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Unformatted text preview: kinsey (cbk323) oldhomework 01 Turner (56725) 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 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 , = 45 , and k e = 8 . 98755 10 9 N m 2 / C 2 . r = 2 L sin = 2 L sin 45 = 2 L 2 2 = L 2 F T,x = F T sin F T,y = F T cos Each sphere is in equilibrium horizontally F electric F T,x = 0 F electric F T sin = 0 and vertically F T,y F g = 0 F T cos F g = 0 F T = F g cos . kinsey (cbk323) oldhomework 01 Turner (56725) 2 From the horizontal equilibrium, F electric = parenleftbigg F g cos parenrightbigg sin F electric = F g tan = F g (tan45 ) = 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 = mg . Thus  q  = radicalBigg r 2 mg 5 k e = radicalBigg ( L 2) 2 mg 5 k e = L radicalbigg 2 mg 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 > and  q 1  >  q 2  attract each other with a force of magnitude 65 . 2 mN, as shown in the figure below....
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This note was uploaded on 11/02/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Charge, Work

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