photoelectric effect

Given this schematic, and that there is an electric field in the downwards direction due to the power supply, when the photon hits the emitter plate, the electrons leave the upper plate and hit the lower plate.

If the electrons succesfully reach the collector plate, doesn't this mean there will be a buildup of negative charge in the lower part of the circuit and a positive charge in the top left part of the circuit?

Also when the electrons successfully reach the collector, how does this create a current if electrons are also coming out of the negative terminal of the power supply?


The plates of the tube make a capacitor. We'll call it a tube-capacitor.

Before the radiation is applied and electrons start flying down to the bottom plate (collector), the tube-capacitor gets charged by the battery to the battery voltage, which puts the circuit in balance, i.e., the battery voltage is exactly opposed by the voltage on the tube-capacitor.

This is no different than a regular capacitor charged by a regular battery, i.e., the charging continues until the capacitor gets charged to the battery voltage.

So, when the radiation is applied and the electrons get knocked out of the top plate, they have to go against the field between the plates of the tube-capacitor, already charged to the battery voltage.

If the field in the tube-capacitor is not too strong (voltage is not too high), most energetic electrons would be able reach the bottom plate. When they do, the voltage on the tube-capacitor becomes greater than the battery voltage, i.e., the balance of the circuit is disturbed.

To restore the balance, the capacitor has to discharge (back to the battery voltage level) and, as it does, the excessive electrons, which have landed on the collector, move around the circuit (back to the top plate that lost them), while the ammeter is measuring the current.

If the battery voltage is raised to a high enough level (stopping voltage), the field in the tube-capacitor, charged by the battery, becomes strong enough to prevent even most energetic electrons, emitted by the top plate, from reaching the collector and, therefore, the current in the circuit becomes zero.

  • $\begingroup$ How do the electrons go around the circuit back up to the top plate, if electrons are coming out of the power source? Also I could you please clarify what you mean by the battery voltage being opposed by the voltage on the tube? To my knowledge the circuit isn't complete until the radiation is applied and a photoelectron manages to reach the other plate. $\endgroup$
    – Jem Bahki
    Aug 23 '18 at 1:57
  • $\begingroup$ @JemBahki I've updated the answer with more details. Think about the tube as a capacitor. Think about how a regular capacitor gets charged. $\endgroup$
    – V.F.
    Aug 23 '18 at 5:02
  • $\begingroup$ @JemBahki You are welcome:) $\endgroup$
    – V.F.
    Aug 24 '18 at 11:17

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