Physhw 2-solutions

Physhw 2-solutions - billing (cab4763) hw 2 opyrchal...

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Unformatted text preview: billing (cab4763) hw 2 opyrchal (121104) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points Consider three charges arranged as shown. + + 4 . 2 cm 3 . 3 cm 3 . 9 C 7 . 8 C 3 . 5 C What is the magnitude of the electric field strength at a point 2 . 3 cm to the left of the middle charge? The value of the Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . Correct answer: 1 . 37962 10 8 N / C. Explanation: Let : q 1 = 7 . 8 C = 7 . 8 10 6 C , q 2 = 3 . 9 C = 3 . 9 10 6 C , q 3 = 3 . 5 C = 3 . 5 10 6 C , r 1 , 2 = 4 . 2 cm = 0 . 042 m , r 2 , 3 = 3 . 3 cm = 0 . 033 m , x = 2 . 3 cm = 0 . 023 m , and k e = 8 . 98755 10 9 N m 2 / C 2 . r 1 = r 1 , 2 x = 0 . 042 m . 023 m = 0 . 019 m r 2 = x = 0 . 023 m r 3 = r 2 , 3 + x = 0 . 033 m + 0 . 023 m = 0 . 056 m vector E net = vector E 1 + vector E 2 + vector E 3 E = k e q r 2 Considering the magnitudes of the electric fields at a point 2 . 3 cm to the left of the middle charge, E 1 = k e q 1 r 2 1 is directed away from the charge since q 1 is positive, E 2 = k e q 2 r 2 2 is directed away from the charge since q 2 is positive, and E 3 = k e | q 3 | r 2 3 is directed toward the charge since q 3 is nega- tive. Thus E net = E 1 E 2 + E 3 = k e parenleftbigg q 1 r 2 1 q 2 r 2 2 + | q 3 | r 2 3 parenrightbigg = ( 8 . 98755 10 9 N m 2 / C 2 ) bracketleftbigg (7 . 8 10 6 C) (0 . 019 m) 2 (3 . 9 10 6 C) (0 . 023 m) 2 + (3 . 5 10 6 C) (0 . 056 m) 2 bracketrightbigg = (8 . 98755 10 9 N m 2 / C 2 ) [(0 . 0216066 N / C) (0 . 0073724 N / C) +(0 . 00111607 N / C)] = 1 . 37962 10 8 N / C , directed along the positive x-axis. 002 (part 2 of 2) 10.0 points What is the magnitude of the force on a 3 . 5 C charge placed at this point? Correct answer: 482 . 866 N. Explanation: Let : q = 3 . 5 C = 3 . 5 10 6 C . The electric force is F electric = q E net = 3 . 5 10 6 C (1 . 37962 10 8 N / C) = 482 . 866 N , bardbl vector F electric bardbl = 482 . 866 N . billing (cab4763) hw 2 opyrchal (121104) 2 F electric has a magnitude of 482 . 866 N and is directed along the negative x axis. 003 (part 1 of 2) 10.0 points Three point charges are placed at the vertices of an equilateral triangle. 2 . 3 m 60 3 . 4 C 3 . 4 C 3 . 4 C P Find the magnitude of the electric field vec- tor bardbl vector E bardbl at P . The value of the Coulomb constant is 8 . 9875 10 9 N m 2 / C 2 . Correct answer: 7 . 70195 10 9 N / C....
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This note was uploaded on 03/25/2011 for the course PHYSICS 121 taught by Professor Fayngold during the Spring '10 term at NJIT.

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Physhw 2-solutions - billing (cab4763) hw 2 opyrchal...

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