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# shim (js64948) - Homework 2 - markert - (56870) This print-out should have 27 questions. Multiple-choice questions may continue on the next column or...

Where and how did they get r1 and r2 in question #4???
shim (js64948) – Homework 2 – markert – (56870) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Two small spheres carry equal amounts oF electric charge. There are equally spaced points ( a , b , and c ) which lie along the same line. a b c + What is the direction oF the net electric feld at each point due to these charges? 1. a b c + 2. a b c + 3. a b c + 4. a b c + 5. a b c + correct 6. a b c + 7. a b c + 8. a b c + 9. a b c + 10. a b c + Explanation: Since the feld originates From positive charges and terminates on the negative charges, a b c + 002 (part 1 of 2) 10.0 points Consider three charges arranged as shown. - + + 5 . 5 cm 3 . 8 cm 1 . 6 μ C 7 . 6 μ C 3 . 4 μ C What is the electric feld strength at a point 1 . 7 cm to the leFt oF the middle charge? The Coulomb constant is 8 . 99 × 10 9 N · m 2 / C 2 . Correct answer: 7 . 64863 × 10 6 N / C. Explanation: Let : q 1 = 7 . 6 μ C = 7 . 6 × 10 6 C , q 2 = 1 . 6 μ C = 1 . 6 × 10 6 C , q 3 = 3 . 4 μ C = 3 . 4 × 10 6 C , r 1 , 2 = 5 . 5 cm = 0 . 055 m , r 2 , 3 = 3 . 8 cm = 0 . 038 m , x = 1 . 7 cm = 0 . 017 m , and k C = 8 . 99 × 10 9 N · m 2 / C 2 . r 1 = r 1 , 2 x = 0 . 055 m 0 . 017 m = 0 . 038 m , r 2 = x = 0 . 017 m , and r 3 = r 2 , 3 + x = 0 . 038 m + 0 . 017 m = 0 . 055 m . The electric feld is given by E = k C q r 2 .
shim (js64948) – Homework 2 – markert – (56870) 2 The magnitude of the electric Felds are E 1 = k C q 1 r 2 1 = (8 . 99 × 10 9 N · m 2 / C 2 ) 7 . 6 × 10 6 C (0 . 038 m) 2 = 4 . 73158 × 10 7 N / C , directed away from the charge, since q 1 is positive, and E 2 = k C q 2 r 2 2 = (8 . 99 × 10 9 N · m 2 / C 2 ) 1 . 6 × 10 6 C (0 . 017 m) 2 = 4 . 97716 × 10 7 N / C , directed away from the charge, since q 2 is positive. E 3 = k C | q 3 | r 2 3 = (8 . 99 × 10 9 N · m 2 / C 2 ) 3 . 4 × 10 6 C (0 . 055 m) 2 = 1 . 01045 × 10 7 N / C , directed toward the charge since q 3 is nega- tive. Consider the directions of E 1 , E 2 and E 3 at point x = 1 . 7 cm to the left of q 2 , assuming that the positive direction is to the right. E net = E 1 E 2 + E 3 = 4 . 73158 × 10 7 N / C 4 . 97716 × 10 7 N / C + 1 . 01045 × 10 7 N / C = 7 . 64863 × 10 6 N / C along the positive x -axis. 003 (part 2 of 2) 10.0 points What is the magnitude of the force on a 3 . 4 μ C charge placed at this point? Correct answer: 26 . 0053 N. Explanation: Let : q = 3 . 4 × 10 6 C . F e = q E net = (3 . 4 × 10 6 C)(7 . 64863 × 10 6 N / C) = 26 . 0053 N . which has a magnitude of 26 . 0053 N . 004 10.0 points Two charges are located in the ( x, y ) plane as shown. The Felds produced by these charges are observed at a point p with coordinates (0 , 0). 8 . 9 C 8 . 8 C p 1 . 5 m 1 . 7 m 2 . 3 m 2 . 9 m ±ind the x -component of the electric Feld at p . The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 1 . 49241 × 10 10 N / C. Explanation: Let : ( x p , y p ) = (0 , 0) , ( x 1 , y 1 ) = (2 . 3 m , 1 . 5 m) , ( x 2 , y 2 ) = ( 2 . 9 m , 1 . 7 m) , q 1 = 8 . 9 C , q 2 = 8 . 8 C , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . q 1 q 2 p y 1 y 2 x 1 x 2 Consider the electric Feld vectors:
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