Phy 303l - Kapoor (mk9499) oldhomework 01 Turner (60230)...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Kapoor (mk9499) – oldhomework 01 – Turner – (60230) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd 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 positive charge, is brought near the ball X as shown in the fgure. A conducting wire is then run between Z and X and then removed. ±inally the third ball is removed. X Z + conducting wire When all this is fnished 1. ball Z is positive and ball X is neutral. 2. ball Z is negative and ball X is neutral. 3. balls Z and X are both positive. 4. balls Z and X are both negative, but ball X carries more charge than ball Z . 5. ball Z is negative and ball X is positive. 6. ball Z is positive and ball X is negative. correct 7. balls Z and X are still uncharged. 8. balls Z and X are both negative, but ball Z carries more charge than ball X . 9. ball Z is neutral and ball X is positive. 10. ball Z is neutral and ball X is negative. Explanation: When the conducting wire is run between Z and X , some positive charge ²ows From X to Z under the in²uence oF the positive charge oF the third ball. ThereFore, aFter the wire is removed, Z is charged positive and X is charged negative. 002 10.0 points Three identical point charges hang From three strings, as shown. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 , and the acceleration oF gravity is 9 . 81 m / s 2 . 45 45 F g 15.0 cm 15.0 cm + + + +q +q +q 0.10 kg 0.10 kg 0.10 kg What is the value oF q ? Correct answer: 9 . 9114 × 10 7 C. Explanation: Let : m = 0 . 10 kg , L = 15 . 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 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Kapoor (mk9499) – oldhomework 01 – Turner – (60230) 2 From the horizontal equilibrium, F electric = p F g cos θ P sin θ F electric = F g tan θ = 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 | = r r 2 mg 5 k e = r ( L 2) 2 mg 5 k e = L · R 2 mg 5 k e = (15 cm) p 1 m 100 cm P × r 2(0 . 1 kg)(9 . 81 m / s 2 ) 5(8 . 98755 × 10 9 N · m 2 / C 2 ) = 9 . 9114 × 10 7 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 55 . 6 mN, as shown in the ±gure below.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

Page1 / 8

Phy 303l - Kapoor (mk9499) oldhomework 01 Turner (60230)...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online