# physics-102 - Chapter 15 HW Solution 15.3 F= ke ( 2e)( 79e)...

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Chapter 15 HW Solution 15.3 () ( ) 2 27 9 e ke e F r = ( ) 2 19 2 9 2 2 14 158 1.60 10 C Nm 8.99 10 91 N rep u lsion C 2. 01 0 m × ⎛⎞ = ⎜⎟ ⎝⎠ × 15.8 The magnitude of the repulsive force between electrons must equal the weight of an electron, Thus, 22 ee ke r m g = or ( ) 2 92 2 1 9 2 31 2 8.99 10 N m C 1.60 10 C 5.08 m 9.11 10 kg 9.80 m s e e ke r mg ×⋅ × == = × 15.11 In the sketch at the right, R F is the resultant of the forces 63 that are exerted on the charge at the origin by the 6.00 nC and the –3.00 nC charges respectively. and FF ( )( ) ( ) 99 2 9 6 3 8.99 3.00 F F ±²³³´µ¶ ± ·²³³´µ¶ ¸¹²³³´µ¶ ´³²¹³³´º ³²»³³´º ± ¹ ±² ± ± ± · 2 6 2 9 5 2 6.00 10 C5 . 00 10 C 10 0.300 m 10 N 3.00 10 . 1.35 10 N 0.100 m −− ×× = × 2 2 C C The resultant is 5 1.38 10 N R FFF =+ = × at 1 3 6 tan 77.5 F F θ = or 5 1.38 10 N at 77. 5 below axis R x ° F G

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15.12 Consider the arrangement of charges shown in the sketch at the right. The distance r is () 22 0.500 m0 .
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## physics-102 - Chapter 15 HW Solution 15.3 F= ke ( 2e)( 79e)...

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