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

Answer

D

Explanation

The acceleration of the electron is `a_(e) = (eE)/(m_(e))`
where `m_(e)` is the mass of the electron.
Starting from rest, the time taken by the electron to fall through a distance h is given by `t_(e) = sqrt((2h)/(a_(e))) = sqrt((2hm_(e))/(eE))`
`= sqrt((2 xx 1.5 xx 10^(-2) xx 9.1 xx 10^(-31))/(1.6 xx 10^(-19) xx 2 xx 10^(4))) = 2.9 xx 10^(-9) s`