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, 22 tan10 e r m g kq or () ( ) 2 2 2 92 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 FTm gT m g θ Σ=⇒ = = and ( ) = sin cos sin t xe FF Tm g an m g θθ = = But, 2 2 2 12 5 4s i n sin 2 sin ee e e e e F rr L LL 2 2 =+= + = Thus, 2 5 tan i n e mg L = or i n t a n 5 e Lmg q k = 5 , 0.10 kg, and 0.300 m mL If 4 =° = = then ()() 2 4 0.300 m 0 q = 2 kg 9.80 m s sin 45 .99 10 Nm C .10 58 tan 45 °° or 6 2.0 10 C 2.0 q C µ = ± ± ± ±² ² ³ ² 15.63 (a) When an electron (negative charge) moves distance x in the direction of an electric field, the work done on it is
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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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