An electron falls through a distance of 1.5cm in a uniform electric field of magnitude `2 times 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 fall in each case.

Explanation

Here, electric force, `F_(e) = qE = ma rArr a = (qE)/m`
Linear distance travelledl at the end of time t is,
`d = v_(0)t + 1/2 at^(2)`
`therefore d = 1/2 (qE)/mt^(2), (therefore v_(0) =0 " and " a =(qE)/m)`
`therefore t^(2) = (2md)/(qE)`
`therefore t = sqrt((2md)/(eE))`......(1)
(`therefore` For electron and proton magnitude of charge is q=e)
Linear distance travelled at the end of time t is,
`d=v_(0)t + 1/2 at^(2)`
`therefore d = 1/2((qE)/m)t^(2) (therefore v_(0)=0 " and " a =(qE)/m`)
`therefore t^(2) = (2md)/(qE)`
`therefore t = sqrt((2md)/(eE))`............(1)
(`therefore` For electron and proton magnitude of charge is q = e)
From equation (1), time taken by electron to pass through distance d (opposite to electric field)
`t_(e) = sqrt((2m_(e)d)/(eE))`
`= sqrt((2 xx 9.1 xx 10^(-31) xx 0.015)/(1.6 xx 10^(-19) xx 2 xx 10^(4))`
`=2.92 xx 10^(-9)` s
`therefore t_(e) = 2.92 ns` (Nano second)...........(2)
From equation (1), time taken by proton to pass through distance d (parallel to electric field)
`t_(p) = sqrt((2m_(p)d)/(eE))`
`=sqrt((2 xx 1.67 xx 10^(-27) xx 0.015)/(1.6 xx 10^(-19) xx 2 xx 10^(4))`
`=1.251 xx 10^(-7)`s
=`125.1 xx 10^(-9)`s
`therefore t_(p) = 125.1`ns.............(3)
From equation (2) and (3),
`t_(p) gt t_(e)` (`therefore` Here, `t prop sqrt(m)` and `m_(p) gt m_(e)`)