# HW2 - johnson(rj6247 – hw 2 – Opyrchal –(121014 1...

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Unformatted text preview: johnson (rj6247) – hw 2 – Opyrchal – (121014) 1 This print-out should have 14 questions. Multiple-choice 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). 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: 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 y-axis 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.

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HW2 - johnson(rj6247 – hw 2 – Opyrchal –(121014 1...

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