14_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

14_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

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14 CHAPTER 15 15.16 The required position is shown in the sketch at the right. Note that this places q closer to the smaller charge, which will allow the two forces to cancel. Requiring that gives 6 FF = 3 () 2 2 nC 3.00 nC 00 m ee kq k x = 6.00 0.6 x + q 2 2 2 0.600 m xx =+ , or Solving for x gives the equilibrium position as 0.6 x 00 m 1.45 m beyond the 21 == 3.00 nC charge ±²³³´µ¶ ·¸²³³´µ¶ ± ² ³²±³³´¹ ± ¸ ±² ± ± 15.17 For the object to “float” it is necessary that the electrical force support the weight, or ( ) ( ) 6 3 2 24 10 C 610 N C or 1.5 10 kg 9.8 m s qE qE mg m g × = = × 15.18 (a) Taking to the right as positive, the resultant electric field at point P is given by 132
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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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