In the photoelectric effect, electrons are liberated from the cathode at a certain rate which is independent of the external voltage applied. I understand the relationship between voltage and current if I think of current as the rate of electrons hitting the anode. When increasing voltage, I understand that the rate of electrons hitting the anode remains the same since this is solely determined by the rate of electrons leaving the cathode and not the speed of the electrons.

However, my understanding breaks down when I try to understand current as the flow of electrons through the wires themselves.

  1. By increasing voltage, the electrons moving in the wires will have greater speeds and reach the cathode faster. Hence, how is it possible for the current to be the same in the wires?
  2. When voltage is constant, how is it possible for the current to be the same everywhere in the wires? Where does the extra energy go, since a stronger field has been established?

I think part of the problem may be my conflation of speed vs rate at which electrons leave the cathode. Any help will be greatly appreciated.


One metaphor to illustrate is the folowing:

Imagine a company $C$ which has two departments:

  1. A production department; let’s call it’s production rate $I$.

  2. A shipping/delivery department; let’s call it’s delivery speed $V$. The average daily rate isn’t affected by the the delivery speed, only the instant delivery rate at a given instant in time is affected. For instance, it could result in a peak of delivery at one given hour in the day… but then, that’s it for the rest of the day (in that particular case), however, again: the average daily, monthly, … , rate of deliveries remains unaffected.

Let’s say a supermarket orders goods from company $C$

Changing $V$ won’t modify $I$

If the company makes 1000 units per day, they cannot deliver more products by delivering them faster.


Here is a diagram of the photoelectric effect:


Light of a particular frequency strikes a clean metal surface inside a vacuum chamber. Electrons are ejected from the metal and are counted by a detector that measures their kinetic energy.

If no light hits the cathode, inside the tube it is like a capacitor at the ends of the battery, no current flows.

When light above a certain frequency is shined ,electrons come out, below that frequency, no current.

The voltage does not have to do with the experimental observation:


The circuit is used just to be able to detect the energy of the electrons, and shows that light, instead of all frequencies giving a current, there is a cutoff at low frequencies, different for different conductors.

Note the y axis , it is the maximum photo electron energy , for a given frequency , of the ejected electrons, a sequence of frequencies is needed in order to get the plot.

The battery voltage is irrelevant to the plot.In general, the current depends on the drift velocity of the electrons in the conductor, as seen here.


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