l

**=[kT**_{D}_{ }e_{o}/ 4p n_{ }_{e}e^{2}]^{ }^{1/2}
Where k is Boltzmann’s constant and e is the electronic charge. Substituting the values for those two constants one can write a simpler numerical form:

l

**=10**_{D}^{0.84 }(ÖT/ Ön)
Technically, this applies

*only to electrons*. The ions will have a Debye length peculiar to the species of ion, e.g. for helium ions one will use Z = 2 and Lithium ions Z = 3. For this ion-species based Debye length we may write:
l

_{D,s}=[kT_{ s}e_{o }/ 4p Z_{ s}**n**^{2}_{ s}e^{2}]^{ }^{1/2}
Where the subscript ‘s’ refers to the

*ion species*.*Example Problem*: A helium plasma is at a temperature of 10

^{6 }K, and exhibits a number density of 10

^{10}/m

^{3}. Find the Debye length.

*Solution*:

_{ s}=2. T

_{ s}= T

_{ He}= 10

^{6 }K

n

_{ s}= n_{ }_{He }= 10^{10}/m^{3}
l

_{D,s}=[(1.38 x 10^{-23}(10^{6 }K) (8.85 x 10^{-12}F/m)_{ }/ (4p) 2**(10**^{2}^{10}/m^{3}) e^{2}]^{ }^{½}
l

_{D,s}= 0.30_{ }m**) Thus for r << l**

_{D,s}**the electric field of the test charge is unaffected by the presence of the plasma.**

_{D,s}However, for r >> l

**the E-field is attenuated by an exponential factor (exp –r/ l**

_{D,s}**) The test charge is then said to be shielded by the plasma with the shielding distance measured by l**

_{ D}**.**

_{ D}*.*Given electron and ion distributions:

_{e}= n

**exp [e j/ T**

_{o}_{e}]

*and:*n

_{i}= n

**exp [e j/ T**

_{o}_{i}]

We have:

*Ñ*^{2}^{ }F = e (n_{e - }n_{i})/ e_{ o}Which is

*Poisson’s equation,*so the potential F must be satisfied in S.I. units..

Adjusting the potential for the case for r >> l

**we have:**_{D,s}
j

**= q / 4p e**_{D}_{o }r (exp –r/ l**)**_{ D}Which also goes by the name, “

*Yukawa potential*”. It is of interest to note that because of the exponential factor it decreases much more rapidly than the Coulomb potential. This means the E-field from the test charge q is effectively shielded at distances much larger than l

**.**

_{ D}In addition to the Debye length scale we also need to use the

*plasma parameter*, L, defined:

*L*

*=*n

**l**

_{o }

^{3}_{ D}

This definition implies:

L

*=*n**l**_{o }^{3}_{ D }>> 1Which in turn implies that the number N of particles in a Debye sphere:

N = 4p n

**l**_{o }^{3}_{ D }/ 3**:**

*Problems*
1)For the helium plasma in the example problem, compute the plasma parameter, and the number of particles in the Debye. sphere.

2)Compare your values with the same ones for the solar atmosphere, for which:

n

n

**= 10**_{ o}^{20}/m^{3}and T = 10,000K.
3)Compare the values for a

n

*laser fusion plasma*with a particle density n**= 10**_{ o}^{29}/m^{3}and temperature T = 10^{7 }K) to the plasma of the tail of Earth’s magnetosphere with:n

**= 10**_{ o}^{6}/m^{3}and T = 10^{7 }K
4)Can a plasma with n

**= 10**_{ o}^{6}/m^{3}be maintained at an electron temperature of 100K? (*Hint*: Calculate the density limit using the plasma parameter).
5) In the limit 1/N << 1 and 1/N

**®**0 we say a plasma is “*collisionless*”. Do any of the plasmas cited in the previous problems qualify as collisionless?
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