$q_eV_s = hf - \phi$
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.