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Homework 4 KEY

# Homework 4 KEY - Golshani(ag5873 Homework 04 Yao(57045 This...

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Golshani (ag5873) – Homework 04 – Yao – (57045) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two uncharged metal balls, X and Z , stand on insulating glass rods. A third ball, carrying a positive charge, is brought near the ball Z as shown in the figure. A conducting wire is then run between X and Z and then removed. Finally the third ball is removed. Z X + conducting wire When all this is finished 1. ball X is positive and ball Z is negative. correct 2. ball X is neutral and ball Z is negative. 3. balls X and Z are both positive. 4. balls X and Z are still uncharged. 5. ball X is neutral and ball Z is positive. 6. balls X and Z are both negative, but ball Z carries more charge than ball X . 7. ball X is negative and ball Z is neutral. 8. ball X is negative and ball Z is positive. 9. ball X is positive and ball Z is neutral. 10. balls X and Z are both negative, but ball X carries more charge than ball Z . Explanation: When the conducting wire is run between X and Z , some positive charge flows from Z to X under the influence of the positive charge of the third ball. Therefore, after the wire is removed, X is charged positive and Z is charged negative. 002 10.0 points A strong lightning bolt transfers about 23 C to Earth. How many electrons are transferred? The charge on an electron is 1 . 60218 × 10 19 C. Correct answer: 1 . 43555 × 10 20 . Explanation: Let : q = 23 C and q e = 1 . 60218 × 10 19 C . The charge is proportional to the number of electrons, so q = n | q e | n = q | q e | = 23 C |− 1 . 60218 × 10 19 C | = 1 . 43555 × 10 20 . 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 k q 2 a 2 2. bardbl vector F bardbl = 3 2 k q 2 a 2 3. bardbl vector F bardbl = 2 k q 2 a 2

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Golshani (ag5873) – Homework 04 – Yao – (57045) 2 4. bardbl vector F bardbl = k q 2 a 2 5. bardbl vector F bardbl = 1 2 3 k q 2 a 2 6. bardbl vector F bardbl = 3 2 k q 2 a 2 7. bardbl vector F bardbl = 2 3 k q 2 a 2 8. bardbl vector F bardbl = 2 3 k q 2 a 2 9. bardbl vector F bardbl = 1 2 k q 2 a 2 10. bardbl vector F bardbl = 3 k q 2 a 2 correct 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 = 3 2 k parenleftbigg q 2 a 2 parenrightbigg ˆ ı .
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Homework 4 KEY - Golshani(ag5873 Homework 04 Yao(57045 This...

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