Solution 3 - homework 03 KANUNGO ANIL Due Sep 9 2007 4:00...

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homework 03 – KANUNGO, ANIL – Due: Sep 9 2007, 4:00 am 1 Question 1, chap 23, sect 1. part 1 of 2 10 points Four point charges are placed at the four corners of a square, where each side has a length a . The upper two charges have identical positive 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. j 4. 1 2 ( i j ) 5. j correct 6. i 7. 1 2 ( i + j ) 8. i Explanation: The direction is already clear: all the x - components cancel, and the lower charges at- tract and the top ones repel, so the direction is j . Question 2, chap 23, sect 1. part 2 of 2 10 points Determine the magnitude of the electric field at point P . 1. k e q a 2 2. 2 k e q a 2 3. 4 2 k e q a 2 correct 4. 8 k e q a 2 5. 2 2 k e q a 2 6. 4 k e q a 2 Explanation: The electric field due to a point charge is E = k e q r 2 . The field at P must be found through super- position of fields due to the four point parti- cles. The direction of a field at a point P is the direction in which a positive test charge at that point would feel a force. The top two charges ( q 1 and q 2 ) give rise to repulsive forces on a positive test charge at P . The fields vector E 1 and vector E 2 are directed as in the figure P vector E 2 vector E 1 The x -components cancel. The distance from the center to one of the particles is a 2 , so the magnitudes are E 1 y = E 2 y = 1 2 k e q parenleftbigg a 2 parenrightbigg 2 = 2 k e q a 2 . The bottom two charges are negative, so the forces on a positive test charge at the center are attractive:
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homework 03 – KANUNGO, ANIL – Due: Sep 9 2007, 4:00 am 2 P E 4 E 3 Similarly, E 3 y = E 4 y = 2 k e q a 2 , so the total magnitude at P (since the x - components cancel) is E = 4 E 1 y = 4 2 k e q a 2 . Question 3, chap 23, sect 1. part 1 of 1 10 points A charge of 3 . 39 μ C is located at the origin, and a charge of 3 . 44 μ C is located along the y axis at 1 . 22529 m. The value of the Coulomb constant is 8 . 99 × 10 0 . At what point along the y -axis is the electric field zero? Correct answer: 0 . 610402 m (tolerance ± 1 %). Explanation: Let : q 1 = 3 . 39 μ C , q 2 = 3 . 44 μ C , and d = 1 . 22529 m . Call the point where the fields cancel y . Since the charges are of equal sign, the only place y can be is somewhere between them. The field from the particle q 1 at the origin is E 1 = k e q 1 y 2 pointing down (since q 1 is negative). The field from the charge q 2 at a point d along the y -axis is E 2 = k e q 2 ( d y ) 2 pointing up (since q 2 is negative). Thus we have cancellation at d provided E 1 equals E 2 ,
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