32_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

32_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

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32 CHAPTER 15 Thus, 2 2 tan10 e r mg = ° k q or ( ) ( ) ( ) 2 2 2 9 2 2 0.015 kg 9.8 m s 0.047 m tan10 tan10 8.99 10 N m C e mgr q k ° ° = = × 8 8.0 10 C q = × giving or 7 ~ 10 C q 15.62 Consider the free-body diagram of the rightmost charge given below. 0 cos or cos y F T mg T mg θ θ Σ = = = and ( ) 0 = sin cos sin t x e F F T mg an mg θ θ θ Σ = = = θ But, ( ) ( ) 2 2 2 2 2 2 2 2 2 1 2 5 4 sin sin 2 sin e e e e e e k q k q k q k q k q F r r L L L 2 2 θ θ θ = + = + = Thus, 2 2 2 5 tan 4 sin e k q mg L θ θ = or 2 2 4 sin tan 5 e L mg q k θ θ = 5 , 0.10 kg, and 0.300 m m L If 4 θ = ° = = ( ) ( ) then ( ) ( ) ( ) ( ) 2 4 0.300 m 0 q = 2 2 9 2 2 kg 9.80 m s sin 45 .99 10 N m C × .10 5 8 tan 45 ° ° or 6 2.0 10 C 2.0 q C µ = × = 15.63 (a) When an electron (negative charge) moves distance
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