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An electron (mass `m_(e )`)falls through a distance d in a uniform electric field of magnitude E. , The direction of the field is reversed keeping its magnitudes unchanged, and a proton(mass `m_(p)`) falls through the same distance. If the times taken by the electrons and the protons to fall the distance d is `t_("electron")` and `t_("proton")` respectively, then the ratio `t_("electron")//t_("proton")`. |
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Answer» The acceleration experienced by a charged particle under the action of an electric field E is `a=(eE)/(m)` If it falls through a distance h, stating form rest, `h=(1)/(2)at^(2)` (as u=0) or, `t^(2)=(2h)/(a)=(2h)/(eE//m)=(2mh)/(eE)` or `t prop sqrt(m)` ...(i) From eq (i) the time of fall will be `(t_("electron"))/(t_("proton"))=sqrt((m_(e)/(m_(p)))` Now since `m_(p)gtm_(e)`, so `t_(p)gtt_(e)`, i.e., time of fall through the same distance is greater for a heavier particle, which is in constrant with the situation of free fall under gravity where the time of fall is independent of the mass of the body. |
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