chap21 - 21.1: C 10 3.20 charge and g 00 . 8 9 lead- - = =...

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Unformatted text preview: 21.1: C 10 3.20 charge and g 00 . 8 9 lead- - = = m a) . 10 . 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- = = = 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 . )( 10 57 . 4 ( 4 4 4 1 16 2 21 2 2 2-- = = = = Fr q r q F And therefore, the total number of electrons required is 890. C 10 1.60 C 10 43 . 1 19 16 = = =-- e q n 21.7: a) Using Coulombs Law for equal charges, we find: . C 10 42 . 7 C 10 5.5 m) 150 . ( 4 1 N 220 . 7 2 13 2 2-- = = = = q q F b) When one charge is four times the other, we have: C 10 71 . 3 C 10 375 . 1 m) 150 . ( 4 4 1 N 220 . 7 2 13 2 2-- = = = = q q 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 . kg 0250 . 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 . ( N) 10 ( 4 4 1 N 10 4 2 4 2 2 4- = = = = q r q 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 2 2 2 = = = =-- r r r q 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- 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 . 0400 . 0200 ....
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This homework help was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.

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chap21 - 21.1: C 10 3.20 charge and g 00 . 8 9 lead- - = =...

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