Old Final - Kapoor (mk9499) oldfinal 6 Turner (60230) 1...

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Unformatted text preview: Kapoor (mk9499) oldfinal 6 Turner (60230) 1 This print-out should have 33 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, each of mass 130 g and charge + q , hang from three strings, as in the figure. The acceleration of gravity is 9 . 8 m / s 2 , and the Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . 9 . 8 m / s 2 1 . 7 c m 130 g + q 1 . 7 c m 130 g + q 41 130 g + q If the lengths of the left and right strings are each 10 . 7 cm, and each forms an angle of 41 with the vertical, determine the value of q . Correct answer: 0 . 696975 C. Explanation: Let : = 41 , m = 130 g = 0 . 13 kg , L = 10 . 7 cm = 0 . 107 m , g = 9 . 8 m / s 2 , and k e = 8 . 98755 10 9 N m 2 / C 2 . Newtons 2nd law: summationdisplay F = ma. Electrostatic force between point charges q 1 and q 2 separated by a distance r F = k e q 1 q 2 r 2 . All three charges are in equilibrium, so for each holds summationdisplay F = 0 . Consider the forces acting on the charge on the right. There must be an electrostatic force F acting on this charge, keeping it balanced against the force of gravity mg . The electro- static force is due to the other two charges and is therefore horizontal. In the x-direction F T sin = 0 . In the y-direction T cos mg = 0 . These can be rewritten as F = T sin and mg = T cos . Dividing the former by the latter, we find F mg = tan , or F = mg tan (1) = (0 . 13 kg) (9 . 8 m / s 2 ) tan41 = 1 . 10747 N . The distance between the right charge and the middle charge is L sin , and the distance to the left one is twice that. Since all charges are of the same sign, both forces on the right charge are repulsive (pointing to the right). We can add the magnitudes F = k e q q ( L sin ) 2 + k e q q (2 L sin ) 2 or F = 5 k e q 2 4 L 2 sin 2 . (2) We have already found F , and the other quan- tities are given, so we solve for the squared charge q 2 q 2 = 4 F L 2 sin 2 5 k e (3) Kapoor (mk9499) oldfinal 6 Turner (60230) 2 or, after taking the square root (we know q > 0) and substituting F from Eq. 1 into Eq. 3 and solving for q , we have q = 2 L sin radicalbigg mg tan 5 k e = 2 (0 . 107 m) sin 41 radicalBigg (0 . 13 kg) (9 . 8 m / s 2 ) tan41 5 (8 . 98755 10 9 N m 2 / C 2 ) = 6 . 96975 10 7 C = . 696975 C . 002 10.0 points Two identical conducting spheres, A and B , carry equal charge. They are stationary and separated by a distance much larger than their diameters. A third identical conduct- ing sphere, C , is uncharged. Sphere C is first touched to A , then to B , and finally removed (to a far away distance)....
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Old Final - Kapoor (mk9499) oldfinal 6 Turner (60230) 1...

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