Kirk on the bowl problem

Conducting Spherical Shell with a Circular Orifice Kirk T. McDonald Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544 (September 21, 2002) 1P r o b l e m A conducting spherical shell with a circular orifice of half angle θ0is at electric potential V0. Show that the difference between the charge densities on the inner and outer surfaces is independent of position, and estimate the ratio of the electric charge on the inner surface to that on the outer. Correct results can be inferred from “elementary” arguments based on superposition, and more “exact” derivations can be based on Legendre polynomial expansions, Green’sfunctions, or the method of inversion. 2S o l u t i o n This mixed boundary value problem is taken from the classic essay by G. Green (1828) [1],where it was discussed using what are now called Green’s functions. We first present an “elementary” solution, and then seek confirmation based on an expansion of the potential in a series of Legendre polynomials, and further confirmation via the methods of Green andThomson. 2.1 Elementary Solution via Superposition We wish to relate the problem of the shell with orifice (which we will also call a sphericalbowl in case angle θ 0is large) to the simpler problem of a complete shell that is at potential V0and hence has total charge Q0=aV0(in gaussian units), uniform radial electric field E0=Q0/a2=V0/aat its outer surface, zero electric field in the interior, and uniform surface charge density σ0=Q0/4πa2=V0/4πa. Following a well-known argument, to a first approximation the configuration of the shell with orifice is equivalent to removing a spherical cap from the complete shell, keeping surfacecharge density σ 0on that cap. The electric field above and below the spherical cap due to this charge density is ±2πσ 0=±E0/2. Subtracting this field from that of the initially complete sphere, we find the approximate field in the orifice to be E0/2. The shell with orifice has charge density σ+on its outer surface and σ−on its inner surface, with total charges Q±on these surfaces. All of the field lines that emanate from the charge distribution σ−on the inner surface pass through the orifice of area Aorifice,s ot h e average electric field at the orifice is the same as if charge Q−were distributed across the orifice; namely Eorifice≈4πQ−/Aorifice.E q u a t i n gt h i st o E0/2=Q0/2a2, we find Q−≈Q0Aorifice 8πa2≈Q0θ2 0 8≈aV0θ2 0 8, (1) 1 where the latter forms hold for a circular orifice of half angle θ0,w h o s ea r e ai s Aorifice≈πa2θ2 0. To a first approximation, the charge Q+on the outer surface of the shell with orifice remains Q0. Hence, we estimate that Q− Q+≈Aorifice 8πa2≈θ2 0 8. (2) We can write the charge density on the outer suface of the conducting shell with orifice asσ+=σ0+Δσ+,w h e r eΔ σ+/lessmuchσ0(except for very small values of distance sfrom the edge of the orifice where we expect that σ+varies as 1 /√ s. Similarly, the charge density σ−, whose integral is Q−/lessmuchQ0,o b e y s σ−/lessmuchσ0with the possible exception of very small values of distance s. The electric fields due the densities σ0,Δσ+andσ−on the conducting shell with orifice must sum to zero inside the material of the shell. The electric field due to charge densityσ 0is zero in the interior a complete shell, and remains near zero when that density exists on the shell except in the orifice. Hence, the electric field inside the conducting material of the shell due to the charge densities Δ σ+andσ−must sum to zero, or very