# Electric Potential : First concepts

1. Sep 20, 2007

### 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) = $$\Delta$$V. 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!