Solutions to Chen's Plasma Physics

Now we’ll suppose the φ goes like a decreasing exponential: φ = φ 0 exp(−x/λ D ). Thus, the Laplacian
acting on this is
∇ 2 φ = 1
λ 2 φ = ne2 ( φ +
φ ) ⇒ 1
D
ɛ 0 kT e kT i λ 2 D
= ne2
ɛ 0
( 1
kT e
+ 1
kT i
) ✷ (10)
To show that λ D is determined by the colder species, we suppose first that the electrons are the
colder species: T e ≪ T i . Then,
1
λ 2 D
√
= ne2
kT e ɛ 0
⇒ λ D =
kT e ɛ 0 ne 2 ✷ (11)
Alternatively, if the ions are colder, T i ≪ T e , then a similar analysis yields:
1
λ 2 D
√
= ne2
kT i ɛ 0
⇒ λ D =
kT i ɛ 0 ne 2 ✷ (12)
1-6. An alternative derivation of λ D will give further insight to its meaning. Consider
two infinite, parallel plates at x = ±d, set at potential φ = 0. The space between them
is uniformly filled by a gas of density n of particles of charge q.
a) Using Poisson’s equation, show that the potential distribution between the plates
is
φ = nq
2ɛ 0
(d 2 − x 2 ) (13)
b) Show that for d > λ D , the energy needed to transport a particle from a plate to the
mid plane is greater than the average kinetic energy of the particles.
1-9. A distant galaxy contains a cloud of protons and antiprotons, each with density
n = 10 6 m −3 and temperature T = 100 o K. What is the Debye length?
The Debye length is given by
λ D = ∑ ɛ 0 kT j
n
j j e 2 (14)
j
Plugging in the numbers:
λ D = 8.85 × 10−12 × 1.4 × 10 −23 × 100
10 6 × (1.6 × 10 −19 ) 2 = 0.48 m ✷ (15)
As a check, use the SI unit form for the Debye length given in Chen. If T is in Kelvin, and n is in
cubic meters, then:
√
√
T
λ D = 69
n m = 69 × 10 2
10 6 m = 69 × 10−2 m ✷ (16)
This is the same order of magnitude so we are ok.
1-10. A spherical conductor of radius a is immersed in a plasma and charged to a
potential φ 0 . The electrons remain Maxwellian and move to form a Debye shield, but
the ions are stationary during the time frame of the experiment. Assuming φ 0 ≪ kT e /e,
derive an expression for the potential as a function of r in terms of a, φ 0 , and λ D . (Hint:
Assume a solution of the form e −br /r.)
Let’s assume a solution of this form: φ = Ae −br /r. Then,
∇ 2 φ = 1 ∂ ∂φ
r 2 (r2
∂r ∂r ) = b2 φ = e (n e − n i ) (17)
ɛ 0