nearly so. The usual argument based on Gaussian pillboxes tells us that the electric field just insidecharge density Δ σ +is−4πΔσ+, while that just outside charge density σ−is 4πσ−.T h u s , we conclude that Δ σ+≈σ−for all points not close to the edge of the orifice, and we might suppose this relation holds there as well. Expressing this conclusion as a relation between the densities σ+andσ−,w eh a v e σ+−σ−=σ0+Δσ+−σ−≈σ0=V0 4πa. (3) Assuming eq. (3) to be correct, we can deduce the first correction to the total charge Q+ on the outer surface of the shell with orifice. Namely, Q+≈Q0+Q−−Qcap=Q0−Q−=Q0/parenleftBigg 1−θ2 0 8/parenrightBigg , (4) and hence, Q++Q−≈Q0=aV0. (5) We infer that if a small hole could be created spontaneously in a conducting spherical shell, the charge originally at the place of the hole would redistribute itself half on the outer andhalf on the inner surface of the shell with orifice. The value of the electric potential of theshell would not change during this process. Since the above model does not necessarily reproduce all details of the actual charge distribution on the shell with orifice, it is useful to confirm the results with alternativeanalyses. 2.2 Solution via Legendre Series By expanding the potential in a Legendre series and applying boundary conditions at the sur-face of the shell, we find “exact” confirmation of the result (3) and approximate confirmationof eq. (2). 2 We work in a spherical coordinate system ( r, θ, ϕ) in which the center of the sphere of radius ais at the origin, and the + zaxis passes through the center of the circular orifice. Once we have an expression for the azimuthally symmetric potential V(r, θ), we obtain the electric field as E=−∇Vand the surface charge densities σ±as σ±(θ)=±Er(r=a±,θ) 4π=∓1 4π∂V(r=a±,θ) ∂r, (6) in Gaussian units. The charge density on the outer surface is labeled σ+. As the potential V(r, θ) is finite at the origin and at large r, an appropriate Legendre series expansion is V(r<a ,θ )=/summationdisplay nAn/parenleftbiggr a/parenrightbiggn Pn(cosθ), (7) V(r>a ,θ )=/summationdisplay nAn/parenleftbigga r/parenrightbiggn+1 Pn(cosθ), (8) which is continuous across the conducting spherical shell at r=a. Hence, we obtain the following series expansions for the charges densities (6), σ+(θ)=1 4πa/summationdisplay n(n+1 )AnPn(cosθ), (9) σ−(θ)=1 4πa/summationdisplay nnAnPn(cosθ). (10) The potential V0on the conductor, whose coordinates are ( r=a,θ 0≤θ≤π), provides a partial boundary condition at r=a, V0=/summationdisplay nAnPn(cosθ<cosθ0), (11) With this condition we can confirm relation (3) between the charge densities (9) and (10), σ+−σ−=1 4πa/summationdisplay nAnPn(cosθ<cosθ0)=V0 4πa. (12) Indeed, whatever the shape of the orifice(s) in the conducting shell, condition (11) holds for the remaining conducting region, and hence relation (12) holds also [2]. The boundary condition for the rest of the spherical shell is that the radial component Er=−∂V(r=a,θ < θ 0)/∂rof the electric field is continuous there, which leads to 0=/summationdisplay n(2n+1 )AnPn(cosθ>cosθ0). (13) We can combine the partial conditions (11) and (13) into a condition over the whole range of θ, /summationdisplay nAnPn(cosθ)=⎧ ⎪⎨ ⎪⎩−2/summationtext nnAnPn(cosθ)(θ<θ 0), V0 (θ>θ 0).