An electron and proton are each placed at rest in an electric field of 440 N/C.Calculate the velocity of each?

Question

particle 27 nanoseconds after being released (also denote the direction parallel to the field as positive). Each have a charge of magnitude of 1.60218 x 10^-19 C. What is the velocity of the electron? units of m/s

Answer 1

F = Eq   = ma

so a = Eq/m     me = 9 x 10^-31 kg I think  mp = 1836 x me

vfinal = 0 + at  = a x 27x10^-9