14CHAPTER 15 15.16The required position is shown in the sketch at the right. Note that this places qcloser to the smaller charge, which will allow the two forces to cancel. Requiring that gives 6FF=3()22nC3.00 nC00 meekqkx=6.000.6x+q2220.600 mxx=+, or Solving for xgives the equilibrium position as 0.6x00 m1.45 m beyond the 21==−3.00 nC charge−±²³³´µ¶·¸²³³´µ¶±²³²±³³´¹±¸±²±±15.17For the object to “float” it is necessary that the electrical force support the weight, or ( )( )6322410C610 N Cor 1.510kg9.8 m sqEqEmgmg−−×==×15.18(a) Taking to the right as positive, the resultant electric field at point Pis 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.