(14) 3 For an approximate solution, we keep terms only up to the largest order nfor which Pn(cosθ0)≈1, in which case we can obtain simple analytic results via the usual method of evaluation of coefficient Amby multiplying the boundary condition by Pmand integrating over cos θ. Since the Legendre polynomial Pnhasn−1 zeroes over the interval 0 <θ<π , we must have θ0<∼π/n,o rn<∼π/θ 0. Thus, we approximate the condition (14) as π/θ0/summationdisplay n=0AnPn(cosθ)≈⎧ ⎪⎨ ⎪⎩−2/summationtextπ/θ0 n=0nAn(θ<θ 0), V0P0 (θ>θ 0),(15) using P0=1 . W es e t An=0f o r n>π / θ 0. To isolate Amfor 1<m<π / θ 0we multiply eq. (15) by Pmand integrate from −1t o1 with respect to dcosθto find Am=2m+1 2⎛ ⎝−2π/θ0/summationdisplay n=0nAn/integraldisplay1 cosθ0Pmdcosθ+V0/integraldisplaycosθ0 −1P0Pmdcosθ⎞ ⎠ ≈2m+1 2⎛ ⎝−2π/θ0/summationdisplay n=0nAn/integraldisplay1 cosθ0dcosθ+V0/integraldisplay1 −1P0Pmdcosθ−V0/integraldisplay1 cosθ0dcosθ⎞ ⎠ =V0δ0m−2m+1 2⎛ ⎝2π/θ0/summationdisplay n=0nAn+V0⎞ ⎠(1−cosθ0) ≈V0δ0m−(2m+1 )θ2 0 4⎛ ⎝2π/θ0/summationdisplay n=0nAn+V0⎞ ⎠, (16) recalling that Pm≈1f o rc o s θ>cosθ0. To complete the evaluation of Amwe need the sum/summationtextπ/θ0 n=0nAn, which we find from eq. (16) to obey π/θ0/summationdisplay m=0mAm=−π/θ0/summationdisplay m=0m(2m+1 )θ2 0 4⎛ ⎝2π/θ0/summationdisplay n=0nAn+V0⎞ ⎠ ≈−π3 6θ0⎛ ⎝2π/θ0/summationdisplay n=0nAn+V0⎞ ⎠, (17) since π/θ0/summationdisplay m=0m(2m+1 )≈/integraldisplayπ/θ0 0x2dx=2 3(π/θ 0)3. (18) Solving eq. (17) we have π/θ0/summationdisplay n=0nAn≈−V0π3/6θ0 1+π3/3θ0, (19) and 2π/θ0/summationdisplay n=0nAn+V0≈V0/parenleftBigg 1−π3/3θ0 1+π3/3θ0/parenrightBigg =V01 1+π3/3θ0≈3θ0 π3V0. (20) 4 In sum, we approximate the Fourier coefficients as An≈⎧ ⎪⎨ ⎪⎩V0δ0n−(2n+1)3θ3 0 4π3V0(n<π / θ 0), 0( n>π / θ 0).(21) The potential on the axis θ= 0 inside the spherical shell is V(r<a , 0) =π/θ0/summationdisplay n=0An/parenleftbiggr a/parenrightbiggn Pn(1)≈V0⎛ ⎝1−π/θ0/summationdisplay 0(2n+1 ) 3θ3 0 4π3/parenleftbiggr a/parenrightbiggn⎞ ⎠, (22) which drops from V0at the center of the sphere to V(a,0)≈V0⎛ ⎝1−π/θ0/summationdisplay 0(2n+1 ) 3θ3 0 4π3⎞ ⎠≈V0/parenleftBigg 1−3θ0 4π/parenrightBigg (23) at the center of the circular orifice. The equipotential surfaces, which would be spherical in the absence of the orifice, are deflected towards the orifice, and in some cases the deflectionforms a “bubble” that passes through the orifice into the interior of the spherical shell. Very similar behavior occurs for the case of a circular hole in a conducting plane, as illustrated in Fig. 3.14 of [3]. Associated with this behavior is the presence of some charge on the interiorsurface of the spherical shell. T h ec h a r g ed e n s i t y σ −on the inner surface of the spherical shell follows from eq. (10), σ−≈1 4πaπ/θ0/summationdisplay n=0nAnPn(cosθ)≈V0 4πaπ/θ0/summationdisplay n=0n(2n+1 ) 3θ3 0 4π3Pn(cosθ). (24) Although there is no physical charge in the region θ<θ 0, eq. (24) is formally defined there, and/integraltext1 −1σ−dcosθ= 0. This permits the charge Q−on the inner surface of the spherical shell with a circular orifice of area Aorifice≈πa2θ2 0to be calculated as Q−=/integraldisplaycosθ0 −12πa2σ−dcosθ=2πa2/integraldisplay1 −1σ−dcosθ−2πa2/integraldisplay1 cosθ0σ−dcosθ ≈V0 4πaπ/θ0/summationdisplay n=0n(2n+1 ) 3θ3 0 4π3Aorifice≈V0 8πaAorifice≈aV0θ2 0 8. (25) We see that the result Q−≈V0Aorifice/8πaholds for a small orifice of any shape [2]. The charge density σ+on the outer surface to that on the inner surface by eq. (12), so the charge Q+on the outer surface is given by Q+=Q−+/integraldisplaycosθ0 −12πa2V0 4πadcosθ ≈aV0/parenleftBiggθ2 0 8+1+c o s θ0 2/parenrightBigg ≈aV0/parenleftBiggθ2 0 8+1−θ2 0 4/parenrightBigg ≈aV0/parenleftBigg 1−θ2 0 8/parenrightBigg . (26) The total charge on the spherical shell is Q++Q−≈aV0, (27) 5 which is the same as the charge on a complete conducting sphere of radius aat potential V0. The ratio of the charge on the inner surface to that on the outer surface is Q− Q+≈Aorifice 2Asphere≈θ2 0 8. (28) The approximate result (28) confirms eq. (2), but is based on the truncated set of Fourier coefficients (21). Hence, additional confirmation is still desirable. 