which, for z>o, can be simplified to z The radius of curvature R of this surface Is given by、 72, which yields R2/a2 R 1+4 A13) 2 n) Also, from Fig. 4b, the tip-to-plane distance is d==(R=0,n=n2)=3m (A14) Eqs.(Al3),(A14)give the parameters a and no if Re and d are specified R a= d 7= R The electric field at the tip is E. dn do 2,and using Eq(A12) R=a7= 2/a -2)hn (A16) which can be expressed in terms of Rc, d, when r<<d, as (Al7) 4d Now, in order for the liquid to be electrostatically able to overcome the surface tension forces and start flowing, even with no applied pressure, one needs to have 16.522, Space Propulsion Lecture 23-25 Prof. Manuel Martinez-Sancheaη = R2 + z + a 2 ⎛ ⎝ ⎞ ⎠ 2 − R2 + z − a 2 ⎛ ⎝ ⎞ ⎠ 2 which, for z>o, can be simplified to z = η a2 4 + R2 1 −η 2 . The radius of curvature Rc of this surface is given by 1 Rc = zRR 1 + zR 2 ( )3 / 2 , which yields, Rc = 1 −η2 2η a 1+ 4 R2 / a2 1− η 2 ( )2 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 3/ 2 (A13) Also, from Fig. 4b, the tip-to-plane distance is d = z R = o,η = ηo ( ) = a 2 ηo (A14) Eqs. (A13), (A14) give the parameters a and ηo if Rc and d are specified: a = 2d 1 + Rc d ; ηo = 1 1 + Rc d (A15) The electric field at the tip is Ez = − ∂φ ∂z ⎛ ⎝ ⎞ ⎠ TIP = − dφ dη dη dz ⎛ ⎝ ⎜ ⎞ ⎠ TIP . Now ∂z ∂η ⎛ ⎝ ⎜ ⎞ ⎠ TIP = ∂z ∂η ⎛ ⎝ ⎜ ⎞ ⎠ R= o,η= ηo = a 2 , and using Eq. (A12), ETIP = − 2V / a 1 −ηo 2 ( )th−1 ηo (A16) which can be expressed in terms of Rc, d, when Rc<<d, as ETIP = − 2V / Rc ln 4d Rc ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ (A17) Now, in order for the liquid to be electrostatically able to overcome the surface tension forces and start flowing, even with no applied pressure, one needs to have 16.522, Space Propulsion Lecture 23-25 Prof. Manuel Martinez-Sanchez Page 9 of 36