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# 1 - Copy (2) - tang(nmt352 oldhomework 01 Turner(56705 This...

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tang (nmt352) – oldhomework 01 – Turner – (56705) 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 Three identical point charges hang from three strings, as shown. 45 45 F g 40.0 cm 40.0 cm + + + +q +q +q 0.10 kg 0.10 kg 0.10 kg What is the value of q ? The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 , and the acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 2 . 64304 × 10 6 C. Explanation: Let : m = 0 . 10 kg , L = 40 . 0 cm , θ = 45 , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . r = 2 L sin θ = 2 L sin 45 = 2 L 2 2 = L 2 F T,x = F T sin θ F T,y = F T cos θ Each sphere is in equilibrium horizontally F electric F T,x = 0 F electric F T sin θ = 0 and vertically F T,y F g = 0 F T cos θ F g = 0 F T = F g cos θ . From the horizontal equilibrium, F electric = parenleftbigg F g cos θ parenrightbigg sin θ F electric = F g tan θ = F g (tan 45 ) = F g . For either of the end charges, F electric = k e q 2 r 2 + k e q 2 ( r 2 ) 2 = k e q 2 r 2 + 4 k e q 2 r 2 = 5 k e q 2 r 2 5 k e q 2 r 2 = m g . Thus | q | = radicalBigg r 2 m g 5 k e = radicalBigg ( L 2) 2 m g 5 k e = L · radicalbigg 2 m g 5 k e = (40 cm) parenleftbigg 1 m 100 cm parenrightbigg × radicalBigg 2(0 . 1 kg)(9 . 81 m / s 2 ) 5(8 . 98755 × 10 9 N · m 2 / C 2 ) = 2 . 64304 × 10 6 C . 002 (part 1 of 2) 10.0 points Three point-charges (+ q , q , and + q ) are placed at the vertices of an equilateral triangle (see figure below). a 60 + q + q q ˆ ı ˆ The magnitude of the electric force on the charge at the bottom left-hand vertex of the

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tang (nmt352) – oldhomework 01 – Turner – (56705) 2 triangle due to the other two charges is given by 1. bardbl vector F bardbl = 2 3 k q 2 a 2 2. bardbl vector F bardbl = 1 2 k q 2 a 2 3. bardbl vector F bardbl = 3 2 k q 2 a 2 4. bardbl vector F bardbl = 2 3 k q 2 a 2 5. bardbl vector F bardbl = 1 2 3 k q 2 a 2 6. bardbl vector F bardbl = 1 2 k q 2 a 2 7. bardbl vector F bardbl = 2 k q 2 a 2 8. bardbl vector F bardbl = 3 2 k q 2 a 2 9. bardbl vector F bardbl = k q 2 a 2 correct 10. bardbl vector F bardbl = 3 k q 2 a 2 Explanation: a + q + q q ˆ ı ˆ In this case, each of these forces has a mag- nitude F 21 = F 31 = k q 2 a 2 . The x -component of the net force is then F x = [ F 31 cos 60 + F 21 ] ˆ ı = bracketleftbigg parenleftbigg k q 2 a 2 parenrightbigg cos 60 + parenleftbigg k q 2 a 2 parenrightbiggbracketrightbigg ˆ ı = bracketleftbigg 1 2 parenleftbigg k q 2 a 2 parenrightbigg + parenleftbigg k q 2 a 2 parenrightbiggbracketrightbigg ˆ ı = + 1 2 k parenleftbigg q 2 a 2 parenrightbigg ˆ ı . On the other hand, the y -component is just F y = bracketleftbigg + parenleftbigg k q 2 a 2 parenrightbigg sin 60 bracketrightbigg ˆ = + 3 2 k parenleftbigg q 2 a 2 parenrightbigg ˆ  . Therefore the magnitude of the net force is bardbl vector F bardbl = radicalBig F 2 x + F 2 y = k radicalbigg 1 4 + 3 4 parenleftbigg q 2 a 2 parenrightbigg = k parenleftBig q a parenrightBig 2 .
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