Physics

I have no idea...I suck at physics

If anyone can solve this or at least point me in the right direction I'll give you 1 million internets.

Two charged, parallel, flat conducting surfaces are spaced d=1.00cm apart and produce a potential difference (delta)V = 625V between them.  An electron is projected from one surface directly toward the second.  What is the initial speed of the electron if it stops just at the second surface?

W = qV
W = .5mv^2

qV = .5mv^2
2qV/m = v^2
v = sqrt(2qV/m)

i must be thick because the only thing I know how to solve for in there is m.  What is V and what is q?

also so I don't just write this down without understanding it, where did you get W = .5mv^2?

I understand how you derived everything else

V is the voltage, or 625 volts.  q is the charge of the electron.

.5 m v^2 is the equation for kinetic energy.  The kinetic energy was converted into potential energy (qV).

V is the electric potential, and is measured in Volts.
The reason >>2 was using energy is because the electron moves across an electric potential difference.  By definition, a charged particle moving across a potential difference gains/loses an amount of energy equal to its charge * the difference in electric potential.

q = -1.6x10^-19 C for an electron,
E = V/d = 62500NC^-1

F = Eq = 1x10^-14N

a=F/m = ...
0 = u^2 + 2ad
sub in and do the math