# why stopping potential in the photoelectric effect is negative?

I am reading an introduction article for photoelectric effect. It is said that the variable potential was tuned until no single one photoelectron can reach the electron collection plate. But if we look at the current-voltage diagram, the stopping potential is negative. I have a hard time to understand why this potential is negative. As I learn from the text, the potential energy an electron experiences should be $-eV$, if voltage difference $V$ is negative, the potential energy should be positive. If the potential energy that an electron has is positive, should be electron move by itself spontaneously? So as my understanding, the more negative voltage difference there is, the more positive potential energy will be, so the electron should move towards the collection plate. So why there is stopping potential and why any voltage difference less than the stopping potential doesn't emit any electron?

For photon energies above what is needed to kick out photoelectons, the electron is departing the surface with some excess kinetic energy (or else you'd never detect them since they'd go nowhere). A negatively charged electron is repelled from a negatively charged plate, and to climb 'up' that potential to reach the plate requires $e \times V$ energy (conveniently given in eV!).
So, a positively charge plate will attract all emitted electrons, regardless of their kinetic energy upon leaving the solid. A neutral plate will pick up all electrons that headed towards it (and didn't scatter off any remaining gas in the vacuum system). As you make the plate increasingly negative, some of the emitted electrons just don't have enough kinetic energy to climb up the hill. At the point where the detected current goes to zero you have just counteracted the maximum excess kinetic energy. Then the photoelectron threshold is the energy of the incident photon minus $e \times V$.