Homework Key Ch.1

# Homework Key Ch.1 - villegas(bmv547 Homework 1...

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villegas (bmv547) – Homework 1 – spurlock – (42004) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A strong lightning bolt transfers about 25 C to Earth. How many electrons are transferred? The elementary charge is 1 . 60218 × 10 19 C. Correct answer: 1 . 56038 × 10 20 . Explanation: Let : q = 25 C and q e = 1 . 60218 × 10 19 C . The charge is proportional to the number of electrons, so q = n | q e | n = q | q e | = 25 C | − 1 . 60218 × 10 19 C | = 1 . 56038 × 10 20 . 002(part1of2)10.0points Assume that you have two objects, one with a mass of 5 kg and the other with a mass of 16 kg, each with a charge of 0 . 015 C and separated by a distance of 9 m. What is the electric force that these objects exert on one another? Correct answer: 25000 N. Explanation: Let : q 1 = q 2 = 0 . 015 C , d = 9 m , and k e = 9 × 10 9 N · m 2 / C 2 . F e = k e q 1 q 2 d 2 = ( 9 × 10 9 N · m 2 / C 2 ) ( 0 . 015 C) 2 (9 m) 2 = 25000 N . 003(part2of2)10.0points What is the gravitational force between them? Correct answer: 6 . 61728 × 10 11 N. Explanation: Let : m 1 = 5 kg , m 2 = 16 kg , and G = 6 . 7 × 10 11 N · m 2 / kg 2 . F = G m 1 m 2 d 2 = ( 6 . 7 × 10 11 N · m 2 / kg 2 ) × (5 kg) (16 kg) (9 m) 2 = 6 . 61728 × 10 11 N . 004 10.0points Two electrons exert a force of repulsion of 0 . 99 N on each other. How far apart are they? The elementary charge is 1 . 602 × 10 19 C and Coulomb’s con- stant is 8 . 987 × 10 9 N · m 2 / C 2 . Correct answer: 1 . 52634 × 10 14 m. Explanation: Let : F e = 0 . 99 N and q 3 = 1 . 602 × 10 19 C . The electric force is F e = k e q 1 q 2 d 2 d 2 = k e q 2 e F e d = q e radicalbigg k e F e = (1 . 602 × 10 19 C) × radicalbigg 8 . 987 × 10 9 N · m 2 / C 2 0 . 99 N = 1 . 52634 × 10 14 m .

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villegas (bmv547) – Homework 1 – spurlock – (42004) 2 005 10.0points A particle with charge 7 μ C is located on the x -axis at the point 8 cm , and a second particle with charge 6 μ C is placed on the x -axis at 8 cm . 2 4 6 8 10 2 4 6 8 10 7 μ C 6 μ C 4 μ C x (cm) What is the magnitude of the total elec- trostatic force on a third particle with charge 4 μ C placed on the x -axis at 2 cm ? The Coulomb constant is 8 . 9875 × 10 9 N · m 2 / C 2 . Correct answer: 34 . 7517 N. Explanation: Let : q 1 = 7 μ C = 7 × 10 6 C , q 2 = 6 μ C = 6 × 10 6 C , q 3 = 4 μ C = 4 × 10 6 C , x 1 = 8 cm = 0 . 08 m , x 2 = 8 cm = 0 . 08 m , and x 3 = 2 cm = 0 . 02 m . Coulomb’s law (in vector form) for the elec- tric force exerted by a charge q 1 on a second charge q 3 , is vector F 13 = k e q 1 q 3 r 2 ˆ r 13 , where ˆ r 13 is a unit vector directed from q 1 to q 3 ; i.e. , vectorr 13 = vectorr 3 vectorr 1 . x 13 = x 3 x 1 = ( 2 cm) (8 cm) = 0 . 1 m x 23 = x 3 x 2 = ( 2 cm) ( 8 cm) = 0 . 06 m ˆ x 13 = x 3 x 1 radicalbig ( x 3 x 1 ) 2 = ˆ ı ˆ x 23 = x 3 x 2 radicalbig ( x 3 x 2 ) 2 = +ˆ ı Since the forces are collinear, the force on the third particle is the algebraic sum of the forces between the first and third and the second and third particles: vector F = vector F 13 + vector F 23 = k e bracketleftbigg q 1 r 2 13 ˆ r 13 + q 2 r 2 23 ˆ r 23 bracketrightbigg q 3 = 8 . 9875 × 10 9 N · m 2 / C 2 × bracketleftbigg (7 × 10 6 C) ( 0 . 1 m) 2 ( ˆ ı ) + (6 ×

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