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Unformatted text preview: tang (nmt352) 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 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 , = 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 . 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 = (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 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 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 In this case, each of these forces has a mag nitude F 21 = F 31 = k q 2 a 2 . The xcomponent of the net force is then F x = [ F 31 cos 60 + F 21 ] = bracketleftbigg parenleftbigg k q 2 a 2 parenrightbigg cos 60 + parenleftbigg k q 2 a 2 parenrightbiggbracketrightbigg = bracketleftbigg 1 2 parenleftbigg k q 2 a 2 parenrightbigg + parenleftbigg k q 2 a 2 parenrightbiggbracketrightbigg = + 1 2 k parenleftbigg q 2 a 2 parenrightbigg ....
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 Spring '10
 Lam
 Charge, Work

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