Spring 11 Phys 121 - HW 02

Spring 11 Phys 121 - HW 02 - dzifanu (sd26397) hw 2...

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Unformatted text preview: dzifanu (sd26397) hw 2 opyrchal (121102) 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 . 4 cm 4 . 8 cm 5 C 8 . 6 C 2 . 9 C What is the magnitude of the electric field strength at a point 2 . 1 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: 4 . 96862 10 7 N / C. Explanation: Let : q 1 = 8 . 6 C = 8 . 6 10 6 C , q 2 = 5 C = 5 10 6 C , q 3 = 2 . 9 C = 2 . 9 10 6 C , r 1 , 2 = 4 . 4 cm = 0 . 044 m , r 2 , 3 = 4 . 8 cm = 0 . 048 m , x = 2 . 1 cm = 0 . 021 m , and k e = 8 . 98755 10 9 N m 2 / C 2 . r 1 = r 1 , 2 x = 0 . 044 m . 021 m = 0 . 023 m r 2 = x = 0 . 021 m r 3 = r 2 , 3 + x = 0 . 048 m + 0 . 021 m = 0 . 069 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 . 1 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 (8 . 6 10 6 C) (0 . 023 m) 2 (5 10 6 C) (0 . 021 m) 2 + (2 . 9 10 6 C) (0 . 069 m) 2 bracketrightbigg = (8 . 98755 10 9 N m 2 / C 2 ) [(0 . 0162571 N / C) (0 . 0113379 N / C) +(0 . 000609116 N / C)] = 4 . 96862 10 7 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 2 . 9 C charge placed at this point? Correct answer: 144 . 09 N. Explanation: Let : q = 2 . 9 C = 2 . 9 10 6 C . The electric force is F electric = q E net = 2 . 9 10 6 C (4 . 96862 10 7 N / C) = 144 . 09 N , bardbl vector F electric bardbl = 144 . 09 N . dzifanu (sd26397) hw 2 opyrchal (121102) 2 F electric has a magnitude of 144 . 09 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. 7 . 5 m 60 4 . 8 C 4 . 8 C 4 . 8 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: 1 . 02258 10 9 N / C....
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This note was uploaded on 09/16/2011 for the course PHYS PHYS 121 taught by Professor Prodan during the Spring '09 term at NJIT.

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Spring 11 Phys 121 - HW 02 - dzifanu (sd26397) hw 2...

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