This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: johnson (rj6247) – hw 4 – Opyrchal – (121014) 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 . 2 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 − 4 . 7 × 10 2 J, what is the distance between the charges? Correct answer: 4 . 1507 m. Explanation: Let : q 1 = 6 . 2 μ C , q 2 = − 3 . 5 μ C , U electric = − 4 . 7 × 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 . 2 × 10 6 C)( − 3 . 5 × 10 6 C) − 4 . 7 × 10 2 J = 4 . 1507 m . 002 (part 1 of 2) 10.0 points An object with a charge 9 C and a mass . 1 kg accelerates from rest to a speed of 17 m / s. Calculate the kinetic energy gained. Correct answer: 14 . 45 J. Explanation: Let : m = 0 . 1 kg and v = 17 m / s . The kinetic energy is K = 1 2 mv 2 = 1 2 (0 . 1 kg) (17 m / s) 2 = 14 . 45 J . 003 (part 2 of 2) 10.0 points Through how large a potential difference did the object fall? Correct answer: 1 . 60556 V. Explanation: Let : q = 9 C . The potential difference is Δ V = K q = 14 . 45 J 9 C = 1 . 60556 V . 004 10.0 points A uniform electric field of magnitude 276 V / m is directed in the positive xdirection. Sup pose a 12 μ C charge moves from the origin to point A at the coordinates, (37 cm, 52 cm)....
View
Full
Document
This note was uploaded on 03/11/2010 for the course PHYS 121 taught by Professor Opyrchal during the Spring '08 term at NJIT.
 Spring '08
 Opyrchal
 Charge, Work

Click to edit the document details