679_Physics ProblemsTechnical Physics

679_Physics ProblemsTechnical Physics - Chapter 23 jb e g...

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Chapter 23 19 P23.51 The proton moves with acceleration a qE m p == × × 160 10 640 1673 10 613 10 19 27 10 . . . C N C kg ms 2 ej bg while the e has acceleration aa e p = × × = 9110 10 112 10 1836 19 31 14 . . . N C kg 2 . (a) We want to find the distance traveled by the proton (i.e., da t p = 1 2 2 ), knowing: 400 1 2 1 2 1837 1 2 22 2 . c m =+= F H G I K J at pe p . Thus, t p = 1 2 21 8 2 . . cm m µ . (b) The distance from the positive plate to where the meeting occurs equals the distance the sodium ion travels (i.e., t Na Na = 1 2 2 ). This is found from: 1 2 1 2 c m Na Cl =+ : 1 2 9 9 1 23 5 4 5 . .. cm u u = F H G I K J + F H G I K J eE t eE t . This may be written as 1 2 1 2 0649 165 1 2 2 . cm Na Na Na = F H G I K J a t so t Na Na cm 1.65 cm = 1 2 243 2 . P23.52 (a) The field, E 1 , due to the 400 10 9 . × C charge is in the – x direction. Er i i 1 2 99 2 8 99 10 4 00 10 250 575 ×⋅ × =− kq r e ± . ± . ± Nm C C m NC e j af FIG. P23.52(a) Likewise, E 2 and E 3 , due to the 500 10 9 . × C charge and the 300 10 9 . × C charge are i i 2 2 2 8 99 10 5 00 10 200 11 2 × =
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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