HW1 - yang(ey942 oldhomework 01 Turner(56705 This print-out...

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yang (ey942) – oldhomework 01 – Turner – (56705) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points The diagram shows an isolated, positive charge Q , where point B is twice as Far away From Q as point A . + Q A B 0 10 cm 20 cm The ratio oF the electric feld strength at point A to the electric feld strength at point B is 1. E A E B = 1 1 . 2. E A E B = 1 2 . 3. E A E B = 8 1 . 4. E A E B = 4 1 . correct 5. E A E B = 2 1 . Explanation: Let : r B = 2 r A . The electric feld strength E 1 r 2 , so E A E B = 1 r 2 A 1 r 2 B = r 2 B r 2 A = (2 r ) 2 r 2 = 4 . 002 10.0 points Three identical point charges hang From three strings, as shown. 45 45 F g 14.0 cm 14.0 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: 9 . 25064 × 10 7 C. Explanation: Let : m = 0 . 10 kg , L = 14 . 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 . ±rom the horizontal equilibrium, F electric = p F g cos θ P sin θ F electric = F g tan θ = F g (tan 45 ) = F g .
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yang (ey942) – oldhomework 01 – Turner – (56705) 2 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 = mg . Thus | q | = r r 2 mg 5 k e = r ( L 2) 2 5 k e = L · R 2 5 k e = (14 cm) p 1 m 100 cm P × r 2(0 . 1 kg)(9 . 81 m / s 2 ) 5(8 . 98755 × 10 9 N · m 2 / C 2 ) = 9 . 25064 × 10 7 C . 003 (part 1 of 2) 10.0 points Three point-charges ( q , + q , and q ) are placed at the vertices of an equilateral triangle (see ±gure 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 triangle due to the other two charges is given by 1. b v F b = 1 2 3 k q 2 a 2 2. b v F b = 2 3 k q 2 a 2 3. b v F b = k q 2 a 2 correct 4. b v F b = 1 2 k q 2 a 2 5. b v F b = 3 2 k q 2 a 2 6. b v F b = 2 3 k q 2 a 2 7. b v F b = 2 k q 2 a 2 8. b v F b = 3 2 k q 2 a 2 9. b v F b = 3 k q 2 a 2 10. b v F b = 1 2 k q 2 a 2 Explanation: 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 = b p k q 2 a 2 P cos 60 + p k q 2 a 2 PB ˆ ı = b 1 2 p k q 2 a 2 P + p k q 2 a 2 PB ˆ ı = + 1 2 k p q 2 a 2 P ˆ ı. On the other hand, the y -component is just F y = b + p k q 2 a 2 P sin 60 B ˆ = + 3 2 k p q 2 a 2 P ˆ . Therefore the magnitude of the net force is b v F b = ± F 2 x + F 2 y = k R 1 4 + 3 4 p q 2 a 2 P = k ² q a ³ 2 .
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yang (ey942) – oldhomework 01 – Turner – (56705) 3 004 (part 2 of 2) 10.0 points The direction of the electric force on the charge at the bottom left-hand vertex of the triangle due to the other two charges is Caution: There are two vectors in each quadrant. Compare their relative orienta- tions.
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HW1 - yang(ey942 oldhomework 01 Turner(56705 This print-out...

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