Find the time taken by a particle of mass `10^-18` kg and carrying a charge `3.2 xx 10^-19` C to fall through a distance of 8 m in a uniform electric field of intensity `8 xx 10^2 N C^-1`

Explanation

`m = 10^(-18) kg`
`q = 3.2 xx 10^(-19) C`
`S = 8.0 m`
`E = 8 xx 10^2 NC^(-1)`
Initial velocity of the particle = `u = 0`
When particle falls through the electric field, then it gets accelerated and acceleration produced in it is given by
`a = (qE)/(m)`
Let t = time taken by the particle to fall through a distance of 8.0 m.
Now `S = ut + 1/2 at^2`
`implies 8 = 0 + 1/2 ((qb)/(m)) t^2`
`implies t^2 = 8 xx (2m)/(qE)`
`implies t^2 = (8 xx 2 xx 10^(-18))/(3.2 xx 10^(-19) xx 8 xx 10^2) `
`implies t^2 = 0.625 xx 10^(-1)`
`implies t^2 = 625 xx 10^(-4)`
`implies t = sqrt(625 xx 10^(-4))`
`implies t = 25 xx 10^(-2)S`
`implies t = 0.25 S`