P25_090 - 19 m/s. (d) Using the Pythagorean theorem, we now...

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90. (Fourth problem of Cluster ) (a) At its displaced position, its potential energy (using k =1 / 4 πε 0 )is U i = k qQ d x 0 + k qQ d + x 0 = 2 kqQd d 2 x 2 0 . And at A ,thepotent ia lenergyis U A =2 µ k qQ d . Setting this di±erence equal to the kinetic energy of the particle ( 1 2 mv 2 ) and solving for the speed yields v = r 2( U i U A ) m =2 x 0 s kqQ md ( d 2 x 2 0 ) . (b) It is straightforward to consider small x 0 (more precisely, x 0 /d ¿ 1) in the above expression (so that d 2 x 2 0 d 2 ). The result is v 2 x 0 d r kqQ md . (c) Plugging in the given values (converted to SI units) yields v
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Unformatted text preview: 19 m/s. (d) Using the Pythagorean theorem, we now have U i = 2 k qQ p d 2 + x 2 . Therefore, (with U A in this part equal to the negative of U A in the previous part) v = r 2 ( U i U A ) m = 2 v u u t k q Q m 1 d 1 p d 2 + x 2 . To simplify, the binomial theorem (Appendix E) is employed: 1 p d 2 + x 2 1 d 1 1 2 x 2 d 2 which leads to v x d r 2 k q Q md ....
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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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