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Unformatted text preview: vo (ytv59) – oldhomework 01 – Turner – (56705) 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 Three identical point charges hang from three strings, as shown. 45 ◦ 45 ◦ F g 12.0 cm 12.0 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: 7 . 92912 × 10 − 7 C. Explanation: Let : m = 0 . 10 kg , L = 12 . 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 . 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 = 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 = (12 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 ) = 7 . 92912 × 10 − 7 C . 002 10.0 points There are two identical small metal spheres with charges 82 . 4 μ C and − 50 . 9712 μ C. The distance between them is 6 cm. The spheres are placed in contact then set at their original distance. Calculate the magnitude of the force between the two spheres at the final position. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 616 . 502 N. Explanation: vo (ytv59) – oldhomework 01 – Turner – (56705) 2 Let : q A = 82 . 4 μ C = 8 . 24 × 10 − 5 C , q B = − 50 . 9712 μ C = − 5 . 09712 × 10 − 5 C , and d = 6 cm = 0 . 06 m . When the spheres are in contact, the charges will rearrange themselves until equi- librium is reached. Each sphere will then have half of the original total charge: q = q A + q B 2 = 8 . 24 × 10 − 5 C − 5 . 09712 × 10 − 5 C 2 = 1 . 57144 × 10 − 5 C . The force between the two spheres is F = k q 2 d 2 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × (1 . 57144 × 10 − 5 C) 2 (0 . 06 m) 2 = 616 . 502 N . 003 10.0 points The diagram shows an isolated, positive charge Q , where point B is twice as far away from Q as point A . + Q A B 10 cm 20 cm The ratio of the electric field strength at point A to the electric field strength at point B is 1....
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This note was uploaded on 09/12/2010 for the course PHY 303L taught by Professor Turner during the Fall '10 term at Austin Community College.

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