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82 Chapter 3 Poisson’s equation becomes –e– . derivative of and integrating from 1 = 0 to 1 = : (3.7-39) = – e , kTe o = e o , kTe = x , D –1/2 d 2 = 1+ o d2 This equation can be integrated once by multiplying both sides by the first 2 –1/2 – 0 2 d 1= 0 1+ o – 0e d . (3.7-40) where 1 is a dummy variable. The solution to Eq. (3.7-40) is 1 2 2 1/2 – 2 – = 2 o 1 + – 1 + e – 1 . ( 3 . 7 - 4 1 ) Since the electric field ( d / dx ) is zero away from the sheath where = 0 , =0 o rearrangement of Eq. (3.7-41) yields 1/2 1/2 = 4 o 1+ +2e– –2(2 o–1) . (3.7-42) o To obtain a solution for ( ), Eq. (3.7-42) must be solved numerically. However, as was shown earlier for Eq. (3.7-22), the right-hand side must always be positive or the potential will have an inflection at or near the sheath edge. Expanding the right-hand side in a Taylor series and neglecting the higher-order terms, this equation will also produce the Bohm sheath criterion and specify that the ion velocity at the sheath edge must equal or exceed the ion acoustic (or Bohm) velocity. An examination of Eq. (3.7.42) shows that the Bohm sheath criterion forces the ion density to always be larger than thePDF Image | Fundamentals of Electric Propulsion: Ion and Hall Thrusters
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