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Unformatted text preview: johnson (rj6247) – hw 2 – Opyrchal – (121014) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Three point charges, two positive and one negative, each having a magnitude of 38 μ C are placed at the vertices of an equilateral triangle (38 cm on a side). aq +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: 89 . 8755 N. Explanation: Let : q = 38 μ C , a = 38 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 first quadrant). First, 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 parenleftbigg k e q 2 a 2 parenrightbigg 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 . 8 × 10 − 5 C ) 2 (0 . 38 m) 2 = 89 . 8755 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 . 4 μ C 1 . 4 μ C p 1 . 6 m 3 m 3 m Determine the electric field at p on a verti cal axis as shown in the figure above. Up is the positive direction. Correct answer: 1024 . 43 V / m. Explanation: Let : x = 3 m , y = 1 . 6 m , q = 1 . 4 μ C , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . E E θ q q y x x johnson (rj6247) – hw 2 – Opyrchal – (121014) 2 The distance from each point charge to the point of interest on the yaxis is r = radicalbig x 2 + y 2 = radicalBig (3 m) 2 + (1 . 6 m) 2 = 3 . 4 m . Therefore, the electric field due to one of the point charges is E = k e q r 2 = (8 . 98755 × 10 9 N · m 2 / C 2 )(1 . 4 × 10 − 6 C) (3 . 4 m) 2 = 1088 . 46 V / m ....
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This note was uploaded on 04/03/2009 for the course PHYS PHYS 121 taught by Professor Prodan during the Spring '09 term at NJIT.
 Spring '09
 PRODAN
 Physics, Charge

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