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

Answer

D

Explanation

Here, the direction of field is upward,
So the negatively charged electron experiences a dowanward force.

`therefore` The acceleration of electron is
`a_(e)=(eE)/m_(e)`
The time required by the electron to fall through a distance h is
`t_(e)=sqrt(((2h)/a_(e)))=sqrt((2hm_(e))/(eE))`
`=[(2xx1.5xx10^(-2)xx9.11xx10^(-31))/(1.6xx10^(-19)xx2xx10^(4))]^(1//2)=2.9xx10^(-9)s`