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HW 2 solution - song(shs546 Homework 2 weathers(22202 This...

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song (shs546) – Homework 2 – weathers – (22202) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points 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.i 2. i 3.j 4. 0 5. 1 2 ( i j ) 6. 1 2 ( i j ) 7. 1 2 ( i + j ) 8. jcorrect 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 . 002(part1of2)10.0points Three positive charges are arranged as shown at the corners of a rectangle. + + + 0 . 321 m 0 . 963 m 4 . 4 nC 2 . 2 nC 2 . 35 nC Find the magnitude of the electric field at the fourth corner of the rectangle. The Coulomb constant is 8 . 98755 × 10 9 N m 2 / C 2 . Correct answer: 207 . 862 N / C. Explanation: r 1 r 2 θ ϕ Q 2 Q 1 Q 3 E 1 E 2 E 3 Let : q 1 = 2 . 2 nC = 2 . 2 × 10 9 C , q 2 = 4 . 4 nC = 4 . 4 × 10 9 C , q 3 = 2 . 35 nC = 2 . 35 × 10 9 C , r 2 = 0 . 963 m , r 1 = 0 . 321 m , and k e = 8 . 98755 × 10 9 N m 2 / C 2 . The length of the diagonal is r 3 = radicalBig r 2 1 + r 2 2 = radicalBig (0 . 963 m) 2 + (0 . 321 m) 2 = 1 . 01509 m . The electric field produced by the charge q 1 is along the positive y -axis, so E 1 ,x = 0 E 1 ,y = E 1 = k c q 1 r 2 1 . The electric field produced by the charge q 2 is along the negative x -axis, so E 2 ,x = E 2 = k c q 2 r 2 2 E 2 ,y = 0 .
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song (shs546) – Homework 2 – weathers – (22202) 2 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 ϕ = r 2 r 3 and sin ϕ = r 1 r 3 Thus E 3 ,x = k e q 3 r 2 3 sin ϕ = k e q 3 r 2 r 3 3 E 3 ,y = k C q 3 r 2 3 cos ϕ = 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 ) × 4 . 4 × 10 9 C (0 . 963 m) 2 ( 8 . 98755 × 10 9 N m 2 / C 2 ) × (2 . 35 × 10 9 C) (0 . 963 m) (1 . 01509 m) 3 = 62 . 0879 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 ) × 2 . 2 × 10 9 C (0 . 321 m) 2 + ( 8 . 98755 × 10 9 N m 2 / C 2 ) × (2 . 35 × 10 9 C) (0 . 321 m) (1 . 01509 m) 3 = 198 . 373 N / C , so that E net = radicalBig E net,x 2 + E net, 4 2 E net = bracketleftbig ( 62 . 0879 N / C) 2 +(198 . 373 N / C) 2 bracketrightbig 1 / 2 = 207 . 862 N / C . 003(part2of2)10.0points What is the direction of this electric field (as an angle measured from the positive x -axis, with counterclockwise positive)? Correct answer: 107 . 379 . Explanation: tan θ = E net,y E net,x θ = tan 1 parenleftbigg E net,y E net,x parenrightbigg = tan 1 parenleftbigg 198 . 373 N / C 62 . 0879 N / C parenrightbigg = 107 . 379 .
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