16_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

16_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

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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 ...
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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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