OHW 1 - terry(ect328 – oldhomework 01 – Turner...

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Unformatted text preview: terry (ect328) – oldhomework 01 – Turner – (59130) 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 positive 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 negative and ball Z is neutral. 2. balls X and Z are both negative, but ball Z carries more charge than ball X . 3. balls X and Z are still uncharged. 4. balls X and Z are both negative, but ball X carries more charge than ball Z . 5. ball X is negative and ball Z is positive. 6. balls X and Z are both positive. 7. ball X is positive and ball Z is negative. correct 8. ball X is neutral and ball Z is negative. 9. ball X is neutral and ball Z is positive. 10. ball X is positive and ball Z is neutral. Explanation: When the conducting wire is run between X and Z , some positive charge flows from Z to X under the influence of the positive charge of the third ball. Therefore, after the wire is removed, X is charged positive and Z 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 30.0 cm 30.0 cm + + + +q +q +q 0.10 kg 0.10 kg 0.10 kg What is the value of q ? Correct answer: 1 . 98228 × 10 − 6 C. Explanation: Let : m = 0 . 10 kg , L = 30 . 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 . terry (ect328) – oldhomework 01 – Turner – (59130) 2 From the horizontal equilibrium, F electric = parenleftbigg F g cos θ parenrightbigg 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 | = radicalBigg r 2 mg 5 k e = radicalBigg ( L √ 2) 2 mg 5 k e = L · radicalbigg 2 mg 5 k e = (30 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 . 98228 × 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 57 . 3 mN, as shown in the figure below....
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OHW 1 - terry(ect328 – oldhomework 01 – Turner...

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