A2. 11 CHARGE RELAXATION Consider a conductive liquid with a conductivity K normally due to the motion of ions of both polarities If their concentration is and their mobilities are u,μ((m/s)/(v/m), then n(*+μ)⑤m) (1) Suppose there is a normal field eg applied suddenly to the gas side of the liquid surface. The liquid surface side is initially un-charged but the field draws ions to it (positive if eg points away from the liquid), so a free charge density o builds up over time. at a rate =KE The charge is related to the two fields, Eg, EI from the"pillbox"version of E。E-EEE= From(3), E and substituting in(2) (4) The quantity Eo=t is the Relaxation Time of the liquid. In terms of it, the solution of (4)that satisfies a(0)=0(for a constant En at t>O)is 16.522, Space P pessan Lecture 23-25 Prof. Manuel martinez Page 4 of16.522, Space Propulsion Lecture 23-25 Prof. Manuel Martinez-Sanchez Page 4 of 36 Consider a conductive liquid with a conductivity K, normally due to the motion of ions of both polarities. If their concentration is +- 3 n = n = n (m /s) and their mobilities are + - µ µ, ((m/s)/ (V/m)), then ( ) ( ) + - K = n + Si m µ µ (1) Suppose there is a normal field g En applied suddenly to the gas side of the liquid surface. The liquid surface side is initially un-charged, but the field draws ions to it (positive if g En points away from the liquid), so a free charge density σf builds up over time, at a rate f l n d = KE dt σ (2) The charge is related to the two fields, g l E , E n n from the “pillbox” version of free ∇ . D = ρ JG JG g l ε εε σ 0n 0n f E- E= (3) From (3), g l n f n 0 E E= - σ ε εε , and substituting in (2), f g f n 0 d K K + =E dt σ σ εε ε (4) The quantity 0 = K εε τ is the Relaxation Time of the liquid. In terms of it, the solution of (4) that satisfies ( ) f σ 0 =0 (for a constant l En at t>0) is A2.1.1 CHARGE RELAXATION