Chapter_21_Solutions

Chapter_21_Solutions - 21.1 mlead = 8.00 g and charge...

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21.1: m lead = 8.00 g and charge =− 3.20 × 10 9 C a) n e = 3.20 × 10 9 C 1.6 × 10 19 C = 2.0 × 10 10 . b) n lead = N A × 8.00 g 207 = 2.33 × 10 22 and n e n lead = 8.58 × 10 13

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21.2: current = 20,000 C s and t = 100 µ s = 10 4 s Q = It = 2.00 C n e = Q 1.60 × 10 19 C = 1.25 × 10 19 .
21.3: The mass is primarily protons and neutrons of m = 1.67 × 10 27 kg, so: n p and n = 70.0 kg 1.67 × 10 27 kg = 4.19 × 10 28 About one-half are protons, so n and the charge on the electrons is given by: Q p = 2.10 × 10 28 = n e = (1.60 × 10 19 C) × (2.10 × 10 28 ) = 3.35 × 10 9 C.

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21.4: Mass of gold = 17.7 g and the atomic weight of gold is 197 g mol. So the number of atoms = N A × mol = (6.02 × 10 23 ) × 17.7 g 197 g mol ( ) = 5.41 × 10 22 . a) n p = 79 × 5.41 × 10 22 = 4.27 × 10 24 q = n p × 1.60 × 10 19 C = 6.83 × 10 5 C b) n e = n p = 4.27 × 10 24 .
21.5: 1.80 mol = 1.80 × 6.02 × 10 23 H atoms = 1.08 × 10 24 electrons. charge =− 1.08 × 10 24 × 1.60 × 10 19 C 1.73 × 10 5 C.

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21.6: First find the total charge on the spheres: F = 1 4 πε 0 q 2 r 2 q = 4 0 Fr 2 = 4 0 (4.57 × 10 21 )(0.2) 2 = 1.43 × 10 16 C And therefore, the total number of electrons required is n = qe = 1.43 × 10 16 C 1.60 × 10 19 C = 890.
21.7: a) Using Coulomb’s Law for equal charges, we find: F = 0.220 N = 1 4 πε 0 q 2 (0.150 m) 2 q = 5.5 × 10 13 C 2 = 7.42 × 10 7 C. b) When one charge is four times the other, we have: F = 0.220 N = 1 4 0 4q 2 (0.150 m) 2 q = 1.375 × 10 13 C 2 = 3.71 × 10 7 C So one charge is 3.71 C, and the other is 1.484 × 10 7 × 10 6 C.

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21.8: a) The total number of electrons on each sphere equals the number of protons. n e = n p = 13 × N A × 0.0250 kg 0.026982 kg mol = 7.25 × 10 24 . b) For a force of 1.00 N to act between the spheres, × 10 4 F = 10 4 N = 1 4 πε 0 q 2 r 2 q = 4 0 (10 4 N) (0.80 m) 2 = 8.43 × 10 4 C. ⇒′ n e = qe = 5.27 × 10 15 c) of the total number. n e is 7.27 × 10 10
21.9: The force of gravity must equal the electric force. mg = 1 4 πε 0 q 2 r 2 r 2 = 1 4 0 (1.60 × 10 19 C) 2 (9.11 × 10 31 kg)(9.8 m s ) = 25.8 m 2 r = 5.08 m.

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21.10: a) Rubbing the glass rod removes electrons from it, since it becomes positive. 7.50 nC = (7.50 × 10 9 C) (6.25 × 10 18 electrons C ) = 4.69 × 10 10 electrons (4.69 × 10 10 electrons) (9.11 × 10 31 kg electron ) = 4.27 × 10 20 kg. The rods mass decreases by 4.27 × 10 20 kg. b) The number of electrons transferred is the same, but they are added to the mass of the plastic rod, which increases by 4.27 × 10 20
21.11: r F 2 is in the + x-direction, so r F 1 must be in the x-direction and q 1 is positive. F 1 = F 2 , k q 1 q 3 r 13 2 = k q 2 q 3 r 23 2 q 1 = 0.0200 0.0400 () 2 q 2 = 0.750 nC 1

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