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



Regarding the answer to the question Brian W attempted to answer:

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?

We assume the energy of the electron is conserved in its travels (i.e. we ignore the tiny radiative losses as the electron decelerates). In this case the PE gain equals the initial KE lost in stopping the electron by the retarding electric field. This means the initial KE of the electron is 10 keV. To simplify the math let's keep things dimensionless. So let e = (KE)/(m*c^2), and let b = v/c. This means

e = 1/sqrt(1 - b^2) - 1 .

Solving for b gives

b = sqrt(2*e*(1 + e/2))/(1 + e).

If we neglect the special relativistic corrections this formula simplifies to the Newtonian result

b = sqrt(2*e).

The special relativistic kinematic corrections themselves amount to a factor of

sqrt(1 + e/2)/(1 + e).

Now e = (10 keV)/(510.9989461(31) keV) = 0.01956951198(12) (CODATA 2014).

This means

b = 0.1949856095(6) with the relativisitic corrections, and
b = 0.197835851 without them.

Converting to SI units the initial speed is v = b*c = 58,455.215 km/s.

The relativistic correction is a factor of
sqrt(1 + e/2)/(1 + e) = 0.985592897
(i.e. about a 1.46% decrease from the Newtonian result).

It stands to reason that the Newtonian result should be off by about 1 to 2 % since the initial kinetic energy is about 2% of the rest energy.

BTW, using the New SI unit definitions the answer for the initial speed would be somewhat more precise than the 3.1*10^(-9) relative precision uncertainty given here because the electron mass/rest energy is more precisely known in terms of unified mass units than in terms of electron volts, and the new SI system (2018) numbers makes the conversions between unified mass units and kilograms or joules a matter of definition rather than measurement.

David Bowman