Question

# In a vacuum diode electrons are emitted from a hot grounded cathode (V=0) and they are...

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.

#### Earn Coins

Coins can be redeemed for fabulous gifts.