physics 122 Lecture 02 October 1st 2013_final

0 kte d 2 ne where n n 0 exp x d the

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Unformatted text preview: ength 2 ∂φ ε 0 ∇ φ = ε 0 2 = −e ( ni − ne ) ∂x 2 This comes from: σ ∇⋅E = ε0 E = −∇φ 2 ∇ ⋅ E = ∇ ⋅ (−∇φ ) = −∇ φ Debye Length For a potential energy qφ the electron distribution function is: Integrating we have: Substituting for ni and ne : Expand using a Taylor series: ( "1 + % f (u) = A exp *− $ mu 2 + qφ ' kTe & ) #2 , ne = n∞ exp(eφ kTe ) )" " eφ %% , d 2φ ε 0 2 = en∞ +$ exp $ '' − 1. dx +# # kTe && . * x2 x3 e = 1+ x + + + ... 2! 3! x Keeping only the linear terms: x <1 2 " % d 2φ eφ 1 " eφ % ε 0 2 = en∞ $ + + ... ' $ kT 2 $ kT ' ' dx # e& #e & We can then define the Debye length, λD : So, the solution to the above equation is: 1/ 2 ! ε 0 kTe $ λD = # 2 & " ne % where n = n∞ φ = φ0 exp(− x / λD ) The Debye length is a measure of the shielding distance, or thickness of sheath. Debye Length 1/ 2 ! ε 0 kTe $ λD = # 2 & " ne % where n = n∞ Engineering formulas: " T λD = 69 $ °K−3 $n m # () 1/ 2 % 'm ' & " kTeV λD = 7430 $ $ n m −3 # () 1/ 2 % 'm ' &...
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This document was uploaded on 01/17/2014.

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