[Phys-L] electron velocity in an electric field.

[brian whatcott <betwys1@sbcglobal.net>]

The other day, I strayed from my usual comfort zone to answer the school 
house/homework problem given below on Quora.

I am fairly sure I made one - possibly several mistakes.  Would someone  
set me straight on this one please?

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Q:Two parallel metal plates with a spacing of 1cm have a potential 
difference of 10kv. An electron is projected from one plate directly 
towards second.What is the initial velocity of electron if it comes to 
rest just at the surface of second plate? 
<https://www.quora.com/Two-parallel-metal-plates-with-a-spacing-of-1cm-have-a-potential-difference-of-10kv-An-electron-is-projected-from-one-plate-directly-towards-second-What-is-the-initial-velocity-of-electron-if-it-comes-to-rest-just>

A: Without looking for an optimal solution, let’s just equate the 
electron kinetic energy to its initial energy of 10keV. One*joule*is 
equal to 6.241509⋅1018 electron-volts, so its initial energy is 10keV / 
6.241 *10^18 joules. kinetic energy is given by 1/2 *m* v^2 for m in kg, 
v in m/s

1/2 * 9.109 383 56 (11) × 10 −31 kg * v^2 = 10000 / 6.241*10^18

so v^2 = 10000/6.241*10^18 * 2 / ( 9.11* 10^-31)  =0.730 *10^17

and V = 2.7 * 10^8 m/s !!!

Recall the speed of light in vacuum is 3*10^8 m/s in vacuum. That’s a 
strong suggestion this calculation is invalid because it is not 
accounting for relativistic issues. Then let’s look at the physicality 
of it: 10 kv at 1 cm requires a breakdown voltage of 1kV/mm or more. 
That’s about the value for air - but higher than the breakdown voltage 
in a vacuum, so we have to assume a filler gas. That means the electron 
suffers collisions with gas en-route which disqualifies this free-path 
speed calculation.

Hmmm….looks like you will have to wait for someone who is more 
comfortable dealing in this kind of school house kinematics problem - sorry!


________________

Brian W