Please help with a Physics Parallel Plate Capacitor/Electron travel problem!?

[Jason]

Two 6.0 cm diameter (circular) electrodes are spaced 5.0 mm apart and charged. The charge on the positive plate is +1nC and the charge on the negative plate is -1nC. An electron is released from rest at the surface of the negative electrode. (assume the space between the electrodes is a vacuum.)

A. How long does it take the electron to cross to the positive electrode?
B. What is its speed as it collides with the positive electrode?

 

[MorningGlory]

dia= 6cm
r = 3cm
r = 3 x 10^-2 m
Area, A of circular plate = pi x r^2
A= pi x (3 x 10^-2)^2
A = 9 x pi x 10^-4 m^2

Calculate capacitance
C=(eo x A)/d
Where d, separation distance of plate
C =( 8.85 x 10^-12 x 9 x pi x 10^-4)/5 x 10-3
C = 5.004 x 10^-12 F
C= 5pF

Now, calculate the voltage between the plates
Q = CV
V = Q/C
V = 1 x 10^-9/5.004 x 10^-12
V = 199.8 V
V = 200V

Knowing the voltage we can calculate the electric field between he plates
E = V/d
E = 200/5 x 10_3
E=4 x 10^4 Vm-1

Force on electron Due to electric field is
F = Eq
F = 4 x 10^4 x (-1.6 x 10^-19)
F = 6.4 x 10^-15 N (Taking magnitude only- ignoring sign)

But, Newtons second law states
F = ma
Therefore the acceleration of the electron in the electric field is
a = F/m
a = 6.4 x 10^-15/9.1 x 10^-31
a =7.03 x 10^15 ms-2

Time to travel distance s, (5mm) given by
s= ut + 0.5gt^2
electron starts from rest so u = 0
s = 0.5at^2
t = root[(2 x s)/a]
t = root[(2 x 5 x 10^-3)/7.03 x 10^15]
t = 1.19 x 10^-9 s
t = 1.2ns
This is the time it takes for the electron to cross to the positive electrode

Finally, use v = u +at
u = 0 as before
v = 7.03 x 10^15 x 1.19 x 10^-9
v = 8.3845 x10^6 m/s
v = 8.4 x 10^6 m/s

Speed is the magnitude of velocity (ignoring direction)
speed as it collides with positive plate = 8.4 x 10^6 m/s