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## Homework Statement

This problem is belonging to a book, which is material for Vietnamese students who will take part in IPhO (International Physics Olympiad). Since this problem is written in Vietnamese, I will try my best to translate it to English:

A vacuum diode consists of two parallel conductive plate, each of them has a area of S and their separation is L. The distance between the plates L is very small in comparison to the plates area S. A voltage U is applied to a plate, another plate was grounded. Because of thermal emission of electrons, electrons emitted from cathode will move to the anode (due to the applied voltage U, and the electric field created by another moving electron inside the vacuum space). The thermal emission effect is too great, so that limiting the current flowing thru the diode is impossible (read comment). In this situation we can consider that the electric field at the cathode equals to zero and the electrons velocity at the cathode is also zero. Prove that, the electric potential distribution inside the diode is given by the following equation:

V(x)=U*(x/L)^(4/3)

Find the relationship between U and I (the current flows thru the diode). Do the diode satisfy Ohm's Laws?

The comment (not included in the problem): In my opinion, this means that the diode works in the U/I region before saturation, when the current stays constant even that the applied voltage was increased.

## Homework Equations

V(x)=U*(x/L)^(4/3)

## The Attempt at a Solution

I tried finding the electron distribution inside the diode, then with superposition principle, I add up the electric field created by the charges distribution and the applied voltage U, then integrate to find the electric potential, but without success. I used Gauss's law differential equation, second Newton laws of motion for a single electron and the hint that current everywhere inside the diode is constant. I manage to find a second order differential equation of V, but it didn't give the answer when solving it.