Electric Potential : First concepts

[mitleid]

I submitted this last night, but for some reason it didn't go through... so I don't have time to make it as awesome as I had before.

Through what potential difference would an electron need to be accelerated for it to achieve a speed of 39.0% of the speed of light, starting from rest?

C = 3.00 x 10^8
M(e-) = 9.11x10^-31
abs(e) = 1.60x10^-19 [I have been using this value as q)

Ka + Ua = Kb + Ub where K is kinetic energy and U is the potential. Ka = 0 and the equation can be simplified to something like -Kb = Ub - Ua.

I have all the variables for Kb as well as the charge. What I worked things down to last night was along the lines of Kb/q = (Ub/q - Ua/q) = \DeltaV. The units worked out (I'm pretty certain) and I got a reasonable answer in kV, something like 39.0kV.

Apologies for the lack of depth here, I actually have to run off to my physics class! Any help is appreciated, thanks!

 

[malawi_glenn]

The work done on the electron by the field is q⋅ΔU and this work will increase the kinetic energy of the electron. Now the problem stated that it started at rest, so q⋅ΔU = mv²/2 non-relativistically.

This gives you 38.9 kV

For the relativistic answer, you need to use E_k = (γ-1)mc² where γ = (1-(v/c)²)^-½.

This gives you 44.0 kV