This preview shows pages 1–3. 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: Reilly (mar3978) 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, 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. ball Z is neutral and ball X is negative. 2. balls Z and X are both positive, but ball X carries more charge than ball Z . 3. ball Z is negative and ball X is positive. correct 4. ball Z is negative and ball X is neutral. 5. balls Z and X are still uncharged. 6. ball Z is neutral and ball X is positive. 7. ball Z is positive and ball X is negative. 8. balls Z and X are both negative. 9. balls Z and X are both positive, but ball Z carries more charge than ball X . 10. ball Z is positive and ball X is neutral. 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 Three identical point charges hang from three strings, as shown. 45 45 F g 20.0 cm 20.0 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 . 32152 10 6 C. Explanation: Let : m = 0 . 10 kg , L = 20 . 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 . Reilly (mar3978) 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 = (20 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 . 32152 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 97 . 3 mN, as shown in the figure below....
View
Full
Document
This note was uploaded on 12/01/2010 for the course PHY 60230 taught by Professor Turner,j during the Spring '10 term at University of Texas at Austin.
 Spring '10
 TURNER,J

Click to edit the document details