Question

An electron is accelerqated in the uniform field betwen two parallel charged oplates. The separation of the plates is 1.20 cm

Answer 1

The speed of electron is
`8.7*10^(6)\ m/s`

Step-by-step explanation:

Given that,

Separation of the plate = 1.20 cm

Suppose the field is
`E=1.80*10^(4)\ N/C`.

If the electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate.

What the speed does it leave the hole?

We need to calculate the acceleration

Using formula of electric force

`F = qE`

`ma=qE`

`a=(qE)/(m)`

We need to calculate the speed of electron

Using equation of motion

`v^2=u^2+2as`

`v^2=2as`

Put the value of acceleration in the formula

`v^2=2*(qE)/(m)* s`

Put the value into the formula

`v^2=2*(1.6*10^(-19)*1.80*10^(4)*1.20*10^(-2))/(9.11*10^(-31))`

`v=\sqrt{2*(1.6*10^(-19)*1.80*10^(4)*1.20*10^(-2))/(9.11*10^(-31))}`

`v=8.7*10^(6)\ m/s`

Hence, The speed of electron is
`8.7*10^(6)\ m/s`