How many volts must an electric current have for the electrons traveling within it to travel 90% the speed of light when going through a vacuum How many volts must an electric current have for the electrons traveling within it to travel 90% the speed of light when going through a vacuum, also, what is the mathematical relationship between volts, and velocity in a vacuum?
 A: Note that particle accelerator energies are discussed in electron-Volts (eV), which makes this kind of question straight forward. The energy of a mass $M=0.511$ MeV particle at $v=0.9c$ is:
$$ E = \gamma Mv=\frac {1} {\sqrt{1-v^2}} Mv = 5.26M = 2.7\,{\rm MeV}$$
(with $c=1$). The kinetic energy is:
$$ T = E-M= (2.7-0.511)\,{\rm MeV}= 2.17\, {\rm MeV}$$
Energy to Volts is trivial in these units$^1$:
$$ V = 2,170,000\,{\rm V}$$
[1] Because
$1\,eV = 1\cdot(1.602\times10^{-19}\,{\rm Coulumb})\cdot(1\,{\rm Volt})$ and $Q=e=1.602\times10^{-19}\,$C.
A: If you're moving an electron with mass $M$ and charge $Q$ through a potential difference of $V$, it's going to gain $QV$ in kinetic energy.    Assuming the electron started at rest, its final kinetic energy is $QV$.
At the same time, the final kinetic energy must be $(1/2)Mv^2$, where $v$ is the final velocity.  (Be careful not to confuse $v$ with $V$!!)  
So $QV=(1/2)Mv^2$, and if you want $v=.9$, (after choosing  units where the speed of light is $1$), this gives  $V=.405 M/Q$.  Now, if you're so inclined, you can look up the values of $M$ and $Q$ and plug them in.
Edited to add:  As lmr points out in comments, this really calls for a relativistic calculation, so final kinetic energy is $M(1-\sqrt{1-v^2})/\sqrt{1-v^2}\approx 1.3M$ when $v=.9$.  Now if I set this equal to $QV$ and solve for $V$, I get $V\approx 1.3 M/Q$, which is morethan 3 times what the non-relativistic calculation gives.
