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

Answer:

$v = 4.24 \times 10^8 m/s$

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

$\frac{1}{2}mv^2 = qV$

so here we have

$m = 9.11 \times 10^{-31} kg$

$q = 1.6 \times 10^{-19} C$

$V = 511 kV$

now we have

$\frac{1}{2}(9.11 \times 10^{-31})v^2 = (1.6 \times 10^{-19})(511 \times 10^3)$

$v = 4.24 \times 10^8 m/s$