Basic Plasma Physics - Homework, 3/2/2012 1. Debye...
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Basic Plasma Physics - Homework, 3/2/2012 1. Debye shielding: Consider a test charge q t and compute the net charge of the Debye shielding cloud (Yukawa potential) as a function of radial distance measured in units of the Debye length. Compute the net charge of the shielding cloud for the Yukawa potential. Φ D = q t 4π 0 r exp - r λ D (1) with λ -2 D = n 0 e 2 0 k B 1 T e + 1 T i Poisson’s equations: ∇ 2 Φ= - 1 0 q t δ (r)+ 1 λ 2 D Φ(r)= - 1 0 ρ c This yields for the charge density: ρ c = q t δ (r) - 0 λ 2 D Φ(r) = q t δ (r) - q t 4πλ 2 D 1 r exp - r λ D Integration over total space yields the total charge: Q net = q t - q t 4πλ 2 D Z 2π 0 Z π 0 Z R 0 1 r exp - r λ D r 2 dr sin θdθdφ = q t - q t λ 2 D Z R 0 r exp - r λ D dr = q t - q t λ 2 D -λ D r exp - r λ D - λ 2 D exp - r λ D R 0 = q t - q t λ 2 D -λ D R exp - R λ D - λ 2 D exp - R λ D + λ 2 D = q t R λ D +1 exp - R λ D where the integral over r is solved through integration by parts. Note, an alternative derivation of the charge density can be obtained by computing ∇ 2 Φ D as indicated in class (with the radial electric field given by
Transcript of Basic Plasma Physics - Homework, 3/2/2012 1. Debye...
Basic Plasma Physics - Homework, 3/2/2012
1. Debye shielding:
Consider a test charge qt and compute the net charge of the Debye shielding cloud (Yukawa potential) asa function of radial distance measured in units of the Debye length.
Compute the net charge of the shielding cloud for the Yukawa potential.
ΦD =qt
4πε0rexp
(− r
λD
)(1)
with
λ−2D =
n0e2
ε0kB
(1
Te+
1
Ti
)
Poisson’s equations:
∇2Φ = − 1
ε0qtδ(r) +
1
λ2D
Φ(r) = − 1
ε0ρc
This yields for the charge density:
ρc = qtδ(r)− ε0λ2D
Φ(r)
= qtδ(r)− qt4πλ2
D
1
rexp
(− r
λD
)
Integration over total space yields the total charge:
Qnet = qt −qt
4πλ2D
∫ 2π
0
∫ π
0
∫ R
0
1
rexp
(− r
λD
)r2dr sin θdθdφ
= qt −qtλ2D
∫ R
0r exp
(− r
λD
)dr
= qt −qtλ2D
[−λDr exp
(− r
λD
)− λ2
D exp(− r
λD
)]R0
= qt −qtλ2D
[−λDR exp
(− R
λD
)− λ2
D exp(− R
λD
)+ λ2
D
]= qt
[R
λD+ 1
]exp
(− R
λD
)
where the integral over r is solved through integration by parts.
Note, an alternative derivation of the charge density can be obtained by computing∇2ΦD as indicated inclass (with the radial electric field given by
Er = −∂Φ
∂r=(r
λD+ 1
)qt
4πε0r2exp
(− r
λD
)
Qnet =∫ R
sphereρcd
3r = ε0
∫ R
sphere∇ · Ed3r
= ε0
∫ R
surfaceE · d2s = 4πR2ε0Er (R)
= qt
(R
λD+ 1
)exp
(− R
λD
)
Note be care full to consider that∇(1/r) = r/r3 and ∇ · (r/r3) = 4πδ(r).
