Find the time taken by a particle of mass `2 xx 10^(-18) kg` and carrying a charge of `1.6 xx 10^(-19)C` to fall through a distance of 4.0 m in a uniform electric field of intensity `1.6 xx 10^3 NC^(-1)`

Explanation

`m = 2 xx 10^(-18)` kg
`q = 1.6 xx 10^(-19) C`
`S = 4.0 m`
`E = 1.6 xx 10^(3) NC^(-1)`
Initial velocity of the particle `= u = 0`
When particle falls through the electric field, then it gets accelerated and a = acceleration produced in the particle due to the electric field
`a = (qE)/(m)`
Let t = time taken by the particle to fall through 2.0 m in the electric field.
Now `S = ut + 1/2 at^2`
`S = 0 + 1/2 (qE)/(m) t^2`
`t^2 = (2 mS)/(qE)`
`= (2 xx 2 xx 10^(-18) xx 4)/(1.6 xx 10^(-19) xx 1.6 xx 10^2)`
`t^2 = 0.625 xx 10^(-1)`
`= 625 xx 10^(-4)`
`t = sqrt(625 xx 10^(-4))`
`= 25 xx 10^(-2)`
` = 0.25 s`