HW1 - yang(ey942 – oldhomework 01 – Turner –(56705 1...

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Unformatted text preview: 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 – 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 , 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 . 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 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 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 ˆ ı ˆ In this case, each of these forces has a mag- nitude F 21 = F 31 = k q 2 a 2 ....
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HW1 - yang(ey942 – oldhomework 01 – Turner –(56705 1...

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