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Unformatted text preview: Niu (qn269) Homework #1 antoniewicz (58855) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A particle of mass 85 g and charge 39 C is released from rest when it is 19 cm from a second particle of charge 14 C. Determine the magnitude of the initial ac celeration of the 85 g particle. Correct answer: 1599 . 21 m / s 2 . Explanation: Let : m = 85 g , q = 39 C = 3 . 9 10 5 C , d = 19 cm = 0 . 19 m , Q = 14 C = 1 . 4 10 5 C , and k e = 8 . 9875 10 9 . The force exerted on the particle is F = k e  q 1  q 2  r 2 = ma bardbl vectora bardbl = k e bardbl vectorq bardblbardbl vector Q bardbl md 2 = k e vextendsingle vextendsingle 3 . 9 10 5 C vextendsingle vextendsingle vextendsingle vextendsingle 1 . 4 10 5 C vextendsingle vextendsingle (0 . 085 kg) (0 . 19 m 2 ) = 1599 . 21 m / s 2 . 002 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 Coulombs constant is 8 . 98755 10 9 N m 2 / C 2 . . 1 5 m 6 . 03 kg . 03 kg Find the magnitude of the charge on each sphere. Correct answer: 5 . 81459 10 8 C. Explanation: Let : L = 0 . 15 m , m = 0 . 03 kg , and = 6 . L a m m q q From the right triangle in the figure above, we see that sin = a L . Therefore a = L sin = (0 . 15 m) sin(6 ) = 0 . 0156793 m . The separation of the spheres is r = 2 a = . 0313585 m . The forces acting on one of the spheres are shown in the figure below. m g F T e T sin T cos Because the sphere is in equilibrium, the resultant of the forces in the horizontal and vertical directions must separately add up to zero: summationdisplay F x = T sin F e = 0 summationdisplay F y = T cos mg = 0 . Niu (qn269) Homework #1 antoniewicz (58855) 2 From the second equation in the system above, we see that T = mg cos , so T can be eliminated from the first equation if we make this substitution. This gives a value F e = mg tan = (0 . 03 kg) ( 9 . 8 m / s 2 ) tan(6 ) = 0 . 0309006 N , for the electric force. From Coulombs law, the electric force be tween the charges has magnitude  F e  = k e  q  2 r 2 , where  q  is the magnitude of the charge on each sphere. Note: The term  q  2 arises here because the charge is the same on both spheres. This equation can be solved for  q  to give  q  = radicalBigg  F e  r 2 k e = radicalBigg (0 . 0309006 N) (0 . 0313585 m) 2 (8 . 98755 10 9 N m 2 / C 2 ) = 5 . 81459 10 8 C ....
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This note was uploaded on 11/18/2009 for the course PHY 58855 taught by Professor Antoniewicz during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Antoniewicz
 Charge, Mass, Work

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