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HW 2 - Kapoor(mk9499 homework 02 Turner(60230 This...

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Kapoor (mk9499) – homework 02 – Turner – (60230) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Four point charges, each of magnitude 5 . 27 μ C, are placed at the corners of a square 84 . 8 cm on a side. The value of Coulomb’s constant is 8 . 98755 × 10 9 N · m 2 / C 2 . If three of the charges are positive and one is negative, find the magnitude of the force experienced by the negative charge. Correct answer: 0 . 664448 N. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , d = 84 . 8 cm = 0 . 848 m , and Q = 5 . 27 μ C = 5 . 27 × 10 - 6 C . F 24 F 34 F 14 + 3 + 1 + 2 - 4 The forces are F 34 = F 24 = k e Q 2 d 2 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × ( 5 . 27 × 10 - 6 C ) 2 (0 . 848 m) 2 = 0 . 347113 N F 14 = k e Q 2 ( 2 d ) 2 = 1 2 k e Q 2 d 2 = 1 2 F 34 The vector sum of F 24 and F 34 is in the same direction as F 14 , so F = F 14 + radicalBig F 34 2 + F 24 2 = 1 2 F 34 + radicalBig 2 F 34 2 = parenleftbigg 1 2 + 2 parenrightbigg F 34 = parenleftbigg 1 2 + 2 parenrightbigg (0 . 347113 N) = 0 . 664448 N . 002 10.0 points A charge Q is spread uniformly along the circumference of a circle of radius R . A point charge q is placed at the center of this circle. What is the total force exerted on q as calculated by Coulomb’s law? 1. Use 2 R for the distance. 2. The result of the calculation is zero. cor- rect 3. Use 2 π R for the distance. 4. Use R for the distance. 5. None of these. Explanation: By symmetry and using the fact that the charge is uniformly distributed along the cir- cumference, the total force on q is zero. For example, consider a small element of charge δQ on the circle. The force on q is along the vector connecting δQ and q . But on the ex- act opposite side there is another element of charge δQ which exerts an equal but opposite force on q . This is true for every point on the circle, so the net force is zero. 003 10.0 points A charge q 1 of - 4 . 00 × 10 - 9 C and a charge q 2 of - 2 . 10 × 10 - 9 C are separated by a distance of 41 . 0 cm. Find the equilibrium position for a third charge of +18 . 0 × 10 - 9 C by identifying its distance from q 1 . Correct answer: 23 . 7741 cm. Explanation: Let : q 1 = - 4 . 00 × 10 - 9 C ,

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Kapoor (mk9499) – homework 02 – Turner – (60230) 2 q 2 = - 2 . 10 × 10 - 9 C , r 1 , 2 = 41 . 0 cm , and q 3 = 18 . 0 × 10 - 9 C .
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