10_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

# 10_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

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10 CHAPTER 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 9 12 6.0 10 C net qq q =+= − × () ( ) 2 2 9 2 9 2 22 6.0 10 C 2 Nm 8.99 10 C 4 0.30 m en e t kq F r −×  == ×   or 7 9.0 10 N (repulsion) F 15.10 The forces are as shown in the sketch at the right. ( )( ) 66 2 9 1 2 -2 12 6.00 10 C 1.50 10 C 8.99 89.9 N C 3.00 10 m e kqq F r −− ×× ==× = × ( )( ) 2 13 9 2 2 -2 13 6.00 10 C 2.00 10 C 8.99 43.2 C 5.00 e kq q F r × = × N ( )( ) 2 23 9 3 2 -2 23 1.50 10 C 2.00 10 C 8.99 67.4 C 2.00 e F r × = × 6 C The net force on the µ charge is 612 46.7 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.

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