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24_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

# 24_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

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24 CHAPTER 15 The total charge enclosed by the cylindrical gaussian surface is Q A σ = , where σ is the charge density on the conducting surface. Hence, Gauss’s law gives 0 A EA σ = or o E σ = 15.47 ( )( ) ( ) 2 9 2 19 2 1 2 2 2 2 -15 8.99 10 N m C 1.60 10 C 57.5 N 2.00 10 m e e k q q k e F r r × × = = = = × 15.48 (a) ( )( ) ( ) 2 1 2 2 2 2 9 2 2 19 8 2 10 8.99 10 N m C 1.60 10 C 8.2 10 N 0.53 10 m e e k q q k e r r = = × × = = × F × (b) ( ) 2 e c e a m v r = = F m , so ( )( ) 10 8 6 -31 0.53 10 m 8.2 10 N 2.2 10 m s 9.11 10 kg e r F v m × × = = = × × 15.49 The three contributions to the resultant electric field at the point of interest are shown in the sketch at the right. The magnitude of the resultant field is
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