Question

An electron is accelerated inside a parallel plate capacitor. The electron leaves the negative plate with a negligible initial velocity and then after the acceleration it hits the positive plate with a final velocity β. The distance between the plates is 11.4 cm, and the voltage difference is 115 kV. Determine the final velocity β of the electron using classical mechanics. (The rest mass of the electron is 9.11×10-31 kg, the rest energy of the electron is 511 keV.)

Answer 1

`v = 4.24 * 10^8 m/s`

Step-by-step explanation:

By energy conservation we can say that electrostatic potential energy of the electron while it reaches to other plate will convert into kinetic energy

So here we will say

`(1)/(2)mv^2 = qV`

so here we have

`m = 9.11 * 10^(-31) kg`

`q = 1.6 * 10^(-19) C`

`V = 511 kV`

now we have

`(1)/(2)(9.11 * 10^(-31))v^2 = (1.6 * 10^(-19))(511 * 10^3)`

`v = 4.24 * 10^8 m/s`