A particle of mass `6.4xx10^(-27)` kg and charge `3.2xx10^(-19)C` is situated in a uniform electric field of `1.6xx10^5 Vm^(-1)`. The velocity of the particle at the end of `2xx10^(-2)`m path when it starts from rest is :

Options

A. `2sqrt3 xx 10^(5) ms^(-1)`
B. `8 xx10^5 ms^(-1)`
C. `16 xx10^5 ms^(-1)`
D. `4sqrt2 xx 10^(5) ms^(-1)`

Answer

D

Explanation

`a=f/m=(qE)/m=((2e)E)/(m_(alpha))=(2eE)/(4m_p) = (eE)/(2m_p)`
`u=0 :. v^2=u^2 +2 as v = sqrt(2as)`