InterviewSolution
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InterviewSolution
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Anelectron falls throgh a distance of 1.5 cm in a uniform electric field of magnitude2.0 xx10^(4) N c^(-1)the direction of the fieldis reversed keeping its magnitude unchangedandaproton falls throughthe same distance computethe time of falls in each case contrastthe situation with that of free fall under gravity |
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Answer» Solution :The field is upward so the negatively charged the magnitude of the electricof the ELECTRON is`a_(e )=eE//m_(e )` where `m_(e )` is the mas of the electron distanceh is given by `t_(e )=sqrt(2h)/(a_(e ))=sqrt(2hm_(e ))/(e E)` For e=`1.6 xx10^(-19) C, m_(e )=9.11 xx10^(-31)` kg `t_(e )=2.9n xx10^(-9)` S where `m_(p)` is the mass of the proton `m_(p)=1.67 xx10^(-27)` kg the time of fall for the proton is `t_(p)=sqrt(2h)/(a_(p))=sqrt(2hm_(p))/(eE)=1.3 xx10^(-7)` S `a_(p)=(eE)/(m_(p))` `=1.9xx10^(12) ms^(-2)` which is enormous compared to the value of g (9.8 `ms^(-2)`)the acceleration DUE to GRAVITY thethus the effect of acceleration due to gravity can be ignored in this example |
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