Question

3.12 An infinite, thin, plane sheet of conducting material has a circular hole of radius a cut in it. A thin, flat disc of the same material and slightly smaller radius lies in the plane, filling the hole, but separated from the sheet by a very narrow insulating ring. The disc is maintained at a fixed potential V, while the infinite sheet is kept at zero potential. (a) Using appropriate cylindrical coordinates, find an integral expression involv- ing Bessei functions for the potential at any point above the piane. (b) Show that the potential a perpendicular distance z above the center of the disc 0.42) = v(1-V ) (c) Show that the potential a perpendicular distance z above the edge of the disc D.cz) = [1 - ka Kas] 3.18 where k = 2al(z? + 4a_)/2, and K(k) is the complete elliptic integral of the first kind. The configuration of Problem 3.12 is modified by placing a conducting plane held at zero potential parallel to and a distance L away from the plane with the disc insert in it. For definiteness put the grounded plane at z = 0 and the other plane with the center of the disc on the z axis at z = L. (a) Show that the potential between the planes can be written in cylindrical co- ordinates (z, p, *) as 0(2, p) = v dA ,(1)).(^pla) sinh(Azla) sinh(ALla) (b) Show that in the limit a → with z, p, L fixed the solution of part a reduces to the expected result. Viewing your result as the lowest order answer in an expansion in powers of a ', consider the question of corrections to the lowest order expression if a is large compared to p and L, but not infinite. Are there difficulties? Can you obtain an explicit estimate of the corrections? Consider the limit of L — with (L - z), a and p fixed and show that the results of Problem 3.12 are recovered. What about corrections for L > a, but not L →00? Jo

Answer 1

(a) The general solution of the Laplace equation in cylindrical coordinates with angular symmetry that vanishes at z = 0 is 00 (2, 3) = |A(k)Jo(kg) sinh(k=) dk.

Multiplying both sides by pJ. (k'p) and integrating at z = L yields | PJo( 2) (P, L) dp [(4) sinh (AL){ PJo( ) (+p) do ak - %* 4(k) sinh(kt) {fork – x)} dk 2A(k') sinh(L) so k AK) – Ak) = sinh(KL) JO pJo (kp) (p, L) dp Vera Vk pJo(kp) dp sinh(KL) Jo V pka k sinh(KL) JO uJ(u) du. I worked out this integral earlier, in Problem 3.12: I uJo(u) du = r.J7(1). Then (29) becomes A(k) = k sinh(k:L) · (ka)Ji(ka) and (28) is D(0, 3) = V [ⓇaJi(ka) Jo(kp) =v [J(1)Jo(^p/a) sinh(AL/a) () 1 sinh(kz) al- sinh(KL) sinh(Az/a) dl. [1] (b) For r <1, Jo(a) – 1 – 222 +... and for r < 1 and y<1, sinh(2) +223 +... rſ sinh(y) y + y2 + ... y [ 6 With these approximations we may expand the terms containing a in (1 ): sinh(r) – + t = [1 +=(x2 – ) +0(x4) - = – 1 + -1: +0(x4) sinh(Az/a) Joapa) sinh(XL/a) dolaplay kimlicaz ya = (1-: %)') ) (1+()*22–)) - E -- (a) * (6_22-2) + **) +.]

Then the potential expansion ( 1 ) becomes 312,3) =" Bilan - - [802-1) +++] Lºxtowych +..] The first integral evaluates to 1, so for a infinite the potential becomes simply (2) = Vz/L. This is just what we expect to get for the potential between two infinite sheets, one grounded and the other at potential V. The second integral, unfortunately, has a bit of an infinity problem. It's not hard to see where the problem comes: I derived the expansion above based on the premise that Va is small, but the integral goes over all up to , so for any finite a the expansions eventually become invalid in the integral. I'm still trying to work out a better procedure for estimating corrections for finite a. (c) In this part we're interested in taking and looking at the potential a fixed distance away from the plane with the circular insert. Calling the fixed distance z', the z coordinate of the point we're interested in is L - Z'. We have sinh k(L – z') sinh kL = sinh(KL) cosh(-kz') + cosh(KL) sinh(-kz') sinh kL = cosh(kz') - coth(kl) sinh(kz') (2) Now, coth(KL) differs significantly from 1 only for klas 1, in which region kz! Sz/L < 1, so cosh(kz') 1 and sinh(kz) ~0. By the time k gets big enough that kz' is starting to get significant, coth(KL) has long since started to look like 1, so the two terms in (2) add directly. The result is that, for all k, (2) can be approximated as exp(-kz'). Then (1) becomes ${p, z) = aV | J1(ka)Jo(kp)e-kz' dk as we found in Problem 3.12.

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