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Unformatted text preview: Patel, Kinal Homework 2 Due: Jan 25 2008, 7:00 pm Inst: Weathers 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Four point charges are placed at the four cor ners of a square, where each side has a length a . The upper two charges have identical pos itive charge, and the two lower charges have charges of the same magnitude as the first two but opposite sign. That is, q 1 = q 2 = q and q 3 = q 4 = q where q > 0. q 4 q 2 q 3 q 1 P j i Determine the direction of the electric field at the center, point P. 1. 1 2 ( i j ) 2. 1 2 ( i + j ) 3. 1 2 ( i + j ) 4. 5. j correct 6. i 7. j 8. i 9. 1 2 ( i j ) Explanation: The direction is already clear: all the x components cancel, and the lower charges at tract and the top ones repel, so the answer is j . keywords: 002 (part 1 of 2) 10 points Three positive charges are arranged as shown at the corners of a rectangle. The Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . + + + . 752 m 2 . 256 m 11 . 9 nC 5 . 95 nC 3 . 73 nC Find the magnitude of the electric field at the fourth corner of the rectangle. Correct answer: 100 . 049 N / C. Explanation: r 1 r 2 Q 2 Q 1 Q 3 E 1 E 2 E 3 Let : q 1 = 5 . 95 nC = 5 . 95 10 9 C , q 2 = 11 . 9 nC = 1 . 19 10 8 C , q 3 = 3 . 73 nC = 3 . 73 10 9 C , r 2 = 2 . 256 m , r 1 = 0 . 752 m , and k e = 8 . 98755 10 9 N m 2 / C 2 . The length of the diagonal is r 3 = q r 2 1 + r 2 2 = q (2 . 256 m) 2 + (0 . 752 m) 2 = 2 . 37803 m . The electric field produced by the charge q 1 is along the positive yaxis, so E 1 ,x = 0 E 1 ,y = E 1 = k c q 1 r 2 1 . Patel, Kinal Homework 2 Due: Jan 25 2008, 7:00 pm Inst: Weathers 2 The electric field produced by the charge q 2 is along the negative xaxis, so E 2 ,x = E 2 = k c q 2 r 2 2 E 2 ,y = 0 . The electric field produced by the charge q 3 is upward and to the left along the diagonal of the rectangle, and is directed away from the charge since q 3 is positive, so E 3 = k e q 3 r 2 3 . Consider the direction of E 3 cos = r 2 r 3 and sin = r 1 r 3 Thus E 3 ,x = k e q 3 r 2 3 sin = k e q 3 r 2 r 3 3 E 3 ,y = k C q 3 r 2 3 cos = k e q 3 r 1 r 3 3 , so that E net,x = E 2 ,x + E 3 ,x = k e q 2 r 2 2 k e q 3 r 2 r 3 3 = ( 8 . 98755 10 9 N m 2 / C 2 ) 1 . 19 10 8 C (2 . 256 m) 2 ( 8 . 98755 10 9 N m 2 / C 2 ) (3 . 73 10 9 C) (2 . 256 m) (2 . 37803 m) 3 = 26 . 6379 N / C , and E net,y = E 1 ,y + E 3 ,y = k e q 1 r 2 1 + k e q 3 r 1 r 3 3 = ( 8 . 98755 10 9 N m 2 / C 2 ) 5 . 95 10 9 C (0 . 752 m) 2 + ( 8 . 98755 10 9 N m 2 / C 2 ) (3 . 73 10 9 C) (0 . 752 m) (2 . 37803 m) 3 = 96 . 4379 N / C , so that E net = q E net,x 2 + E net, 4 2 E net = ( 26 . 6379 N / C) 2 +(96 . 4379 N / C) 2 / 1 2 = 100 . 049 N / C ....
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This note was uploaded on 10/24/2010 for the course PHYS PHYS 2220 taught by Professor Weathers during the Spring '08 term at North Texas.
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