An electron falls through distance of 1.5 cm in a uniforth electric field of magnitude `2.0xx 10 ^(4) NC^(-1) ` The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance compute the time of falls in each case . Contrast the situation with that of 'free fall under gravity'

Explanation

The field is upward so the negatively charged the magnitude of the electric of the electron is `a_(e )=eE//m_(e )`
where `m_(e )` is the mas of the electron
distance h 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 the thus the effect of acceleration due to gravity can be ignored in this example