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Unformatted text preview: terry (ect328) oldhomework 03 Turner (59130) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 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. 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. i 2. 1 2 ( i j ) 3. i 4. j 5. 1 2 ( i + j ) 6. 1 2 ( i j ) 7. j correct 8. 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 10.0 points A uniformly charged insulating rod of length 13 . 5 cm is bent into the shape of a semicircle as in the figure. The value of the Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . 1 3 . 5 c m 5 . 86 C O If the rod has a total charge of 5 . 86 C, find the horizontal component of the electric field at O , the center of the semicircle. Define right as positive. Correct answer: 1 . 81573 10 7 N / C. Explanation: Let : L = 13 . 5 cm = 0 . 135 m and q = 5 . 86 C = 5 . 86 10 6 C . Call the length of the rod L and its charge q . Due to symmetry E y = integraldisplay dE y = 0 and E x = integraldisplay dE sin = k e integraldisplay dq sin r 2 , where dq = dx = r d , so that b y x E x = k e r integraldisplay 3 / 2 / 2 cos d terry (ect328) oldhomework 03 Turner (59130) 2 = k e r (sin ) vextendsingle vextendsingle vextendsingle vextendsingle 3 / 2 / 2 = 2 k e r , where = q L and r = L . Therefore, E x = 2 k e q L 2 = 2 (8 . 98755 10 9 N m 2 / C 2 ) (0 . 135 m) 2 ( 5 . 86 10 6 C) = 1 . 81573 10 7 N / C . Since the rod has negative charge, the field is pointing to the left (towards the charge dis tribution). A positive test charge at O would feel an attractive force from the semicircle, pointing to the left. 003 (part 1 of 2) 10.0 points Consider a square with side a . Four charges + q , + q , q , and + q are placed at the corners A , B , C , and D , respectively. + + + D C A B a The magnitude of the electric field at D due to the charges at A , B , and C is given by 1. bardbl vector E bardbl = 5 4 k q a 2 2. bardbl vector E bardbl = 2 k q a 2 3. bardbl vector E bardbl = 7 2 k q a 2 4. bardbl vector E bardbl = 3 4 k q a 2 5. bardbl vector E bardbl = k q a 2 6. bardbl vector E bardbl = 2 k q a 2 7. bardbl vector E bardbl = 5 2 k q a 2 8. bardbl vector E bardbl = 3 2 k q a 2 correct 9. bardbl vector E bardbl = 3 k q a 2 10. bardbl vector E bardbl = 9 4 k q a 2 Explanation: The magnitudes of the electric fields at...
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 Fall '09
 Turner

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