2.3 Solution via Green’s Functions Green [1] introduced what are now called Green’s functions to provide a (then) new derivation of Poisson’s integral for a sphere of radius a, namely that the potential at an interior point (b,θ,φ) is given in terms of the potential on the surface Sas V(r, θ, φ)=a2−b2 4πa/integraldisplayV(r/prime=a,θ/prime,φ/prime) R3dS/prime, (29) where Ris the distance from the interior point to area element on the surface [4]. He then considered the case of a shell at potential V0with a circular orifice of half angle θ0/lessmuch1, and imagined the effect of the orifice is the same as the superposition of the spherical cap of half angle θ0on a complete shell and a conducting disk of radius aθ0of charge opposite to that of the cap. After some very clever analysis he found the charge densities σ+andσ−to be related by eq. (3), with σ−given by σ−(θ)≈V0 4π2a/parenleftBigg√ 2θ0/2 √ 1−cosθ−tan−1√ 2θ0/2 √ 1−cosθ/parenrightBigg . (30) The total charge Q−on the inner surface is therefore, Q−=/integraldisplaycosθ0 −12πa2σ−dcosθ =aV0 2π/integraldisplaycosθ0 −1/parenleftBigg√ 2θ0/2 √ 1−cosθ−tan−1√ 2θ0/2 √ 1−cosθ/parenrightBigg dcosθ =aV0 2π/integraldisplay1 θ0/2/parenleftBig x−tan−1x/parenrightBigθ2 0 x3dx =aV0θ2 0 2π/bracketleftbigg −1 x+1 2/parenleftbigg 1+1 x2/parenrightbigg tan−1x+1 2x/bracketrightbigg1 θ0/2 ≈aV0θ2 0 2π/parenleftbiggπ 4−1 2/parenrightbigg =aV0θ2 0 8/parenleftbigg 1−2 π/parenrightbigg ≈aV0θ2 0 22. (31) As we will see in the following section, the “exact” form for σ−, which in effect includes terms to all orders in θ0, changes the factor√ 1−cosθto√ cosθ0−cosθ. This has the effect of shifting the upper limit of the xintegration from 1 to ∞, which in turn changes the factor π/4−1/2t oπ/4. It is surprising that higher-order terms could change the result by a factor of 3, which shows the delicacy of Green’s approximations. 6 For completeness, we review Green’s derivation of eq. (30). We write the total charge density σat radius r=aas σ(θ)=σ++σ−=σ0+Δσ, (32) where σ0=V0/4πarelates the uniform charge distribution σ0on a complete shell that is at potential V0, which is also the potential of the conducting shell with orifice. In that orifice, the total charge distribution vanishes, so Δ σ(θ<θ 0)=−σ0. The potential V(r, θ, ϕ) in the interior of the shell due to charge distribution (32) can be written V(r, θ, ϕ)=V0+ΔV, (33) Thus, the potential Δ Vis due to the charge distribution Δ σ. Since the potential of the shell with orifice is V0, we obtain the partial boundary condition that Δ V(r=a,θ > θ 0)=0 . To complete a solution, we need to determine the charge distribution Δ σthat is in- duced on a grounded, conducting shell with orifice of half angle θ0due to a uniform charge distribution −σ0on the spherical cap of the orifice. This solution could be readily implemented if we knew the charge distribution induced on a grounded, conducting shell with orifice by a unit charge at an arbitrary point on the spherical cap of the orifice. While this approach is now associated with the name of Green, he did not in fact use this approach in 1828, and it was W. Thomson who first used thismethod to provide a solution valid for