q3_sol - . 5 / (6 . 02 10 23 ) = 10 . 5 10-23 g = 1 . 05...

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Quiz 3 Phys. 2306 F09 CRN 94850 You may use your text book, notes, and a calculator. In order to illustrate that matter is electrically neutral to a very high degree, consider two Cu spheres of radii 0.1 cm that are 1 m apart. Thus they look like point charges. Even if the charge magnitudes of the proton and electron differ by 1 part in 10 8 , a measureable force would be observed, as below. You may use the following: the atomic number of Cu is 29, the atomic mass is 63.5 g/mol, the mass density is 8.9 g/cm 3 , Avogadro’s number is 6 . 02 × 10 23 mole - 1 , the proton charge is e= 1 . 60 × 10 - 19 C, and 1 4 π± 0 = 8 . 988 × 10 9 N(m/C) 2 . 1) What is the mass of each sphere in kg? a) 0.37 b)0 . 37 × 10 - 1 c)0 . 37 × 10 - 3 d)none of these M(sphere)=mass density x volume = (8 . 9)((4 / 3) π 0 . 1 3 ) g = 0 . 037 × 10 - 3 kg. So the correct choice is (d) none of these. 2) What is the mass of 1 Cu atom in kg? a) 1.1 b) 1 . 1 × 10 - 26 c) 1 . 1 × 10 - 13 d) none of these From the atomic mass, 1 mole of Cu has a mass of 63.5 g. However, Avogadro’s num- ber is the number of Cu atoms/mole. So, m(atom)= 63
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Unformatted text preview: . 5 / (6 . 02 10 23 ) = 10 . 5 10-23 g = 1 . 05 10-25 kg. So the correct choice is (d) none of these. 3) How many Cu atoms are in each sphere? a) 3 . 5 10 18 b)3 . 5 10 20 c) 3 . 5 10 22 d)none of these N(atoms)m(atom)=M(sphere), so N= 0 . 037 10-3 / 1 . 05 10-25 = 3 . 5 10 20 atoms, (b). 1 4) If the electron charge is =-(1 . 00-10-8 )e, what is the net charge on a sphere in C? a) 1 . 6 10-3 b) 1 . 6 10-5 c) 1 . 6 10-1 d) none of these Since the atomic number is 29, that is the number of protons and electrons. Thus, the total charge on the sphere is (3 . 5 10 20 )(29)[1 . 00-(1 . 00-10-8 )](1 . 60 10-19 ) = 1 . 62 10-5 = 1 . 6 10-5 C, (b). 5) What force would one sphere exert on the other in N? a) 2.4 b) 2 . 4 10 3 c) 2 . 4 10 5 d) none of these Use Coulombs law, F= ( q/D ) 2 4 = 8 . 988 10 9 (1 . 62 10-5 / 1) 2 = 2 . 36 = 2 . 4 N, (a). 2...
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q3_sol - . 5 / (6 . 02 10 23 ) = 10 . 5 10-23 g = 1 . 05...

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