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Unformatted text preview: Version 055 – Exam 01 – Chiu – (58295) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two uncharged metal balls, Z and X , stand on insulating glass rods. A third ball, carrying a negative charge, is brought near the ball X as shown in the figure. A conducting wire is then run between Z and X and then removed. Finally the third ball is removed. X Z − conducting wire When all this is finished 1. balls Z and X are both positive, but ball Z carries more charge than ball X . 2. ball Z is neutral and ball X is negative. 3. ball Z is neutral and ball X is positive. 4. balls Z and X are still uncharged. 5. balls Z and X are both negative. 6. ball Z is positive and ball X is neutral. 7. ball Z is positive and ball X is negative. 8. ball Z is negative and ball X is positive. correct 9. ball Z is negative and ball X is neutral. 10. balls Z and X are both positive, but ball X carries more charge than ball Z . Explanation: When the conducting wire is run between Z and X , some negative charge flows from X to Z under the influence of the negative charge of the third ball. Therefore, after the wire is removed, Z is charged negative and X is charged positive. 002 10.0 points Two charged particles of equal magnitude (+ Q and + Q ) are fixed at opposite corners of a square that lies in a plane (see figure be low). A test charge − q is placed at a third corner. + Q − q + Q What is the direction of the force on the test charge due to the two other charges? 1. 2. correct 3. 4. 5. 6. 7. 8. Explanation: The force between charges of the same sign is repulsive. The force between charges with opposite signs is attractive. Version 055 – Exam 01 – Chiu – (58295) 2 + Q − q + Q The resultant force is the sum of the two vectors in the figure above. Therefore the correct choice is 003 10.0 points Four point charges are placed at the four cor ners of a square. Each side of the square has a length L . b b b b b 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 . Given L = 1 m and q = 1 . 35 μ C. 1. 0.0403483 2. 0.0249351 3. 0.0172143 4. 0.0512627 5. 0.0431955 6. 0.0245697 7. 0.0227834 8. 0.0507383 9. 0.0496976 10. 0.0375981 Correct answer: 0 . 0245697 N. Explanation: F 1 F 4y F 4 F 3 F 4x q 2 From the above figure, we see that F 1 = − k q 2 L 2 ˆ ı = − ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × (1 . 35 × 10 − 6 C) 2 (1 m) 2 = − . 0163798 N F 3 = − k q 2 L 2 ˆ = − ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × (1 . 35 × 10 − 6 C) 2 (1 m) 2 = − . 0163798 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 . 35 × 10 − 6 C) 2 √ 2 = 0 . 00579114 N 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 . 35 × 10 − 6 C) 2 √ 2 = − . 00579114 N bardbl vector F bardbl 2 = ( F 1...
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This note was uploaded on 04/26/2009 for the course PHY 58310 taught by Professor Chiu during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Chiu
 Physics, Charge

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