HW #1-solutions - zapata(jz883 – HW#1 – Antoniewicz...

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Unformatted text preview: zapata (jz883) – HW #1 – Antoniewicz – (56495) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A particle of mass 63 g and charge 46 μ C is released from rest when it is 78 cm from a second particle of charge − 18 μ C. Determine the magnitude of the initial ac- celeration of the 63 g particle. Correct answer: 194 . 151 m / s 2 . Explanation: Let : m = 63 g , q = 46 μ C = 4 . 6 × 10 − 5 C , d = 78 cm = 0 . 78 m , Q = − 18 μ C = − 1 . 8 × 10 − 5 C , and k e = 8 . 9875 × 10 9 . The force exerted on the particle is F = k e | q 1 || q 2 | r 2 = ma bardbl vectora bardbl = k e bardbl vectorq bardblbardbl vector Q bardbl md 2 = k e vextendsingle vextendsingle 4 . 6 × 10 − 5 C vextendsingle vextendsingle vextendsingle vextendsingle − 1 . 8 × 10 − 5 C vextendsingle vextendsingle (0 . 063 kg) (0 . 78 m 2 ) = 194 . 151 m / s 2 . 002 10.0 points Two small silver spheres, each with a mass of 75 g, are separated by 0 . 92 m. The num- ber of electrons per atom of silver is 47, and the number of atoms per gram is Avogadro’s number divided by the molar mass of silver, 107 . 87 g / mol. Calculate the fraction of the electrons in one sphere that must be transferred to the other in order to produce an attrac- tive force of 10000 N (about a ton) be- tween the spheres. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 and the Avogadro’s number is 6 . 02214 × 10 23 mol − 1 . Correct answer: 3 . 07785 × 10 − 10 . Explanation: Let : m = 75 g , r = 0 . 92 m , n Ag = 47 , M Ag = 107 . 87 g / mol , f = 10000 N , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . The force between the spheres is F = k e q 2 r 2 q = r radicalbigg F k e . The total charge is q = nq e n transferred = q q e = r q e radicalbigg F k e = . 92 m 1 . 60218 × 10 − 19 C × radicalBigg 10000 N 8 . 98755 × 10 9 N · m 2 / C 2 = 6 . 05699 × 10 15 where q e is the charge of one electron. There are n atoms = m M Ag N A atoms in each sphere, where N A is Avogadro’s number, so the total number of electrons is n electrons = n Ag n atoms = n Ag mN A M Ag = 47(75 g) 107 . 87 g / mol × (6 . 02214 × 10 23 mol − 1 ) = 1 . 96793 × 10 25 . Thus the ratio is R = n transferred n electrons = 6 . 05699 × 10 15 1 . 96793 × 10 25 = 3 . 07785 × 10 − 10 . zapata (jz883) – HW #1 – Antoniewicz – (56495) 2 The fraction of electrons needed to create a significant force is very tiny. When we talk about the “charge” of macro- scopic objects (like the silver spheres in this problem), we generally mean the surplus charge, which corresponds to only a tiny frac- tion of the total number of electrons in the material....
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This note was uploaded on 10/01/2010 for the course PHYSICS PHY 303 L taught by Professor Antoniewicz during the Spring '10 term at Texas Pan American.

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HW #1-solutions - zapata(jz883 – HW#1 – Antoniewicz...

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