In the uniform electric field of `E = 1 xx 10^(4) NC^(-1)` an electron is accelerated from rest . The velocity of the electron when it has travelled a distance of `2 xx 10^(-2)` m is nearly … `ms^(-1)`
(`e/m` of electron = `1.8 xx 10^(11) C kg^(-1)`)

Options

A. `1. 6 xx 10^(6)`
B. `0.85 xx 10^(6)`
C. `0.425 xx 10^(6)`
D. `8.5 xx 10^(6)`

Answer

D

Explanation

Force on electron in uniform electric field is F = eE
The acceleration of the electron is `a = (F)/(m) = (eE)/(m)`
where m is the mass of the electron .
Starting from rest , the velocity of the electron when it has travelled distance s is `v = sqrt(2as) = sqrt (2 "" (eE)/(m) "" s)`
Here , `(e)/(m) = 1.8 xx 10^(11) C kg^(-1)`
`E =1 xx 10^(4) NC^(-1)`
`s = 2 xx 10^(-2)` m
`therefore v = sqrt(2 xx 1.8 xx 10^(11) xx 1 xx 10^(4) xx 2 xx 10^(-2))`
`= 8.5 xx 10^(6) ms^(-1)`