hw3 solution - lam(kl9432 – hw3 – Shneidman –(12108 1...

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Unformatted text preview: lam (kl9432) – hw3 – Shneidman – (12108) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 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. − j correct 2. j 3. − i 4. − 1 √ 2 ( i + j ) 5. 1 √ 2 ( i + j ) 6. − 1 √ 2 ( i − j ) 7. i 8. 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 (part 2 of 2) 10.0 points Determine the magnitude of the electric field at point P . 1. √ 2 k e q a 2 2. k e q a 2 3. 4 k e q a 2 4. 2 √ 2 k e q a 2 5. 4 √ 2 k e q a 2 correct 6. 8 k e q a 2 Explanation: The electric field due to a point charge is E = k q r 2 . The field at P must be found through super- position of fields due to the four point par- ticles. Recall that 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 E 1 and E 2 are directed as in the figure P E 2 E 1 We see that the x-components cancel. The distance from the center to one of the particles is a √ 2 . Therefore, the magnitudes are E 1 y = E 2 y = 1 √ 2 k q parenleftbigg a √ 2 parenrightbigg 2 = √ 2 k q a 2 The bottom two charges are negative, so the forces on a positive test charge at the center are attractive: lam (kl9432) – hw3 – Shneidman – (12108) 2 P E 4 E 3 Similarly, E 3 y = E 4 y = √ 2 k q a 2 so the total magnitude at P (since the x- components cancel) is E = 4 E 1 y = 4 √ 2 k q a 2 . 003 (part 1 of 3) 10.0 points Coulomb constant is 8 . 98755 × 10 9 N m 2 / C 2 . The 1 . 18 μ C charge is at the origin and a − 5 . 35 μ C charge is 10 cm to the right, as shown in the figure. x O I II III y 1 . 18 μ C − 5 . 35 μ C 10 cm Identify the direction of vector E in the region II (0 < x < 10 cm, along the x-axis). 1. left 2. None of these 3. up 4. right correct 5. down 6. all possibilities: right, left, or zero Explanation: Let : q 1 = 1 . 18 μ C , q 2 = − 5 . 35 μ C , and a = 10 cm . The direction of the electric field at a point P is the direction that a positive charge would move if placed at P . A positive charge placed in region II would be attracted to q 2 and repelled by q 1 . Thus the direction is to the right....
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hw3 solution - lam(kl9432 – hw3 – Shneidman –(12108 1...

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