University Physics 12e Solutions (Ch. 21)

University Physics Vol 2 (Chapters 21-37) (12th Edition)

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E LECTRIC C HARGE AND E LECTRIC F IELD 21.1. (a) IDENTIFY and SET UP: Use the charge of one electron 19 ( 1.602 10 C) −× to find the number of electrons required to produce the net charge. EXECUTE: The number of excess electrons needed to produce net charge q is 9 10 19 3.20 10 C 2.00 10 electrons. 1.602 10 C/electron q e == × −− × (b) IDENTIFY and SET UP: Use the atomic mass of lead to find the number of lead atoms in 3 8.00 10 kg × of lead. From this and the total number of excess electrons, find the number of excess electrons per lead atom. EXECUTE: The atomic mass of lead is 3 207 10 kg/mol, × so the number of moles in is 3 8.00 10 kg × 3 tot 3 8.00 10 kg 0.03865 mol. 207 10 kg/mol m n M × = × A N (Avogadro’s number) is the number of atoms in 1 mole, so the number of lead atoms is The number of excess electrons per lead atom is 23 22 A (0.03865 mol)(6.022 10 atoms/mol) = 2.328 10 atoms. Nn N × × 10 13 22 2.00 10 electrons 8.59 10 . 2.328 10 atoms × × EVALUATE: Even this small net charge corresponds to a large number of excess electrons. But the number of atoms in the sphere is much larger still, so the number of excess electrons per lead atom is very small. 21.2. IDENTIFY: The charge that flows is the rate of charge flow times the duration of the time interval. SET UP: The charge of one electron has magnitude 19 1.60 10 C. e EXECUTE: The rate of charge flow is and 20,000 C/s 4 100 s 1.00 10 s. t µ × The number of electrons is 4 (20,000 C/s)(1.00 10 s) 2.00 C. Q = 19 e 19 1.25 10 . 1.60 10 C Q n × × EVALUATE: This is a very large amount of charge and a large number of electrons. 21.3. IDENTIFY: From your mass estimate the number of protons in your body. You have an equal number of electrons. SET UP: Assume a body mass of 70 kg. The charge of one electron is 19 1.60 10 C. EXECUTE: The mass is primarily protons and neutrons of The total number of protons and neutrons is 27 1.67 10 kg. m 28 p and n 27 70 kg 4.2 10 . 1.67 10 kg n × × About one-half are protons, so 28 p e 2.1 10 nn = ×= . The number of electrons is about The total charge of these electrons is 28 2.1 10 . × 19 28 9 ( 1.60 10 C/electron)(2.10 10 electrons) 3.35 10 C. Q =− × × × EVALUATE: This is a huge amount of negative charge. But your body contains an equal number of protons and your net charge is zero. If you carry a net charge, the number of excess or missing electrons is a very small fraction of the total number of electrons in your body. 21.4. IDENTIFY: Use the mass m of the ring and the atomic mass M of gold to calculate the number of gold atoms. Each atom has 79 protons and an equal number of electrons. SET UP: . A proton has charge + e . 23 A 6.02 10 atoms/mol N EXECUTE: The mass of gold is 17.7 g and the atomic weight of gold is 197 gm o l . So the number of atoms is 23 22 A 17.7 g (6.02 10 atoms/mol) 5.41 10 atoms 197 g mol ⎛⎞ ⎜⎟ ⎝⎠ . The number of protons is . 22 24 p (79 protons/atom)(5.41 10 atoms) 4.27 10 protons n = × 19 5 p ( )(1.60 10 C/proton) 6.83 10 C Qn = × .
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This note was uploaded on 05/05/2008 for the course PHYSICS 240 taught by Professor Davewinn during the Spring '08 term at University of Michigan.

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University Physics 12e Solutions (Ch. 21) - ELECTRIC CHARGE...

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