Photoelectric effect stopping potential $q_eV_s = hf - \phi$
My question is... suppose we are testing the photoelectric effect. One plate is illuminated. We have applied the stopping potential. Suppose an electron leaves one plate with $KE = hf - \phi$ Now by the time it reaches the other plate, it should have 0 kinetic energy right? 
My question is about current... current should be 0 in this case, but charge has traveled from one plate to the other... so I'm confused as to how current can be 0.
Also, I'm having trouble thinking about the electron barely reaching the other side... vs being pulled back to the first plate before it can... are these situations somehow equivalent current wise?
Appreciate any clarification.
 A: You are right - the electron might make it all the way to the other plate. But then it will have zero velocity, and it still feels the electric field. So it will "fall back" to the plate it came from - and the net current is zero.
In reality the velocity distribution of the electrons is continuous (they don't all travel directly to the other plate with the maximum possible velocity) so you will see a drop in current as you increase the retarding potential.
A: Some notes:


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*this Kinetic Energy that you use is the Maximum amount of kinetic energy among the electrons. it means maybe just a few of them has this amount of energy, other photo electrons with lower kinetic energies (therefor lower Velocity) just accelerate back to the plate that they been shot from.

*the Kinetic energy actually gets to Zero at the next plate. it means the $V=0$ therefor the electron can not enter the circuit. so even the Highest energy photo electron couldn't reach the other plate to enter the circuit and create a current therefor $I=0$ and that voltage is called stoppage Voltage.
