- In photoelectric cells, a current is detected when photoelectrons reach the electrode on the opposite side of the tube after being emitted. But shouldn't current be detected when photoelectrons leave the first electrode and not just when they reach the second electrode? Because this would create a positive charge on the first electrode which they are emitted from, so a redistribution of electrons in the external wire and therefore a current.
I don't know if I've explained it well enough, so here is a diagram to show what I mean...
Also, I don't understand how a current is detected when electrons reach the electrode on the opposite side. Is it because of an excess negative charge it gains?
Why does a potential difference need to be applied to a photoelectric cell if you're not investigating stopping potential? The electrons would reach the other side with their own kinetic energy, so what is the point of the battery in the diagram above?