Chapter 21 - 21.1: a) n e b) n lead m lead 8 . 00 g and...

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21.1: C 10 3.20 charge and g 00 . 8 9 lead - × - = = m a) . 10 0 . 2 C 10 6 . 1 C 10 20 . 3 10 19 9 e × = × - × - = - - n b) . 10 58 . 8 and 10 33 . 2 207 g 00 . 8 13 lead e 22 lead - × = × = × = n n N n A 21.2: s 10 s 100 and s C 000 , 20 current 4 - = = = μ t Q = It = 2.00 C . 10 25 . 1 C 10 60 . 1 19 19 e × = × = - Q n 21.3: The mass is primarily protons and neutrons of 27 10 67 . 1 - × = m kg, so: 28 27 n and p 10 19 . 4 kg 10 1.67 kg 70.0 × = × = - n About one-half are protons, so e 28 p 10 10 . 2 n n = × = and the charge on the electrons is given by: . C 10 35 . 3 ) 10 10 . 2 ( C) 10 60 . 1 ( 9 28 19 × = × × × = - Q 21.4: Mass of gold = 17.7 g and the atomic weight of gold is 197 mol. g So the number of atoms ( 29 . 10 41 . 5 ) 10 02 . 6 ( mol 22 mol g 197 g 7 . 17 23 × = × × = × A N a) 24 22 p 10 27 . 4 10 41 . 5 79 × = × × = n C 10 6.83 C 10 60 . 1 5 19 p × = × × = - n q b) . 10 27 . 4 24 p e × = = n n 21.5: . electrons 10 1.08 atoms H 10 6.02 1.80 mol 80 . 1 24 23 × = × × = C. 10 1.73 C 10 60 . 1 10 1.08 charge 5 19 24 × - = × × × - = - 21.6: First find the total charge on the spheres: C 10 43 . 1 ) 2 . 0 )( 10 57 . 4 ( 4 4 4 1 16 2 21 0 2 0 2 2 0 - - × = × = = = πε Fr πε q r q πε F And therefore, the total number of electrons required is 890. C 10 1.60 C 10 43 . 1 19 16 = × × = = - - e q n
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21.7: a) Using Coulomb’s Law for equal charges, we find: . C 10 42 . 7 C 10 5.5 m) 150 . 0 ( 4 1 N 220 . 0 7 2 13 2 2 0 - - × = × = = = q q πε F b) When one charge is four times the other, we have: C 10 71 . 3 C 10 375 . 1 m) 150 . 0 ( 4 4 1 N 220 . 0 7 2 13 2 2 0 - - × = × = = = q q πε F So one charge is 7 10 71 . 3 - × C, and the other is C. 10 484 . 1 6 - × 21.8: a) The total number of electrons on each sphere equals the number of protons. . 10 25 . 7 mol kg 026982 . 0 kg 0250 . 0 13 24 p e × = × × = = A N n n b) For a force of 4 10 00 . 1 × N to act between the spheres, . C 10 43 . 8 m) 08 . 0 ( N) 10 ( 4 4 1 N 10 4 2 4 0 2 2 0 4 - × = = = = πε q r q πε F 15 e 10 27 . 5 × = = e q n c) 10 e 10 7.27 is - × n of the total number. 21.9: The force of gravity must equal the electric force. . m 08 . 5 m 8 . 25 ) s m 8 . 9 ( kg) 10 11 . 9 ( C) 10 60 . 1 ( 4 1 4 1 2 31 2 19 0 2 2 2 0 = = × × = = - - r πε r r q πε mg 21.10: a) Rubbing the glass rod removes electrons from it, since it becomes positive. kg. 10 27 . 4 ) electron kg 10 11 . 9 ( electrons) 10 4.69 ( electrons 10 69 . 4 ) C electrons 10 25 . 6 ( C) 10 7.50 ( nC 50 . 7 20 31 10 10 18 9 - - - × = × × × = × × = The rods mass decreases by kg. 10 27 . 4 20 - × b) The number of electrons transferred is the same, but they are added to the mass of the plastic rod, which increases by kg. 10 27 . 4 20 - ×
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21.11: positive. is and direction - the in be must so direction, - the in is 1 1 2 q x x - + F F ( 29 nC 750 . 0 0400 . 0 0200 . 0 , 2
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Chapter 21 - 21.1: a) n e b) n lead m lead 8 . 00 g and...

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