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# 2 - kavo(ak22862 HW02 Janow(12121 This print-out should...

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kavo (ak22862) – HW02 – Janow – (12121) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 15.0 points Forces between two objects which are in- versely proportional to the square of the dis- tance between the objects include which of the following? I) Gravitational force between two celestial bodies II) Electrostatic force between two electrons III) Nuclear force between two neutrons. 1. III only 2. II and III only 3. I only 4. I, II, and III 5. I and II only correct Explanation: I) Gravitational force 1 r 2 . II) Electrostatic force 1 r 2 . III) The nuclear force is very strong com- pared to the gravitational force and electro- static force over a very short range; outside of that range, however, it is negligible and cannot be proportional to 1 r 2 002 15.0 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 . 08 m 6 0 . 04 kg 0 . 04 kg Find the magnitude of the charge on each sphere. Correct answer: 3 . 58086 × 10 8 C. Explanation: Let : L = 0 . 08 m , m = 0 . 04 kg , and θ = 6 . 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 . 08 m) sin(6 ) = 0 . 00836228 m . The separation of the spheres is r = 2 a = 0 . 0167246 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

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kavo (ak22862) – HW02 – Janow – (12121) 2 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 . 04 kg) ( 9 . 8 m / s 2 ) tan(6 ) = 0 . 0412009 N , for the electric force. From Coulomb’s 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 . 0412009 N) (0 . 0167246 m) 2 (8 . 98755 × 10 9 N m 2 / C 2 ) = 3 . 58086 × 10 8 C . 003 (part 1 of 2) 15.0 points Three charges are arranged as shown in the figure. The Coulomb cosnstant is 8 . 98755 × 10 9 N · m 2 / C 2 . x y 3 . 9 nC 4 . 1 m 0 . 6 nC 1 . 2 nC 1 . 9 m Find the magnitude of the electrostatic force on the charge at the origin. Correct answer: 2 . 18596 nN. Explanation: Let : ( x 0 , y 0 ) = (0 m , 0 m) , q 0 = 0 . 6 nC , ( x 1 , y 1 ) = (4 . 1 m , 0 m) , q 1 = 3 . 9 nC , ( x 2 , y 2 ) = (0 m , 1 . 9 m) , and q 2 = 1 . 2 nC .
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2 - kavo(ak22862 HW02 Janow(12121 This print-out should...

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