An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude `2.0xx10^(4) N C^(-1)`. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance. Compute the time of fall 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