Voltage between metal plates...?

Question

I'm copying this right out of the book; I'll put the details of the alpha particle at the bottom.

A voltage is applied between two parallel metal plates separated by 30cm. The setup results in an electric field of 1500 N/C between the plates. An alpha particle is placed near the positive p late and released so that it accelerates toward the other plate. Calculate the kinetic energy and speed of the alpha particle when it passes through a small hole in the other plate. Calculate the voltage between the plates.

Alpha particle:

mass: 6.64 * 10 ^ -27 kg

charge: +2e

e= 1.602 * 10 ^ -19 C

If anyone can answer ANY part of this question, it would be such an awesome help to me! Thanks!

Answer 1

Since E=1500 N/C

V=voltage difference = E*0.3 meter

=1500*0.3=450 volts = work/unit charge

D=30/100=0.3 meter

charge=+2q=2*1.602*10 ^-19 C

work=energy=V*q

=450 * 2 * 1.60200 * (10^(-19)) = 1.4418 × 10-16 joules

K.E.=work

K.E=  1.4418 × 10-16 joules = (1/2)*m*v^2

v=sqr((2 * 1.4418 * 10^-16) / (6.64 * 10^-27))= 208,393 m/s

Answer 2

For the voltage, E = -V / d (forget the negative)

V= Ed = 1500 x .30 = 450 Volts

For the speed, all the potential energy goes into the kinetic energy

U= qV = (1/2)m v^2

solve for velocity you get

v= 208,393.1641 m/s