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Hand In 2 Solutions(07)

# Hand In 2 Solutions(07) - Solutions for Hand-In Set#2 PHYS...

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Solutions for Hand-In Set #2 PHYS 212 — Spring 2007 Problem A5 (Problem 2 on E-fields program) The signs are (1) positive; (2) negative; (3) negative; (4) negative; (5) positive. The way to figure this out: bring the positive test charge very close to each num- bered charge, so that the field from this charge domi- nates. If the electric field vector points away from the numbered charge, then it is a positive charge. If the vector points toward the numbered charge, then it is a negative charge. Problem A6 — Addition of electric fields P c P b P a q q The directions and relative magnitudes of the fields from the two charges ( q 1 = q 2 = q = 5 × 10 - 6 C) are illustrated in the figure. a) At y = - 3 m the fields from both charges point in the negative y direction. E = - kq r 2 1 + kq r 2 2 ˆ j = - kq 1 3 2 + 1 9 2 ˆ j = ( - 5 . 55 × 10 3 N/C) ˆ j. b) At y = 3 m the fields from the charges point in op- posite directions. Because the point is equidistant from both charges both fields have the same magnitude, and the vector sum is zero. c) At y = 9 m the fields from both charges point in the positive y direction. E = kq r 2 1 + kq r 2 2 ˆ j = kq 1 3 2 + 1 9 2 ˆ j = (5 . 55 × 10 3 N/C) ˆ j. Problem A7 — Determining ~ E using integrals x y dq r d ~ E θ θ The magnitude of the field due to the infinitesimal charge dq located at (0 , y ) is given by dE = k dq r 2 = kλ dy a 2 + y 2 . The direction of the vector d ~ E is indicated in the figure.

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Hand In 2 Solutions(07) - Solutions for Hand-In Set#2 PHYS...

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