12_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

12_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

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12 CHAPTER 15 Thus, Σ= and The resultant force on the 6.00 nC charge is then () 7 23 cos45.0 3.81 10 N x FFF + ° = × 8 sin45.0 7.63 10 N y Σ= − ° = × ()() 2 2 7 3.89 10 N Rx y =Σ + Σ = × at 1 tan 11.3 y x F F θ Σ  = =− °  Σ  or 7 3.89 10 N at 11.3 below + ax R x ° F G is 15.13 The forces on the 7.00 µ C charge are shown at the right. ( )( ) ( 66 2 9 1 2 2 2 9 2 2 2 7.00 10 C 2.00 10 C Nm 8.99 10 C 0.500 m 0.503 N 7.00 10 C 4.00 10 C 8.99 C 0.500 m 1.01 N F F −− ×× = = ( Thus, ) 12 cos60.0 0.755 N x Σ= + ° = sin60.0 0.436 N y ° = and The resultant force on the 7.00
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

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