Whenever a book talks about the photoelectric effect, it talks about how we can use a battery to attain a $V_{stop}$. Basically, you are making the metal plate on which you don't shine light have a negative charge. This prevents the electrons from going from one plate to another.

Now, I do not understand why there is no current at $V_{stop}$. After all, the metal plate on which you shine light is positive. The electrons, being excited, would preferentially go to the positive side than to the negative side. So, should the direction of the current not be reversed rather than inexistent?

  • $\begingroup$ In other words, how come the electrons don't travel in the opposite way? Why do they always have to come out of the surface? $\endgroup$
    – yolo123
    Nov 5, 2014 at 4:09
  • $\begingroup$ If when excited they do not have enough energy to go to the negative plate, falling back will balance the charge of the plate to where it was, thus no current. If they have enough energy to cross the charge on the unshined plate changes and current has to flow to reach the capacitor's equilibrium. $\endgroup$
    – anna v
    Nov 5, 2014 at 5:31

1 Answer 1


The thing is that the plates are already charged, see the picture below, hence no current flows in absence of light (charged capacitor). The sum of the voltages around the loop is already zero. Of course initially there must be some current when you connect the plates with the battery. That will charge the plates and the final charge will depend on the capacitance of the configuration.

Assuming that the plates are already charged, the incoming photons with enough energy will allow some electrons (the quantity depends on the intensity of the light) to reach the collector, which will increase the potential difference between the plates. As a result a current will flow to restore the voltage to the one imposed by the battery.

You can check this link if you still have doubts.

enter image description here


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