HW 05-solutions - cooper(mlc4285 HW 05 florin(55245 This...

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cooper (mlc4285) – HW 05 – florin – (55245) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Three small spheres carry equal amounts of electric charge. They are equally spaced and lie along the same line. + + What is the direction of the net electric force on each charge due to the other charges? 1. + + 2. + + correct 3. + + 4. + + 5. + + 6. + + 7. + + 8. + + 9. + + 10. + + Explanation: Since like charges repel and unlike charge attract, + + 002(part1of2)10.0points Object A and object B are initially uncharged and separated by a distance of 2 meters. Sup- pose 10,000 electrons are removed from object A and placed on object B, creating an electric force between A and B. The electric force is 1. repulsive. 2. attractive. correct 3. zero. Explanation: Object A becomes positively charged be- cause negatively charged electrons are re- moved. B becomes negatively charged be- cause electrons are added. Since opposite charges attract, object A and B experience an attractive force. 003(part2of2)10.0points An additional 10,000 electrons are removed from A and placed on B. By what factor does the electric force change? Correct answer: 4. Explanation: When an additional 10,000 electrons are moved from A to B the charge on each object doubles. F e = k e q 1 q 2 r 2 q 1 q 2 . The factor is (2 q 1 ) (2 q 2 ) = 4 ( q 1 q 2 ) .
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cooper (mlc4285) – HW 05 – florin – (55245) 2 004 10.0points Four point charges are placed at the four cor- ners of a square. Each side of the square has a length L . q 1 = q q 3 = q q 2 = q q 4 = q P L L Find the magnitude of the electric force on q 2 due to all three charges q 1 , q 3 and q 4 for L = 1 m and q = 1 . 77 μ C. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 0 . 0422356 N. Explanation: F 1 F 4y F 4 F 3 F 4x q 2 From the figure, F 1 x = k q 2 L 2 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × (1 . 77 × 10 6 C) 2 (1 m) 2 = 0 . 0281571 N , F 3 y = k q 2 L 2 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × (1 . 77 × 10 6 C) 2 (1 m) 2 = 0 . 0281571 N , F 4 x = k q 2 2 L 2 1 2 = 8 . 98755 × 10 9 N · m 2 / C 2 2 · (1 m) 2 × (1 . 77 × 10 6 C) 2 2 = 0 . 00995504 N , and F 4 y = k q 2 2 L 2 1 2 = 8 . 98755 × 10 9 N · m 2 / C 2 2 · (1 m) 2 × (1 . 77 × 10 6 C) 2 2 = 0 . 00995504 N . For the net force, F x = F 1 x + F 4 x = 0 . 0281571 N + 0 . 00995504 N = 0 . 0182021 N and F y = F 3 y + F 4 y = 0 . 0281571 N + ( 0 . 00995504 N) = 0 . 0381121 N , so bardbl vector F bardbl 2 = F 2 x + F 2 y = ( 0 . 0182021 N) 2 + ( 0 . 0381121 N) 2 = 0 . 00178385 N 2 bardbl vector F bardbl = 0 . 00178385 N 2 = 0 . 0422356 N .
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