16_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

# 16_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

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

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

Unformatted text preview: −19 F qE ( 1.60 × 10 C ) ( 640 N C ) a= = = = 6.12 × 1010 m s 2 m mp 1.673 × 10 -27 kg (a) (b) t = 1.20 × 10 6 m s ∆v = = 1.96 × 10 −5 s = 19.6 µ s a 6.12 × 1010 m s 2 ∆x = (c) 2 v 2 − v0 f (d) KE f = 2 6 10 2 11.8 m 2 1 1 mp v 2 = ( 1.673 × 10 −27 kg )( 1.20 × 106 m s ) = 1.20 × 10 −15 J f 2 2 q1 = 3.00 nC The altitude of the triangle is h = ( 0.500 m ) sin 60.0° = 0.433 m 00 15.24 2a (1.20 × 10 m s ) − 0 = = 2 ( 6.12 × 10 m s ) m 15.23 CHAPTER 15 0.5 16 and the magnitudes of the fields due to each of the charges are q2 = 8.00 nC E1 = 9 2 2 −9 k e q1 ( 8.99 × 10 N ⋅ m C ) ( 3.00 × 10 C ) = 2 h2 ( 0.433 m ) 60.0° 0.250 m = 144 N C 9 2 2 −9 k e q2 ( 8.99 × 10 N ⋅ m C ) ( 8.00 × 10 C ) E2 = 2 = = 1.15 × 10 3 N C 2 r2 0.250 m ) ( and E3 = k e q3 r32 ( 8.99 × 10 = 9 N ⋅ m 2 C 2 ) ( 5.00 × 10 −9 C ) ( 0.250 m ) 2 = 719 N C h u r E2 u r u E q2 = 5.00 nC r E1 3 ...
View Full Document

## This document was uploaded on 10/20/2011.

Ask a homework question - tutors are online