2. Plasma properties:
(a) Calculate the electron thermal speed, Debye length, and the plasma parameter for
• a tokamak plasma with Te = 108 K, n0 =1019 m−3
• the tail magnetosphere with Te = 107 K, n0 =106 m−3
• the ionosphere with Te = 103 K, n0 =1012 m−3
• the solar atmosphere with Te = 104 K, n0 =1020 m−3
with kB = 1.38 · 10−23 JK−1, ε0 = 8.85 · 10−12 F/m, me = 9.11 · 10−31 kg, and e = 1.60 · 10−19 C
Solution: Electron thermal speed:
vthe =
(kBTeme
)1/2
= 3.9 · 103T 1/2e m/s
Electron Debye length:
λDe =
(ε0kBTen0e2
)1/2
=
(8.85 · 10−12 · 1.38 · 10−23
1.62 · 10−38
)1/2 (Ten0
)1/2
m
= 69.1 ·(Ten0
)1/2
m
Electron plasma parameter:
Λe = n0λ3De = 3.3 · 105T
3/2e
n1/20
Te / K n0 / m−3 vthe / m/s λDe / m Λe
Tokamak 108 1019 3.9 · 107 2.2 · 10−4 1.04 · 108
Magnetotail 107 106 1.2 · 107 220 1.04 · 1013
Ionosphere 103 1012 1.2 · 105 2.2 · 10−3 1.04 · 104
Solar atmosphere 104 1020 3.9 · 105 6.9 · 10−7 33
(b) Compute collision frequency and mean free path for these plasmas.
Solution: Plasma frequency:
ωpe =
(n0e
2
meε0
)1/2
(2)
ωpe =
(n0e
2
meε0
)1/2
= 1.60 · 10−19(
n0
9.11 · 10−318.85 · 10−12
)1/2
=1.60 · 10−19
2.84 · 10−21n
1/20 = 56.34n
1/20
Collision frequency:
νc =
√π
2
1
32π
ωpΛ
ln [12πΛ]
= 1.247 · 10−2ωpΛ
ln (37.7Λ)
Mean free path:
Lc = vthe/νc
=
√2
π32π
(kBTeme
)1/2 (meε0n0e2
)1/2 Λ
ln [12πΛ]=
√2
π32π
λDeΛ
ln [12πΛ]
n0 / m−3 Λe vthe / m/s ωp / s−1 νc / s−1 ln (37.7Λ) Lc / mTokamak 1019 1.04 · 108 3.9 · 107 1.78 · 1011 4.71 · 102 22.09 8.28 · 104
Magnetotail 106 1.04 · 1013 1.2 · 107 5.63 · 104 2.27 · 10−9 33.60 5.29 · 1015
Ionosphere 1012 1.04 · 104 1.2 · 105 5.63 · 107 8.69 · 102 12.88 1.38 · 102
Solar Atmosph. 1020 33 3.9 · 105 5.63 · 1011 1.52 · 109 7.13 2.5 · 10−4
3. Plasma Definition
Can a fully ionized plasma be maintained at temperatures of Te = 100 K (Hint: derive a condition for thedensity in relation to the temperature). How important is recombination for such a cold plasma?
Solution:
Plasma definition:
Λ = nλ3D = n−1/2
(ε0kBTee2
)3/2
� 1
Taking the square
n−1
(ε0kBTee2
)3
= n−1
(8.85 · 10−12 · 1.38 · 10−23 · 102
1.62 · 10−38
)3 [m−3
]= n−1
(4.8 · 105
)3 [m−3
]= n−1 · 1.1 · 1017
[m−3
]or taking number density as particle per cm−3
√n� 3.3 · 105
which implies that n could be as small as 107or even 108 cm−3 to satisfy the plasma definition, i.e., thatthe potential energy is much smaller than the kinetic energy of the charged particles.
While this is true there is the additional aspect of how long such a plasma could be maintained. Eventho a sufficiently low density gas of charged particles can satisfy the plasma definition, recombinationcaused by collisions can remove the the charged particles. For long life time of such a plasma it wouldbe necessary to have a very low level of collisions that could result in recombinations. In other wordsthe collision frequency must be sufficiently low and with a comparatively small value of the plasmaparameter of for instance 102 the collision frequency is still quite large and would lead to relatively fastrecombination leading to a loss of this plasma within a short time frame.
