10−9 N (repulsion)
(b)
Fg =
Gm1m2
r2 =
6.67 × 10−11 N ⋅ m2 kg2 ( )(1.67 × 10− 27 kg)2
(3.80 × 10−10 m)2 = 1.29 × 10− 45 N
The electric force is larger by 1.24 × 1036 times
(c) If
ke
q1q2
r2 = G
m1m2
r2 with q1 = q2 = q and m1 = m2 = m, then
q
m
=
G
ke
=
6.67 × 10−11 N ⋅ m2 /kg2
8.99 × 109 N ⋅ m2 /C2 = 8.61 × 10−11 C/kg
23.3If each person has a mass of ≈ 70 kg and is (almost) composed of water, then each person
contains
N ≈
70,000 grams
18 grams mol



 6.02 × 1023 molecules
mol



 10
protons
molecule



 ≈2.3 × 1028 protons
With an excess of 1% electrons over protons, each person has a charge
q = (0.01)(1.6 × 10−19 C)(2.3 × 1028) = 3.7 × 107 C
So
F = ke
q1q2
r2 = (9 × 109 )(3.7 × 107 )2
0.62 N = 4 × 1025 N ~ 1026 N
This force is almost enough to lift a "weight" equal to that of the Earth:
Mg = (6 × 1024 kg)(9.8 m s2) = 6 × 1025 N~ 1026 N
2 Chapter 23 Solutions
23.4 We find the equal-magnitude charges on both spheres:
F = ke
q1q2
r 2 = ke
q2
r 2 so
q = r
F
ke
= 1.00 m ( ) 1.00 × 104 N
8.99 × 109 N ⋅m2/C2 = 1.05 × 10−3 C
The number of electrontransferred is then
Nxfer = 1.05 × 10−3 C ( )1.60 × 10−19 C/e− ( )= 6.59 × 1015 electrons
The whole number of electrons in each sphere is
Ntot = 10.0 g
107.87 g /mol



 6.02 × 1023 atoms/mol ( )47 e− /atom ( )= 2.62 × 1024 e−
The fraction transferred is then
f =
Nxfer
Ntot
=





 6.59 × 1015
2.62 × 1024 = 2.51 × 10–9 = 2.51 charges in every billion
23.5
F = ke
q1q2
r2=
8.99 × 109 N⋅m2 C2 ( )1.60 × 10−19 C ( )2
6.02 × 1023 ( )2
2(6.37 × 106 m) [ ]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 = keq1q2
r2 = 8.99 × 109
N⋅m2
C2




12.0 × 10−9 C ( )18.0 × 10−9 C ( )
(0.300 m)2 = 2.16 × 10− 5 N
(b) The net charge of − × − 6 00 10 9 . C willbe equally split between the two spheres, or
− 3.00 × 10−9 C on each. The force is one of repulsion, and its magnitude is
F = keq1q2
r2 = 8.99 × 109
N⋅m2
C2




3.00 × 10−9 C ( )3.00 × 10−9 C ( )
(0.300 m)2 = 8.99 × 10−7 N
Chapter 23 Solutions 3
© 2000 by Harcourt, Inc. All rights reserved.
23.7
F1 = ke
q1q2
r2 = (8.99 × 109 N ⋅m2/C2 )(7.00 × 10−6 C)(2.00 × 10−6 C)
(0.500m)2 = 0.503 N
F2 = ke
q1q2
r2 = (8.99 × 109 N ⋅m2 /C2 )(7.00 × 10−6 C)(4.00 × 10−6 C)
(0.500 m)2 = 1.01 N
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 point charges are located at the corners of an equilateral triangle as shown in Figure P23.7.
Calculate the net electric force onthe 7.00− µC charge.
G : Gather Information: The 7.00− µC charge experiences a repulsive force F1 due to the 2.00− µC
charge, and an attractive force F2 due to the −4.00− µC charge, where F2 = 2F1. If we sketch these
force vectors, we find that the resultant appears to be about the same magnitude as F2 and is
directed to the right about 30.0° below the horizontal.
O : Organize : We can findthe net electric force by adding the two separate forces acting on the
7.00− µC charge. These individual forces can be found by applying Coulomb’s law to each pair of
charges.
A : Analyze: The force on the 7.00− µC charge by the 2.00− µC charge is
F1 =
8.99 × 109 N ⋅m2/C2 ( )7.00 × 10−6 C ( )2.00 × 10−6 C ( )
0.500 m ( )2 cos60°i + sin60°j ( )= F1 = 0.252i + 0.436j ( ) N
Similarly, the...
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