What is the slowest speed the electron can have if it is to reach the negative plate?

Question

A parallel-plate capacitor is formed of two 19.0 cm × 19.0 cm plates and spaced L = 4.50 cm apart. The plates are charged to ±2.80 nC. An electron is shot through a very small hole in the positive plate.

Answer 1

Potential difference will do work to stop electron.

Capacitance of parallel plate capacitor C = e0*A / d

C = 8.854 x 10^-12 x (0.19 x 0.19) / 0.045

C = 7.1028756e-12 F

now using

Q = CV

2.8 x 10^-9 = 7.1028756e-12 x V

V = 394.21 Volt

Initial K.E. of electron = work done on electron

mv^2 /2 = qV

9.109 x 10^-31 x v^2 /2 = 1.6 x 10^-19 x 394.21

v = 11768021.6 m/s