any angle θ 0, as decribed in to following section. Rather, the approach of Green in his Essay [1] was to consider that part of the the potential Δ Vdue to the uniform charge distribution Δ σ=−σ0on the spherical cap ( r= a,θ,θ 0) to be equivalent to that of a uniform flat disk of radius aθ0/lessmuchawhich carries charge density −σ0. By the usual argument using a Gaussian pillbox, the charge density on that disk is related to the electric field and the potential by −∂ΔV(r=a−) ∂r=ΔEr(r=a−)=2πσ 0. (34) Thus we have a mixed boundary value problem, with knowledge of the potential over part of the boundary and knowledge of the normal derivative of the potential. We can apply Poisson’s integral (29) to the potential Δ V,i nw h i c hc a s ew el e a r nt h a t ΔV(r, θ, φ)=a2−b2 4πa/integraldisplay capΔV(r/prime=a,θ/prime,φ/prime) R3dS/prime, (35) since Δ V= 0 on the conducting shell. (More to come...) 2.4 Solution by Inversion for a Spherical Bowl Apparently, W. Thomson (Lord Kelvin) first deduced the surface charge distribution on aconducting spherical shell with a circular orifice of any size (henceforth called a spherical bowl) in 1847, using his method of inversion [5] starting from the charge density (38) onthe surface of a thin conducting disk, but he published the result only in 1872 [6]. In 7 his discussion of Thomson’s calculation [7], Maxwell seems unaware of Green’s prior work. Thomson’s solution by inversion for the spherical bowl is also discussed by Jeans [8]. In the notation of the present paper, Thomson found the charge density σ−on the inner surface of the spherical bowl to be. σ−(θ)=V0 4π2a⎛ ⎝/radicalBigg 1−cosθ0 cosθ0−cosθ−tan−1/radicalBigg 1−cosθ0 cosθ0−cosθ⎞ ⎠. (36) We see that Green’s result (30) is the small-angle limit of eq. (36). The total charge Q−on the inner surface follows from eq. (36) as Q−=/integraldisplaycosθ0 −12πa2σ−dcosθ =aV0 2π/integraldisplaycosθ0 −1⎛ ⎝/radicalBigg 1−cosθ0 cosθ0−cosθ−tan−1/radicalBigg 1−cosθ0 cosθ0−cosθ⎞ ⎠dcosθ =aV0 2π/integraldisplay∞ √ 1−cosθ0/√ 1+cos θ0/parenleftBig x−tan−1x/parenrightBig4sin2(θ0/2) x3dx =2aV0sin2(θ0/2) π/bracketleftbigg −1 x+1 2/parenleftbigg 1+1 x2/parenrightbigg tan−1x+1 2x/bracketrightbigg∞ √ 1−cosθ0/√ 1+cos θ0 =aV0 2π/bracketleftBig πsin2(θ0/2) + sin θ0−θ0/bracketrightBig ≈aV0θ2 0 8, (37) where the approximation holds for small θ0. This derivation is the firmest evidence we offer in support of the “elementary” result (2). As angle θ0approaches π, the spherical bowl approaches a thin conducting disk of radius b=a(π−θ0)/lessmucha. In this limit the charge density (36) becomes σ−(r)=V0 2π2√ b2−r2,and Q−=bV0 2π, (38) which is the well-known result for the charge density on one side of a conducting disk, r being the distance from the center of the disk (see the Appendix for a highly geometricderivation of eq. (38)). Also in this limit, the charge distribution σ 0=V0/4πais small compared to σ−. Hence, the charge distribution on the other side of the thin conducting disk is σ+=σ0+σ−=σ−. As expected, the charge distribution is the same on both sides of a conducting disk. We now present details of a derivation leading to eq. (36), following Thomson [6]. The starting point is based on the discussion of eqs. (32)-(33), that the difference between thecharge distribution on a conducting sphercial bowl and the uniform charge distribution σ 0on a complete sphere at the same potential is the same as that induced on a grounded sphericalbowl by charge distribution −σ 0on the spherical cap that completes the spherical bowl. Thomson solved this problem by first finding the charge distribution induced on a grounded spherical bowl by a unit charge at an arbitrary point on the spherical cap, using his methodof inversion. 