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Unformatted text preview: 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. b b b b b 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 xcomponents 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: 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 C is located at the origin, and a charge of 3 . 44 C is located along the y axis at 1 . 22529 m. The value of the Coulomb constant is 8 . 99 10 . At what point along the yaxis is the electric field zero? Correct answer: 0 . 610402 m (tolerance 1 %). Explanation: Let : q 1 = 3 . 39 C , q 2 = 3 . 44 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....
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This homework help was uploaded on 04/02/2008 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Charge, Work

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