hw02b - aljabr(faa335 Hw02 Ross(89251 This print-out should...

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aljabr (faa335) – Hw02 – Ross – (89251) 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 A sphere is charged with electrons to - 1 × 10 5 C. The charge of an electron is - 1 . 6 × 10 19 C. How many electrons are there on the sphere? Correct answer: 6 . 25 × 10 13 . Explanation: Let : q = - 1 × 10 5 C and q e = - 1 . 6 × 10 19 C . The total charge is q = n q e n = q q e = - 1 × 10 5 C - 1 . 6 × 10 19 C = 6 . 25 × 10 13 . 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 Coulomb’s constant is 8 . 98755 × 10 9 N m 2 / C 2 . 0 . 21 m 7 0 . 02 kg 0 . 02 kg Find the magnitude of the charge on each sphere. Correct answer: 8 . 37574 × 10 8 C. Explanation: Let : L = 0 . 21 m , m = 0 . 02 kg , and θ = 7 . L a θ m m q q From the right triangle in the figure above, we see that sin θ = a L . Therefore a = L sin θ = (0 . 21 m) sin(7 ) = 0 . 0255926 m . The separation of the spheres is r = 2 a = 0 . 0511851 m . The forces acting on one of the spheres are shown in the figure below. θ θ m g F T e T sin θ T cos θ 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 θ - F e = 0 summationdisplay F y = T cos θ - m g = 0 . From the second equation in the system above, we see that T = m g cos θ , so T can be eliminated from the first equation if we make this substitution. This gives a value F e = m g tan θ = (0 . 02 kg) ( 9 . 8 m / s 2 ) tan(7 ) = 0 . 0240658 N ,
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aljabr (faa335) – Hw02 – Ross – (89251) 2 for the electric force.
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