Solution 1 - homework 01 KANUNGO ANIL Due Sep 4 2007 4:00...

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homework 01 – KANUNGO, ANIL – Due: Sep 4 2007, 4:00 am 1 Question 1, chap 22, sect 1. part 1 of 1 10 points A plastic rod is rubbed against a wool shirt, thereby acquiring a charge of 4 μ C. The charge on an electron is 1 . 6 × 10 19 C. How many electrons are transferred from the wool shirt to the plastic rod? Correct answer: 2 . 5 × 10 13 (tolerance ± 1 %). Explanation: Let : q = 4 μ C = 4 × 10 6 C and e = 1 . 6 × 10 19 C . The total charge is q = n e e n e = q e = 4 × 10 6 C 1 . 6 × 10 19 C = 2 . 5 × 10 13 . Question 2, chap 22, sect 2. part 1 of 1 10 points A particle of mass 26 g and charge 44 μ C is released from rest when it is 83 cm from a second particle of charge 22 μ C. Determine the magnitude of the initial ac- celeration of the 26 g particle. Correct answer: 485 . 719 m / s 2 (tolerance ± 1 %). Explanation: Let : m = 26 g = 0 . 026 kg , q = 44 μ C = 4 . 4 × 10 5 C , r = 83 cm = 0 . 83 m , Q = 22 μ C = 2 . 2 × 10 5 C , and k e = 8 . 9875 × 10 9 N · m 2 / C 2 . The force exerted on the particle is F = k e | q | | Q | r 2 = m | a | | a | = k e | q | | Q | m r 2 = (8 . 9875 × 10 9 N · m 2 / C 2 ) × vextendsingle vextendsingle 4 . 4 × 10 5 C vextendsingle vextendsingle vextendsingle vextendsingle 2 . 2 × 10 5 C vextendsingle vextendsingle (0 . 026 kg) (0 . 83 m) 2 = 485 . 719 m / s 2 . Question 3, chap 22, sect 2. part 1 of 2 10 points Three charges are arranged in the ( x, y ) plane (as shown in the figure below, where the scale is in meters). 9 nC 9 nC 3 nC y (m) 0 1 2 3 4 5 6 7 8 9 10 x (m) 0 1 2 3 4 5 6 7 8 9 10 What is the magnitude of the resulting force on the 9 nC charge at the origin? Correct answer: 12 . 1332 nN (tolerance ± 1 %). Explanation: Let : q o = 9 nC = 9 × 10 9 C , ( x o , y o ) = (0 m , 0 m) , q a = 9 nC = 9 × 10 9 C , ( x a , y a ) = (10 m , 0 m) , q b = 3 nC = 3 × 10 9 C , and ( x b , y b ) = (0 m , 5 m) . The 9 nC charge is on the x -axis, so r oa = x a and Coulomb’s Law for q o and q a is F oa x = k e q o q a x 2 a = (8 . 98755 × 10 9 N C 2 / m 2 ) × (9 × 10 9 C) (9 × 10 9 C) (10 m) 2 = 7 . 27992 × 10 9 N .
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