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# solution_pdf - kinsey(cbk323 oldhomework 01 Turner(56725...

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kinsey (cbk323) – oldhomework 01 – Turner – (56725) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two uncharged metal balls, X and Z , stand on insulating glass rods. A third ball, carrying a negative charge, is brought near the ball Z as shown in the figure. A conducting wire is then run between X and Z and then removed. Finally the third ball is removed. Z X conducting wire When all this is finished 1. ball X is positive and ball Z is negative. 2. ball X is negative and ball Z is positive. correct 3. ball X is negative and ball Z is neutral. 4. ball X is neutral and ball Z is negative. 5. balls X and Z are both positive, but ball X carries more charge than ball Z . 6. ball X is neutral and ball Z is positive. 7. balls X and Z are both positive, but ball Z carries more charge than ball X . 8. balls X and Z are both negative. 9. ball X is positive and ball Z is neutral. 10. balls X and Z are still uncharged. Explanation: When the conducting wire is run between X and Z , some negative charge flows from Z to X under the influence of the negative charge of the third ball. Therefore, after the wire is removed, X is charged negative and Z is charged positive. 002 10.0 points Three identical point charges hang from three strings, as shown. 45 45 F g 28.0 cm 28.0 cm + + + +q +q +q 0.10 kg 0.10 kg 0.10 kg What is the value of q ? The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 , and the acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 1 . 85013 × 10 6 C. Explanation: Let : m = 0 . 10 kg , L = 28 . 0 cm , θ = 45 , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . r = 2 L sin θ = 2 L sin 45 = 2 L 2 2 = L 2 F T,x = F T sin θ F T,y = F T cos θ Each sphere is in equilibrium horizontally F electric F T,x = 0 F electric F T sin θ = 0 and vertically F T,y F g = 0 F T cos θ F g = 0 F T = F g cos θ .

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kinsey (cbk323) – oldhomework 01 – Turner – (56725) 2 From the horizontal equilibrium, F electric = parenleftbigg F g cos θ parenrightbigg sin θ F electric = F g tan θ = F g (tan 45 ) = F g . For either of the end charges, F electric = k e q 2 r 2 + k e q 2 ( r 2 ) 2 = k e q 2 r 2 + 4 k e q 2 r 2 = 5 k e q 2 r 2 5 k e q 2 r 2 = m g . Thus | q | = radicalBigg r 2 m g 5 k e = radicalBigg ( L 2) 2 m g 5 k e = L · radicalbigg 2 m g 5 k e = (28 cm) parenleftbigg 1 m 100 cm parenrightbigg × radicalBigg 2(0 . 1 kg)(9 . 81 m / s 2 ) 5(8 . 98755 × 10 9 N · m 2 / C 2 ) = 1 . 85013 × 10 6 C . 003 (part 1 of 2) 10.0 points Two identical small metal spheres with q 1 > 0 and | q 1 | > | q 2 | attract each other with a force of magnitude 65 . 2 mN, as shown in the figure below. F 21 F 12 2 . 45 m q 1 q 2 r 1 = 20 μ m r 2 = 20 μ m Figure 1: Before touching The spheres are then brought together until they are touching. At this point, the spheres are in electrical contact so that the charges can move from one sphere to the other until both spheres have the same final charge, q .
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solution_pdf - kinsey(cbk323 oldhomework 01 Turner(56725...

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