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Chapter 23 Solutions
  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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