611_PartUniversity Physics Solution

611_PartUniversity Physics Solution - 21-4 Chapter 21...

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21-4 Chapter 21 21.13. IDENTIFY: Apply Coulomb’s law. The two forces on 3 q must have equal magnitudes and opposite directions. SET UP: Like charges repel and unlike charges attract. EXECUTE: The force 2 F G that 2 q exerts on 3 q has magnitude 23 2 2 2 qq Fk r = and is in the + x direction. 1 F G must be in the x direction, so 1 q must be positive. 12 FF = gives 13 22 kk rr = . 2 2 1 2 2.00 cm (3.00 nC) 0.750 nC 4.00 cm r r ⎛⎞ == = ⎜⎟ ⎝⎠ . EVALUATE: The result for the magnitude of 1 q doesn’t depend on the magnitude of 2 q . 21.14. IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on Q . SET UP: The force that 1 q exerts on Q is repulsive, as in Example 21.4, but now the force that 2 q exerts is attractive. EXECUTE: The x -components cancel. We only need the y -components, and each charge contributes equally. 66 2 0 1 (2.0 10 C) (4.0 10 C) sin 0.173 N (since sin 0.600). 4 (0.500 m) yy αα π −− ×× = = P Therefore, the total force is 20 . 3 5 N , F = in the -direction y . EVALUATE: If 1 q is 2.0 C μ and 2 q is 2.0 C + , then the net force is in the + y -direction. 21.15. IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on 1 q . SET UP: Like charges repel and unlike charges attract, so 2 F G and 3 F G are both in the + x -direction. EXECUTE: 54 12 13 6.749 10 N, 1.124 10 N × × . 4 1.8 10 N FF F =+=× . 4 1.8 10 N F and is in the + x -direction. EVALUATE: Comparing our results to those in Example 21.3, we see that 1 on 3 3 on 1 =− G G , as required by Newton’s third law. 21.16. IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on 2 q . SET UP: 2 on 1 F G is in the + y -direction. EXECUTE: () 92 2 6 6 2on 1 2 (9.0 10 N m C ) (2.0 10 C) (2.0 10 C) 0.100 N 0.60 m F ×⋅ × × . ( ) 2 on 1 0 x F = and 2 on 1 0.100 N y F =+ . on 1 Q F is equal and opposite to 1 on Q F (Example 21.4), so ( ) on 1 0.23N Q x F and on 1 0.17 N Q y F = . 2 on 1 on 1 0.23 N xQ x x F = . ( ) 2 on 1 on 1 0.100 N 0.17 N 0.27 N yQ y y F = . The magnitude of the total force is 0.23 N 0.27 N 0.35 N. F = 1 0.23 tan 40 0.27 , so F G is 40 ° counterclockwise from the + y axis, or 130 ° counterclockwise from the + x axis. EVALUATE: Both forces on 1 q are repulsive and are directed away from the charges that exert them. 21.17. IDENTIFY and SET UP: Apply Coulomb’s law to calculate the force exerted by 2 q and 3 q on 1 . q Add these forces as vectors to get the net force. The target variable is the x -coordinate of 3 . q EXECUTE: 2 F G is in the x -direction. 2 12 3.37 N, so 3.37 N x F r = + and 7.00 N xxx x F = 32 7.00 N 3.37 N 10.37 N xx x F =− = = For 3 x F to be negative, 3 q must be on the x -axis.
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611_PartUniversity Physics Solution - 21-4 Chapter 21...

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