Two charged metal plates in vacuum are 10cm apart. A uniform electric field of intensity `(45//16)xx10^(3)NC^(-1)` is applied between the plates. An electron is released between the plates from rest at a point just outside the negative plate.
(i) Calculate how long (t) will the electron take to reach the other plate?
(ii) At what velocity (v) will it be electron take to hits the other plate?

Options

A. (i) `2xx10^(-6)s` (ii) `10^(5) ms^(-1)`
B. (i) `3xx10^(-8)s` (ii) `10^(6) ms^(-1)`
C. (i) `2xx10^(-8)s` (ii) `10^(7) ms^(-1)`
D. (i) `3xx10^(-6)s` (ii) `10^(7) ms^(-1)`

Answer

C

Explanation

a. The force F experienced by the electrons towards the positive plate is given by
`F=qE=1.6xx10^(-19)xx(45//16)xx10^(3)`
`=4.5xx10^(-16)N`
Hence, the acceleration experienced by the electron is given by
`a=F/m=(4.5xx10^(-16)N)/(9.0xx10^(-31)kg)=5.0xx10^(14)ms^(-2)`
Now, the electrons is released from the negative plate and is traveling towards the positve plate. Hence, we have u=0, x=0.10m, and `a=5.0xx10^(14)ms^(-2)`
According to eq. `x=ut+1/2at^(2)`
`t=sqrt((2x)/(a))=sqrt((2xx0.10)/(5.0xx10^(14))``=2xx10^(-8)s`
b. `v=u+at=0+5.0xx10^(14)xx2.0xx10^(-8)=10^(7) ms^(-1)`.