An electron initially at rest falls a distance of 1.5 cm in a uniform electric field of magnitude `2 xx10^(4) N//C`. The time taken by the electron to fall this distance is

Options

A. `1.3 xx10^2 s`
B. `2.1 xx10^(-12) s`
C. `1.6 xx 10^(-10) s`
D. `2.9 xx 10^(-9) s`

Answer

D

Explanation

`u=0 , S = 1.5 cm =1.5 xx10^(-2)m`
`a=(eE)/(m)=(1.6 xx10^(-19)xx2 xx10^4)/(9.1 xx10^(-31))`
`=3.5 xx10^(15) m//s^2`
`S=1/2 "at"^2`
`implies t = sqrt((2S)/a)=sqrt((2 xx1.5 xx10^(-2))/(3.5 xx10^(15)))=2.93 xx10^(-9) sec `