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Exam _1 Solutions

# Exam _1 Solutions - Physics 2213 1 4 6 All F G(a 2 A 5(a...

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Physics 2213 Prelim #1 – Solutions Spring 2009 1. All F 2. A 3. (a) - (b) induction (c) ± , + , contact (d) ² , - , induction 4. G 5. (a) 1 (b) 2, 3 (c) #1: C #3: A #4: B #5: B 6. (a) I 13 = ³ R 1 + R 3 = 20 V 10 ´ + 30 ´ = 0.50 A V 3 = I 3 R 3 = (0.50 A)(30 ´ ) = 15 V = V 4 I 24 = V 4 R 4 = 15 V 40 ´ = 0.375 A I = I 13 + I 24 = 0.50 A + 0.375 A = 0.875 A (b) V +- = V 3 - ³ = 15 V - 20 V = - 5 V 7. (a) F 1 = kQ 2 x 2 = F 2 F 3 = kQ 2 x 2 + D 2 = F 4 Net force F(x) = - F 1 - F 2 + F 3 cos µ + F 4 cos µ = - 2kQ 2 x 2 + 2kQ 2 x 2 + D 2 x x 2 + D 2 = - 2kQ 2 x 2 + 2kQ 2 x ( ) x 2 + D 2 3/2 = 2kQ 2 x 2 · · ¸ ¹ º º » - 1 + x 3 ( ) x 2 + D 2 3/2 (b) ± (c) [F(x)] = (N-m 2 /C 2 )(C) 2 m 2 + (N-m 2 /C 2 )(C) 2 (m) (m 2 ) 3/2 = N + N As Q ¼ , F(x) ¼ . As x ¼ , F(x) ½ . As D ½ , F(x) ½ . (d) F 1 and F 2 get large because their +Q and -Q from different dipoles get close together, while F 3 and F 4 remain finite because their Qs remain ± D apart. Furthermore, F 3 and F 4 point in nearly opposite directions so they nearly cancel. So the net force is attraction between two pairs of ±Q. ¾ F(x) ± - 2kQ 2 x 2 Our F(x) ± 2kQ 2 x 2 · ¸ ¹ º » -1 + x 3 D 3 ± - 2kQ 2 x 2 (e) E ( Charge-dipole force ± 1/x 3

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