8 We begin with a conducting disk of radius bat potential V0. Then, from eq. (38) and the geometry shown in Fig. 1, the charge density σdiskon each side of the disk can be written σdisk=V0 2π2/radicalBig (b+r)(b−r)=V0 2π2√ EP·PD=V0 2π2√ AP·PB, (39) where APB is any chord that passes through point PamdECD is the diameter that contains P. Figure 1: Point Pon diameter ECD is at distance rfrom the center of a circle of radius b.APB is any chord that contains P. Triangles ADP and BEP are similar since /negationslash DAP =/negationslash DAB =(/negationslash BCD)/2= /negationslash BED =/negationslash BEP. Hence EP/PB =AP/PD ,a n dAP·PB=EP·PD=(b+r)(b−r)=b2−r2. Figure 2: The inverse of a disk with respect to a sphere of radius scentered at point Ois a spherical bowl that lies on a sphere that contains point O.T h e plane OA/primeB/primeis not necessarily perpendicular to the plane of the disk, and in general A/primeB/primeis not a diameter of the disk, but only a chord. 9 Next, we invert the disk with respect to a sphere of radius swhose center Ois not in the plane of the disk, as shown in Fig. 2. The plane that contains the disk inverts into a spherethat passes through the center of inversion O, and the disk inverts into a spherical bowl the occupies part of that sphere. The distance OPfrom the center of inversion to a point Pon the bowl is related to the distance OP /primeof the inverse point on the disk by OP·OP/prime=s2. (40) The principle of the method of inversion is that if we relate the charge dq=σbowldS in area element dSabout point Pon the bowl to charge dq/prime=σdisk(V0)dS/primein element dS/prime about point P/primeon the conducting disk whose potential is V0according to dq=−dq/prime(OP/s), then the charge distribution on the spherical bowl is that for the case that the bowl is agrounded conductor in the presence of charge sV 0at point O. Since the conducting disk has the same charge distribution σdiskon both of its sides, the method of inversion tells us that both the inner and outer surfaces of a grounded, conductingspherical bowl have the same charge distributions induced by a charge placed anywhere onthe spherical cap that completes the bowl. As the case of a conducting spherical bowl atpotential V 0is the superposition of a complete shell of charge density σ0=V0/4πaand a grounded conducting bowl when charge density −σ0covers the spherical cap, we have another confirmation of relation (3) that σ+−σ−=σ0. Area element dSabout point Pon the bowl is the inverse of element dS/primeabout P/primeon the disk. Hence, dS dS/prime=/parenleftbiggOP OP/prime/parenrightbigg2 =(OP)4 s4, (41) using eq. (39). Following the spirit of Green, we desire the charge distribution at point Pon each side of a grounded conducting bowl induced by unit charge at point O, which we obtain from eqs. (40)-(41) as σbowl(P,V=0,qO=1 ) =dq dS=−1 sV0dq/primeOP ss4 dS/prime·(OP)4=−σdisk(V0) V0s2 (OP)3 =−s2 2π2(OP)3√ A/primeP/prime·P/primeB/prime. (42) Expression (42) will be more useful if we can replace distances A/primeP/primeandP/primeB/primemeasured on the disk by quantities related to the spherical bowl. Referring to Fig. 2, we see that triangles APO andA/primeP/primeOare similar, so that A/primeP/prime AP=OA/prime OP=s2 OA·OP, (43) sinceAandA/primeare inverse points with respect to the sphere of radius sabout