11764-515-M = RL-2226 011764-515-M 6 December 1973 MEMO TO: FROM: SUBJECT: File T. B. A. Senior More about the surface field on a constant impedance half plane In Memorandum 011764-514-M it was suggested that K+(k) ---- 1 as |1 -> o. This is incorrect. From a more detailed consideration of the split functions (to be reported later), it can be shown that K +() ) os8 (1+cos 0) 2 l1+sin(x+O) + cos X+ sin 0 cos X+ sin 1 -sin(- ) 1/4 cos X - sin 0 ex + OD O exp+ 6 *^ r j""""""""~8 v dv cosv v (1) where cos = 1/ and cos O = /k. For large | [, sin 0. k e -~ i log 2 k and, hence, if rl 0, K+(9) cos)2 (i2) 4 exp i- x) v dv COs v As 1 — o o, both integrals tend to zero, giving DISTRIBUTION Knott Liepa Hiatt/File 1

011764-515-M - 1/2 K+(O) ~ r 0; (2) K+(k) thus K+(k) 2 K (3) thus (3) as I c — + o, from which we obtain 1/2 EZ(0,+0)= /2 K+(k). (4) z-' K (k) r / exp(- ) and in this case we now have E (0, +0) exp (-(5) When r = 8, corresponding to a normalised electrical resistivity of 4, eq. (5) gives E (0, +0)= 0.9606 Z - which is fantastically close to the value obtained from the computed data for electrically resistive sheets. 2