Acceleration of electron

This question requires two answers. An electron is 1.5 cm from a -23.0 nC charge.

a. What will be the magnitude of the acceleration of the electron?

b. If released from rest at 1.5 cm, what will its speed be at 5 cm?

a)
Acceleration of electron
= force of attraction / mass
= [(9 x 10^9 x 1.6 x 10^-19 x 23 x 10^-9) / (0.015)^2] / (9.1 x 10^-31)
= 1.61758 x 10^17 m/s^2.

b)
Acceleration at a distance x cm
= 1.61758 x 10^17 * (0.015/x)^2 m/s^2
= 3.639555 x 10^13 / x^2 m/s^2
=> velocity at x cm = - 7.27911 x 10^13 / x^3 m/s^2
plugging x = 0.05 m,
magnitude of velocity at 5 cm
= 7.27911 x 10^13 /(0.05)^3 m/s
= 58233 m/s.