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Physics I – Exam 3 – Spring 2003 Answer Key Part A – 1. E, 2. C, 3. E, 4. F There are some partial credits allowed for the following wrong answers: 2. D = 2 points for knowing the field is perpendicular to equipotential lines. 3. A = 2 points for correct magnitude of KE change but wrong sign. 3. D = 2 points for correct sign of KE change but wrong magnitude. 4. E = 2 points for knowing F is not in the plane of v and B. B-1 g G M (6.67 10 11 )( 2.0 10 30 ) 1.33 10 12 m / s 2 R2 (10.0 10 3 ) 2 B-2 D, E, C, A, B B-3 r m v (9.1 10 31 )(1.4 10 7 ) 1.4 m q B (1.6 10 19 )(5.7 10 5 ) B-4 V V 0V y y d V d E y (0.05)(300) 15 Volts Ey C-1 Analysis of the directions of E from each charge reveals that E will be zero somewhere in Region III, to the right of -4q or x > d. 1 9q 1 4q 0 2 40 x 40 ( x d ) 2 9 4 x 3d 2 x ( x d) 2 C-2 The + charge makes +V and the – charge makes –V. Analysis of the contributions from each charge reveals that V will be zero somewhere in Region II (between charges) and also in Region III. Only one is required for the question; either is OK. Region II: Region III: 1 9q 1 4q 1 9q 1 4q 0 0 40 x 40 (d x ) 40 x 40 ( x d ) 9 4 9 4 x 139 d 0.6923 d x 95 d 1.8 d x (d x ) x ( x d) C-3 Calculate the X & Y components of E from each charge, then add. (9q is charge 1.) 1 9q 1 1 E1 î ĵ 2.86 10 5 î 2.86 10 5 ĵ V / m 2 40 2 d 2 2 1 4q E2 0 î 1 ĵ 0 î 3.60 10 5 ĵ V / m 2 40 d E total E1 E 2 2.86 105 î 0.74 105 ĵ V / m Calculate and add V from each charge; V is a scalar. V 1 40 qi r i 1 9q 4q 1 q 9 4 2.1 10 3 Volts 40 2d d 40 d 2