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Unformatted text preview: patel (dp23452) – Homework 1 – Spurlock – (14423) 1 This print-out should have 43 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Calculate the net charge on a substance con- sisting of a combination of 8 . 1 × 10 13 protons and 7 . 6 × 10 13 electrons. Correct answer: 8 × 10 − 7 C. Explanation: Let: N p = 8 . 1 × 10 13 charges , q p = 1 . 60 × 10 − 19 C , N e = 7 . 6 × 10 13 charges and q e = − 1 . 60 × 10 − 19 C . Q net = N p q p + N e q e Q net = (8 . 1 × 10 13 )(1 . 6 × 10 − 19 C) + (7 . 6 × 10 13 )( − 1 . 6 × 10 − 19 C) = 8 × 10 − 7 C 002 10.0 points A particle of mass 11 g and charge 44 μ C is released from rest when it is 81 cm from a second particle of charge − 18 μ C. Determine the magnitude of the initial ac- celeration of the 11 g particle. Correct answer: 986 . 283 m / s 2 . Explanation: Let : m = 11 g , q = 44 μ C = 4 . 4 × 10 − 5 C , d = 81 cm = 0 . 81 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 . 4 × 10 − 5 C vextendsingle vextendsingle vextendsingle vextendsingle − 1 . 8 × 10 − 5 C vextendsingle vextendsingle (0 . 011 kg) (0 . 81 m 2 ) = 986 . 283 m / s 2 . 003 10.0 points Three charges are arranged in the ( x, y ) plane (as shown in the figure below, where the scale is in meters). − 6 nC 8 nC − 6 nC y ( m ) 0 1 2 3 4 5 6 7 8 9 10 x ( m ) 1 2 3 4 5 6 7 8 9 10 What is the magnitude of the resulting force on the − 6 nC charge at the origin? Correct answer: 44 . 9378 nN. Explanation: Let : q o = − 6 × 10 − 9 C , ( x o , y o ) = (0 m , 0 m) , q a = 8 × 10 − 9 C , ( x a , y a ) = (4 m , 0 m) , q b = − 6 × 10 − 9 C , and ( x b , y b ) = (0 m , 3 m) . Coulomb’s Law for q o and q a is bardbl vector F oa bardbl = k e q o q a radicalbig ( x a − x o ) 2 + ( y a − y o ) 2 = k e q o q a radicalbig x 2 oa + y 2 oa = 8 . 98755 × 10 9 N C 2 / m 2 patel (dp23452) – Homework 1 – Spurlock – (14423) 2 × ( − 6 × 10 − 9 C) (8 × 10 − 9 C) radicalbig (4 m) 2 + (0 m) 2 = − 2 . 69627 × 10 − 8 N . F oa x = − k e q o q a r 2 oa cos θ oa = − k e q o q a r 2 oa x oa r oa = − (8 . 98755 × 10 9 N C 2 / m 2 ) × ( − 6 × 10 − 9 C) (8 × 10 − 9 C) (4 m) 2 × (4 m) (4 m) = 2 . 69627 × 10 − 8 N . F oa y = − k e q o q a r 2 oa sin θ oa = − k e q o q oa r 2 oa y oa r oa = − (8 . 98755 × 10 9 N C 2 / m 2 ) × ( − 6 × 10 − 9 C) (8 × 10 − 9 C) (4 m) 2 × (0 m) (4 m) = 0 N ....
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This note was uploaded on 02/15/2010 for the course PHYS 1442 taught by Professor O'donnell during the Spring '08 term at UT Arlington.

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solution_pdf - patel (dp23452) – Homework 1 – Spurlock...

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