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Unformatted text preview: taherisefat (mt23852) – hw 4 – Opyrchal – (121012) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two charges are located along the xaxis. One has a charge of 6 . 5 μ C, and the second has a charge of − 3 . 5 μ C. The Coulomb constant is k e = 8 . 98755 × 10 9 N m 2 / C 2 . The acceleration of gravity is 9 . 81 m / s 2 . If the electrical potential energy associated with the pair of charges is − 5 . 5 × 10 2 J, what is the distance between the charges? Correct answer: 3 . 71859 m. Explanation: Let : q 1 = 6 . 5 μ C , q 2 = − 3 . 5 μ C , U electric = − 5 . 5 × 10 2 J , and k e = 8 . 99 × 10 9 N · m 2 / C 2 . U electric = k e q 1 q 2 r r = k e q 1 q 2 U electric = (8 . 99 × 10 9 N · m 2 / C 2 ) × (6 . 5 × 10 6 C)( − 3 . 5 × 10 6 C) − 5 . 5 × 10 2 J = 3 . 71859 m . 002 (part 1 of 2) 10.0 points An object with a charge 9 C and a mass . 2 kg accelerates from rest to a speed of 23 m / s. Calculate the kinetic energy gained. Correct answer: 52 . 9 J. Explanation: Let : m = 0 . 2 kg and v = 23 m / s . The kinetic energy is K = 1 2 mv 2 = 1 2 (0 . 2 kg) (23 m / s) 2 = 52 . 9 J . 003 (part 2 of 2) 10.0 points Through how large a potential difference did the object fall? Correct answer: 5 . 87778 V. Explanation: Let : q = 9 C . The potential difference is Δ V = K q = 52 . 9 J 9 C = 5 . 87778 V . 004 10.0 points A uniform electric field of magnitude 365 V / m is directed in the positive xdirection. Sup pose a 20 μ C charge moves from the origin to point A at the coordinates, (32 cm, 58 cm)....
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This note was uploaded on 01/31/2011 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.
 Spring '08
 moro
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