Question

The electron gun in a television tube accelerates electrons (mass = 9.11x10^-31 kg, charge = 1.6 x 10^-19C) from rest to 3 x10^7 m/s within a distance o 2 cm. What electric field is required?

Answer 1

Electric field acting on the electron is 127500 N/C.

Step-by-step explanation:

It is given that,

Mass of an electron,
`m=9.11* 10^(-31)\ kg`

Charge on electron,
`q=1.6* 10^(-19)\ C`

Initial speed of electron, u = 0

Final speed of electron,
`v=3* 10^7\ m/s`

Distance covered, s = 2 cm = 0.02 m

We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :

`a=(v^2-u^2)/(2s)`

`a=((3* 10^7)^2-0)/(2* 0.02)`

`a=2.25* 10^(16)\ m/s^2`

According to Newton's law, force acting on the electron is given by :

F = ma

`F=9.1* 10^(-31)* 2.25* 10^(16)`

`F=2.04* 10^(-14)\ N`

Electric force is given by :

F = q E, E = electric field

`E=(F)/(q)`

`E=(2.04* 10^(-14))/(1.6* 10^(-19))`

E = 127500 N/C

So, the electric field is 127500 N/C. Hence, this is the required solution.