CH 15 - Electric Forces & Fields-solutions

CH 15 - Electric Forces & Fields-solutions - This print-out...

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Unformatted text preview: This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points We want to find how much charge is on the electrons in a nickel coin. Follow this method. A nickel coin has a mass of about 5 . 5 g. Each mole (6 . 02 × 10 23 atoms) has a mass of about 58 . 2 g. Find the number of atoms in a nickel coin. Correct answer: 5 . 689 × 10 22 atoms. Explanation: Let : N a = 6 . 02 × 10 23 atoms , M = 58 . 2 g , and m = 5 . 5 g . Mass is proportional to the number of atoms in a substance, so for m grams in N atoms in the nickel coin and M grams in N a atoms in one mole, we have m M = N N a N = m M N a = (5 . 5 g) (58 . 2 g) (6 . 02 × 10 23 atoms) = 5 . 689 × 10 22 atoms . 002 (part 2 of 3) 10.0 points Find the number of electrons in the coin. Each nickel atom has 28 electrons / atom. Correct answer: 1 . 59292 × 10 24 electrons. Explanation: Let : n Ni = 28 electrons / atom . If n Ni electrons are in each Nickel atom, then the total number of electrons n e in the coin is n e = N n Ni = ( 5 . 689 × 10 22 atoms ) × (28 electrons / atom) = 1 . 59292 × 10 24 electrons . 003 (part 3 of 3) 10.0 points Find the magnitude of the charge of all these electrons. Correct answer: 2 . 55214 × 10 5 C. Explanation: Let : q e =- 1 . 60218 × 10- 19 C / electron . The total charge q for the n e electrons is q = n e q e = (1 . 59292 × 10 24 electrons) × (- 1 . 60218 × 10- 19 C / electron ) =- 2 . 55214 × 10 5 C , which has a magnitude of 2 . 55214 × 10 5 C . 004 10.0 points A particle of mass 86 g and charge 24 μ C is released from rest when it is 67 cm from a second particle of charge- 12 μ C. Determine the magnitude of the initial ac- celeration of the 86 g particle. Correct answer: 67 . 0476 m / s 2 . Explanation: Let : m = 86 g , q = 24 μ C = 2 . 4 × 10- 5 C , d = 67 cm = 0 . 67 m , Q =- 12 μ C =- 1 . 2 × 10- 5 C , and k e = 8 . 9875 × 10 9 ....
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This note was uploaded on 10/03/2010 for the course PH 102 taught by Professor Staff during the Fall '08 term at Alabama.

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CH 15 - Electric Forces & Fields-solutions - This print-out...

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