32_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

32_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 10/20/2011.

Ask a homework question - tutors are online