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# OHW 3 - terry(ect328 oldhomework 03 Turner(59130 This...

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terry (ect328) – oldhomework 03 – Turner – (59130) 1 This print-out should have 10 questions. Multiple-choice 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. 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. 0 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 μ C O If the rod has a total charge of 5 . 86 μ 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 μ 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 θ = k e integraldisplay dq sin θ r 2 , where dq = λ dx = λ r dθ , so that y x θ E x = k e λ r integraldisplay 3 π/ 2 π/ 2 cos θ dθ

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terry (ect328) – oldhomework 03 – Turner – (59130) 2 = k e λ r (sin θ ) vextendsingle vextendsingle vextendsingle vextendsingle 3 π/ 2 π/ 2 = 2 k e λ r , where λ = q L and r = L π . Therefore, E x = 2 k e q π L 2 = 2 (8 . 98755 × 10 9 N m 2 / C 2 ) (0 . 135 m) 2 × ( 5 . 86 × 10 6 C) π = 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 D due to A and C are E A = E C = k q a 2 since they are at a distance a from d , whereas E B = k q ( a 2) 2 = k q 2 a 2 since B is at a distance 2 a from d .
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