121HW2 - lew (dl9564) hw 2 Opyrchal (121006) This print-out...

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lew (dl9564) – hw 2 – Opyrchal – (121006) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Three point charges, two positive and one negative, each having a magnitude oF 34 μ C are placed at the vertices oF an equilateral triangle (22 cm on a side). a -q +q +q The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . What is the magnitude F oF the electro- static Force on one oF the positive charges? Correct answer: 214 . 661 N. Explanation: Let : q = 34 μ C , a = 22 cm , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . The magnitude oF electrostatic Force is F = k e | q 1 | | q 2 | r 2 The answer is independent oF the exact orien- tation oF the triangle, but let’s choose a par- ticular orientation ( i.e. , choose a coordinate system) to use For our calculations. Consider the arrangement oF charges where the nega- tive point charge is positioned at the origin and one oF the positive charges is located at x = a . Then consider the electrostatic Force on the upper positive charge, which we call A ; ( i.e. , A be in the frst quadrant). ±irst, note that by symmetry and by the Fact that neg- ative charge exerts an attractive Force while the positive charge exerts a repulsive Force, F y = 0 . This leaves us with only the x component oF the Force on A . Using the principle oF super- position, and recalling that For an equilateral triangle, θ = 60 : F = 2 p k e q 2 a 2 P cos 60 = k q 2 a 2 since cos 60 = 1 2 , so F = k e q 2 a 2 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × ( 3 . 4 × 10 5 C ) 2 (0 . 22 m) 2 = 214 . 661 N . 002 (part 1 oF 2) 10.0 points Two charges are located on a horizontal axis. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 1 . 9 μ C 1 . 9 μ C p 1 . 5 m 2 m 2 m Determine the electric feld at p on a verti- cal axis as shown in the fgure above. Up is the positive direction. Correct answer: 3278 . 66 V / m. Explanation: Let : x = 2 m , y = 1 . 5 m , q = 1 . 9 μ C , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . E θ q q y x - x
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lew (dl9564) – hw 2 – Opyrchal – (121006) 2 The distance from each point charge to the point of interest on the y -axis is r = r x 2 + y 2 = R (2 m) 2 + (1 . 5 m) 2 = 2 . 5 m . Therefore, the electric Feld due to one of the point charges is E = k e q r 2 = (8 . 98755 × 10 9 N · m 2 / C 2 )(1 . 9 × 10 6 C) (2 . 5 m) 2 = 2732 . 22 V / m . Now, we need to vector add the contribu- tions from each charge. By symmetry, the x -components cancel, and we only have twice the y -components left. To Fnd the y compo- nents, we note that the angle θ is θ = arcsin p 1 . 5 m 2 . 5 m P = 36 . 8699 , so the total electric Feld is E tot = 2 (2732 . 22 V / m) sin36 . 8699 = 3278 . 66 V / m , where E tot points upward.
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121HW2 - lew (dl9564) hw 2 Opyrchal (121006) This print-out...

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