Question

In a vacuum diode electrons are emitted from a hot grounded
cathode (V=0) and they are accelerated towards the anode at
potential V_{0} (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:
d^{2}ϕ/dx^{2} = 4 π ρ e, with boundary condition
ϕ(x=0)=0 and ϕ(x=d)=V_{0}.

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:
(V_{0}) ^{3/2}. Notice that the current in a diode
does not obey Ohm’s Law.

Answer #1

Suppose a diode consists of a cylindrical cathode with a radius
of 6.200×10?2 cm , mounted coaxially within a
cylindrical anode with a radius of 0.5580 cm . The potential
difference between the anode and cathode is 280 V . An electron
leaves the surface of the cathode with zero initial speed
(vinitial=0). Find its speed vfinal when it
strikes the anode.

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