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 1 2 6.0 10 C net q q q = + = − × ( ) ( ) ( ) 2 2 9 2 9 2 2 2 6.0 10 C 2 N m 8.99 10 C 4 0.30 m e net k q F r × = = × or 7 9.0 10 N (repulsion) F = × 15.10 The forces are as shown in the sketch at the right. ( )( ) ( ) 6 6 2 9 1 2 1 2 2 2 -2 12 6.00 10 C 1.50 10 C N m 8.99 10 89.9 N C 3.00 10 m e k q q F r × × = = × = × ( )( ) ( ) 6 6 2 1 3 9 2 2 2 2 -2 13 6.00 10 C 2.00 10 C N m 8.99 10 43.2 C 5.00 10 m e k q q F r × × = = × = × N
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