A charged particle of mass m and charge q is released from rest in an uniform electric field `vecE`. Neglecting the effect of gravity, the kinetic energy of the charged particle after 't' second is

Options

A. `(2E^(2)t^(2))/(mq)`
B. `(Eq^(2)m)/(2t^(2))`
C. `(Eqm)/(t)`
D. `(E^(2)q^(2)t^(2))/(2m)`

Answer

D

Explanation

Force on the particle
`F=qE`
Acceleration, `alpha=(F)/(m)=(qE)/(m)`
Now, v=u+`alphat=0+(qE)/(m)t`
`K.E.=(1)/(2)mxx((qE)/(m)t)^(2)=(q^(2)E^(2)t^(2))/(2m)`