A proton and $($:$q=1\mathrm{.}6{10}^{-19}C\}$ with a speed of $2\mathrm{.}0{10}^5\frac{m}{s}$ falls through a potential difference $V$ and thereby increases its speed to $4\mathrm{.}0{10}^5\frac{m}{s}.$ Through what potential difference did the proton fall?

$-626V$

$626V$

$210V$

$-210V$

$840V$