P23_036 - x we Fnd ∆ K K i = ∆ 1 2 m e v 2 1 2 m e v 2...

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36. (a) The initial direction of motion is taken to be the + x direction (this is also the direction of ~ E ). We use v 2 f v 2 i =2 a x with v f =0and ~a = ~ F/m = e ~ E/m e to solve for distance ∆ x : x = v 2 i 2 a = m e v 2 i 2 eE = (9 . 11 × 10 31 kg)(5 . 00 × 10 6 m / s) 2 2(1 . 60 × 10 19 C)(1 . 00 × 10 3 N / C) =7 . 12 × 10 2 m . (b) Eq. 2-17 leads to t = x v avg = 2∆ x v i = 2(7 . 12 × 10 2 m) 5 . 00 × 10 6 m / s =2 . 85 × 10 8 s . (c) Using ∆ v 2 =2 a x with the new value of ∆
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Unformatted text preview: x , we Fnd ∆ K K i = ∆( 1 2 m e v 2 ) 1 2 m e v 2 i = ∆ v 2 v 2 i = 2 a ∆ x v 2 i = − 2 eE ∆ x m e v 2 i = − 2(1 . 60 × 10 − 19 C)(1 . 00 × 10 3 N / C)(8 . 00 × 10 − 3 m) (9 . 11 × 10 − 31 kg)(5 . 00 × 10 6 m / s) 2 = − 11 . 2% ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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