P25_049 - ) m e v 2 i = 2(1 . 6 10 19 C) 2 8 . 99 10 9 N m...

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49. Let the distance in question be r . The initial kinetic energy of the electron is K i = 1 2 m e v 2 i ,wh e r e v i =3 . 2 × 10 5 m / s. As the speed doubles, K becomes 4 K i .Thu s U = e 2 4 πε 0 r = K = (4 K i K i )= 3 K i = 3 2 m e v 2 i , or r = 2
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Unformatted text preview: ) m e v 2 i = 2(1 . 6 10 19 C) 2 8 . 99 10 9 N m 2 C 2 3(9 . 11 10 19 kg)(3 . 2 10 5 m / s) 2 = 1 . 6 10 9 m ....
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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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