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Unformatted text preview: Kapoor (mk9499) – oldmidterm 01 – Turner – (60230) 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two spheres, fastened to “pucks”, are rid ing on a frictionless airtrack. Sphere “1” is charged with 1 nC, and sphere “2” is charged with 3 nC. Both objects have the same mass. 1 nC is equal to 1 × 10 − 9 C. As they repel, 1. sphere “2” accelerates 3 times as fast as sphere “1”. 2. sphere “1” accelerates 9 times as fast as sphere “2”. 3. sphere “2” accelerates 9 times as fast as sphere “1”. 4. they have the same magnitude of acceler ation. correct 5. they do not accelerate at all, but rather separate at constant velocity. 6. sphere “1” accelerates 3 times as fast as sphere “2”. Explanation: The force of repulsion exerted on each mass is determined by F = 1 4 π ǫ Q 1 Q 2 r 2 = ma where r is the distance between the centers of the two spheres. bardbl vector F 12 bardbl = bardbl vector F 21 bardbl Since both spheres have the same mass and are subject to the same force, they have the same acceleration. 002 10.0 points Three equal charges of 3 μ C are in the x y plane. One is placed at the origin, another is placed at (0.0, 99 cm), and the last is placed at (85 cm, 0.0). The Coulomb constant is 9 × 10 9 N m 2 / C 2 . Calculate the magnitude of the force on the charge at the origin. Correct answer: 0 . 13928 N. Explanation: Let : q = 3 μ C = 3 × 10 − 6 C , ( x 1 , y 1 ) = (0 , 99 cm) = (0 , . 99 m) , ( x 2 , y 2 ) = (85 cm , 0) = (0 . 85 m , 0) and ( x , y ) = (0 , 0) . The electric fields are E y = k e q x 2 1 + y 2 1 = ( 9 × 10 9 N m 2 / C 2 )( 3 × 10 − 6 C ) 0 + (0 . 99 m) 2 = 27548 . 2 N / C , and E x = k e q x 2 2 + y 2 2 = ( 9 × 10 9 N m 2 / C 2 )( 3 × 10 − 6 C ) (0 . 85 m) 2 + 0 = 37370 . 2 N / C , Thus E = radicalBig E 2 x + E 2 y and F = q E = q radicalBig E 2 x + E 2 y = ( 3 × 10 − 6 C ) · radicalBig (27548 . 2 N / C) 2 + (37370 . 2 N / C) 2 = . 13928 N . 003 10.0 points Two identical small charged spheres hang in equilibrium with equal masses as shown in the figure. The length of the strings are equal and the angle (shown in the figure) with the vertical is identical. The acceleration of gravity is 9 . 8 m / s 2 and the value of Coulomb’s constant is 8 . 98755 × 10 9 N m 2 / C 2 . Kapoor (mk9499) – oldmidterm 01 – Turner – (60230) 2 . 1 5 m 5 ◦ . 03 kg . 03 kg Find the magnitude of the charge on each sphere. Correct answer: 4 . 4233 × 10 − 8 C. Explanation: Let : L = 0 . 15 m , m = 0 . 03 kg , and θ = 5 ◦ . L a θ m m q q From the right triangle in the figure above, we see that sin θ = a L ....
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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.
 Fall '08
 Turner
 Physics

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