Answer
$v = 1.96\times 10^5~m/s$
Work Step by Step
We can find the force on the proton:
$F = q~E$
$F = (1.6\times 10^{-19}~C)(2.00\times 10^4~N/C)$
$F = 3.2\times 10^{-15}~N$
We can find the acceleration:
$F = ma$
$a = \frac{F}{m}$
$a = \frac{3.2\times 10^{-15}~N}{1.67\times 10^{-27}~kg}$
$a = 1.92\times 10^{12}~m/s^2$
We can find the speed attained by the proton:
$v^2 = v_0^2+2ax$
$v^2 = 0+2ax$
$v = \sqrt{2ax}$
$v = \sqrt{(2)(1.92\times 10^{12}~m/s^2)(1.00\times 10^{-2}~m)}$
$v = 1.96\times 10^5~m/s$
