Question

An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a vertical distance of 1.0 μm, what is its speed? (Do not neglect the gravitational force on the electron.) mm/s

Answer 1

Step-by-step explanation:

It is given that,

An electron is released from rest in a weak electric field of,
`E=2.3* 10^(-10)\ N/C`

Vertical distance covered,
`s=1\ \mu m=10^(-6)\ m`

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

`v^2-u^2=2as`

`v^2=2as`.............(1)

Electric force is
`F_e` and force of gravity is
`F_g`. As both forces are acting in downward direction. So, total force is:

`F=mg+qE`

`F=9.1* 10^(-31)* 9.8+1.6* 10^(-19)* 2.3* 10^(-10)`

`F=4.57* 10^(-29)\ N`

Acceleration of the electron,
`a=(F)/(m)`

`a=(4.57* 10^(-29)\ N)/(9.1* 10^(-31)\ kg)`

`a=50.21\ m/s^2`

Put the value of a in equation (1) as :

`v=√(2as)`

`v=\sqrt{2* 50.21* 10^(-6)}`

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.