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Unformatted text preview: yang (ey942) oldhomework 01 Turner (56705) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find 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 10 cm 20 cm The ratio of the electric field strength at point A to the electric field 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 field 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 , = 45 , and k e = 8 . 98755 10 9 N m 2 / C 2 . r = 2 L sin = 2 L sin 45 = 2 L 2 2 = L 2 F T,x = F T sin F T,y = F T cos Each sphere is in equilibrium horizontally F electric F T,x = 0 F electric F T sin = 0 and vertically F T,y F g = 0 F T cos F g = 0 F T = F g cos . From the horizontal equilibrium, F electric = parenleftbigg F g cos parenrightbigg sin F electric = F g tan = F g (tan 45 ) = F g . 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  = radicalBigg r 2 mg 5 k e = radicalBigg ( L 2) 2 mg 5 k e = L radicalbigg 2 mg 5 k e = (14 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 ) = 9 . 25064 10 7 C . 003 (part 1 of 2) 10.0 points Three pointcharges ( q , + q , and q ) are placed at the vertices of an equilateral triangle (see figure below). a 60 q q + q The magnitude of the electric force on the charge at the bottom lefthand vertex of the triangle due to the other two charges is given by 1. bardbl vector F bardbl = 1 2 3 k q 2 a 2 2. bardbl vector F bardbl = 2 3 k q 2 a 2 3. bardbl vector F bardbl = k q 2 a 2 correct 4. bardbl vector F bardbl = 1 2 k q 2 a 2 5. bardbl vector F bardbl = 3 2 k q 2 a 2 6. bardbl vector F bardbl = 2 3 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 = 3 k q 2 a 2 10. bardbl vector F bardbl = 1 2 k q 2 a 2 Explanation: a q q + q In this case, each of these forces has a mag nitude F 21 = F 31 = k q 2 a 2 ....
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 Spring '10
 Turner

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