Electric Field Question?

Question

An electron and a proton are each placed at
rest in an electric field of 520 N/C. Calculate
the speed of each particle 48.0 ns after
being released.

Answer 1

The magnitude of force is the same for both particles - they have equal but opposite charges.

|F| = E*e = 520 N/C*1.6x10^-19 C = 8.32*10^-17 N

Now speed is v = acceleration x time = at

And a = force/mass = F/m

For the proton, m = 1.6x10^-27 kg

v = 8.32*10^-17*48x10^-9/1.6*10^-27 = 2496 m/s

For the electron, m = 9.11x10^-31

v = 4.38x10^6 m/s

Answer 2

Force equal to the product of field strength and the charge placed in that field
F=Eq

We know that
F=ma
and that velocity is
V=at then it follows
V=[F/m]t
V=[Eq/m]t
_____________________________
Case 1 electron
q=-1.602 E−19 C
mass= 9.109 E–31 kg

V=[Eq/m]t = (520 x 1.602 E−19 /9.109 E–31 ] 48E-9
V=4.4 E+6 m/s

Case 2 proton
q= 1.602 E−19 C
mass= 1.672 E−27 kg

V=[Eq/m]t = (520 x 1.602 E−19 /1.672 E−27 ] 48E-9
V=2.4 E+3 m/s