Electron Gun Equation Question

[rorygolledge]

I'm having trouble understanding how the electron gun equation can predict final velocity of an electron when accelerated between plates when no "distance travelled" variable is included,

"In an electron gun, an electron is accelerated by a potential difference of 28 kV.

At what speed will the electrons exit the assembly?"

I understand to derive the question, we let u = 0 and change in kinetic energy to equal work done.

(1/2)*mv^2 = qV

so v = sqrt((2qv)/m)

My problem is that shouldn't the final velocity given be different based on how long the electron is in the field for?

It just seems to fall apart for me as I see that if it is only in the electric field for 1cm, it has the same final velocity when exiting as if it were in the field for 100m when using this equation.