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Quiz01_solutions - 3.027 10 N 0.30 N tan tan 265 2.495 10 N...

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Physics 222 Fall 2011 Quiz 1 (Solutions for A, B and C) Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m. The charges are C, 0 . 8 C, 0 . 4 μ μ - + and C. 0 . 6 μ - Calculate the magnitude and direction of the net force on charge C 0 . 4 μ + . Solution The forces on each charge lie along a line connecting the charges. Let the variable d represent the length of a side of the triangle. Since the triangle is equilateral, each angle is 60 o . First calculate the magnitude of each individual force. ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 6 6 9 2 2 1 2 12 2 2 21 6 6 9 2 2 1 3 13 2 2 31 4.0 10 C 8.0 10 C 8.988 10 N m C 1.20m 0.1997 N 4.0 10 C 6.0 10 C 8.988 10 N m C 1.20m 0.1498 N Q Q F k d F Q Q F k d F - - - - × × = = × × = = × × = = × × = = ( 29 ( 29 ( 29 ( 29 6 6 9 2 2 2 3 23 32 2 2 8.0 10 C 6.0 10 C 8.988 10 N m C 0.2996 N 1.20m Q Q F k F d - - × × = = × × = = Now calculate the net force on each charge and the direction of that net force, using components. ( 29 ( 29 ( 29 ( 29 o o 2 1 12 13 o o 1 1 12 13 1 1 2 2 1 1 o 1 1 1 1 2 1 0.1997 N cos60 0.1498 N cos60 2.495 10 N 0.1997 N sin 60 0.1498 N sin 60 3.027 10 N 3.027
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Unformatted text preview: 3.027 10 N 0.30 N tan tan 265 2.495 10 N x x x y y y y x y x F F F F F F F F F F F θ------= + = -+ = -× = + = --= -×-× = + = = = =-× ( 29 ( 29 ( 29 o 1 2 21 23 o 1 2 21 23 1 2 2 2 1 1 o 2 2 2 2 1 2 0.1997 N cos60 0.2996 N 1.998 10 N 0.1997 N sin 60 1.729 10 N 1.729 10 N 0.26 N tan tan 139 1.998 10 N x x x y y y y x y x F F F F F F F F F F F------= + =-= -× = + = + = × × = + = = = =-× 1 Q 2 Q 3 Q 12 F r d 13 F r d d 32 F r 31 F r 21 F r 23 F r Physics 222 Fall 2011 ( 29 ( 29 ( 29 o 1 3 31 32 o 1 3 31 32 1 3 2 2 1 1 o 3 3 3 3 1 3 0.1498 N cos60 0.2996 N 2.247 10 N 0.1498 N sin 60 1.297 10 N 1.297 10 N 0.26 N tan tan 30 2.247 10 N x x x y y y y x y x F F F F F F F F F F F θ------= + = -+ = × = + = = × × = + = = = = × + Physics 222 Fall 2011...
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