Question

In a vacuum diode electrons are emitted from a hot grounded cathode (V=0) and they are accelerated towards the anode at potential V₀ (see Griffiths, prob.2.48). The clouds of emitted electrons build up until they reduce the electric field at the cathode to zero. From then on a steady current I flows between the plates. Let the anode and cathode be much larger than the distance d between them, so that the potential ϕ, electron density ρ and electron velocity v are all function of x only (the distance to the cathode). Show that:

a) The potential satisfies the following equation: d²ϕ/dx² = 4 π ρ e, with boundary condition ϕ(x=0)=0 and ϕ(x=d)=V₀.

b) Find the velocity v of the electrons at position x as a function of the potential ϕ(x) at that point.

c) From the constant electron current I = e ρ v S (where S is the surface of the anode/cathode), find the relation between the density ρ and the potential ϕ.

d) Inserting that relation into the equation for ϕ (see (a) above) solve the differential equation for the potential ϕ(x). Then find ϕ(x) and v(x). Show that the current I is proportional to: (V₀) ^3/2. Notice that the current in a diode does not obey Ohm’s Law.

Answer 1