Chapter 2319P23.51The proton moves with accelerationaqEmp==××=×−−160 106401673 10613 10192710...C NCkgms2ejbgwhile the e−has accelerationaaep=××=−−9110 10112 101836193114...NCkg2.(a)We want to find the distance traveled by the proton (i.e., datp=122), knowing:4001212183712222. cm=+=FHGIKJatpep.Thus,tp=1221 82..cmmµ.(b)The distance from the positive plate to where the meeting occurs equals the distance thesodium ion travels (i.e., tNaNa=122). This is found from:1212cmNaCl=+:1299123545...cmu u=FHGIKJ+FHGIKJeEteEt.This may be written as12120649165122.cmNaNaNa=FHGIKJa tsotNaNacm1.65cm=122432.P23.52(a)The field, E1, due to the 400 109.×−C charge is in the –xdirection.Erii129928 99 104 00 10250575×⋅−×=−−kqre±.±.±Nm CCmNCejafFIG. P23.52(a)Likewise, E2and E3, due to the 500 109.×−C charge and the 300 109.×−C charge areii2228 99 105 00 1020011 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 .