10CHAPTER 15 (b) When the spheres are connected by a conducting wire, the net charge will divide equally between the two identical spheres. Thus, the force is now 9126.010Cnetqqq−=+=− ×()( )229292226.010C2Nm8.9910 C4 0.30 menetkqFr−−×⋅==×or 79.010N (repulsion)F−=×15.10The forces are as shown in the sketch at the right. ( )( )662912-2126.0010C1.5010C8.9989.9 NC3.0010 mekqqFr−−××⋅==×=×( )( )213922-2136.0010C2.0010C8.9943.2C5.00ekq qFr⋅×=×N ( )( )223932-2231.5010C2.0010C8.9967.4C2.00eFr⋅×=×6CThe net force on the µcharge is 61246.7 N (to the left)F=−=5
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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.