O. Likewise, similar triangles BPO andB/primeP/primeOlead to P/primeB/prime PB=OB/prime OP=s2 OB·OP. (44) 10 Thus, A/primeP/prime·P/primeB/prime=OB/prime OP=s4 (OP)2AP·PB OA·OB. (45) If we keep points OandP/primefixed then point P/primeis fixed also, but we can vary the chord A/primeP/primeB/prime and consequently the location of points AandBas well. Under such variation the ratio s4/(OP)2remains constant, and the product A/primeP/prime·P/primeB/primealso remains constant according to the logic of eq. (39) and Fig. 1. Hence, we obtain the peculiar theorem that the ratio (AP·PB)/(OA·OB) is also constant during such variation, which result Thomson attributes to Liouville. In any case, we see that eq. (42) for the charge density at point Pcan also be written σbowl(P,V=0,qO=1 )= −1 2π2(OP)2√ OA·OB √ AP·PB. (46) Figure 3: When the center of inversion, O, lies in the plane of the original disk whose center is C/prime, then the inverse of that disk is another disk with center atCin the same plane. While point C/primeis not the inverse of point C[9], all other primed points are the inverses of their unprimed partner. When thechord A /primeP/primeB/primelies along the line OP/prime, triangles OBD andOAE are similar, and hence OA·OB=OD·OE=(a−b)(a+b)=a2−b2. This result is remarkable in that it does not depend on the radius sof the sphere of inversion nor (directly) on the radius of the spherical bowl. In particular, we can take point Oto lie in the plane of the original disk and outside its bounding circle, in which case the spherical bowl degenerates into another disk (which lies between point Oand the original disk). Hence, charge distribution (46) also holds for the case of a grounded, conducting diskin the presence of unit charge at point O. We write the distance from point Oto the center of the grounded disk as a, the radius of the disk as b, the distance from the center of the disk to pointPasr, and the distance OPasR. We have seen that the ratio ( OA·OB)/(AP·PB)i s independent of the choice of chord A /primeP/primeB/primefor fixed points O,PandP/prime. We can conveniently evaluate this ratio of the case that the chord A/primeP/primeBlies along the line OP/prime,a ss h o w ni n 11 Fig 3. By the argument in the caption of Fig. 1, AP·PB=b2−r2, and by a very similar argument OA·OB=a2−b2. Thus, eq. (46) tells us that the charge distribution induced on each side of a grounded, conducting circular disk of radius bby unit charge in the plane of the disk at distance a>b is σdisk(r)=−1 2π2R2√ a2−b2 √ b2−r2, (47) where Ris the distance between the exterior unit charge and the point of interest on the disk. This result was first given by Green, p. 181 of [1]. We can now complete the calculation of the charge distribution induced on a grounded, conducting spherical bowl of radius aby uniform charge distribution −σ0on the spherical cap that completes the bowl. We will calculate the distribution σ−(θ) on the inner surface of the bowl, and obtain the distribution σ+on the outer surface via relation (3). Integrating eq.(46) over points O=(a,θ/prime<θ 0,ϕ/prime) on the spherical cap, for point P=(a,θ > θ 0,0) we have σ−(θ>θ 0,V=0,σcap=−σ0)=/integraldisplay1 cosθ0a2dcosθ/prime/integraldisplay2π 0dϕ/primeσ0 2π2(OP)2√ OA·OB √ AP·PB.(48) Points A,B,OandPin the integrand of eq. (48) all lie in the same plane, but according to the theorem of Liouville proved above, we are free to chose for this any plane that containspoints OandP.P o i n t s AandBare then the intersection of this plane with the rim of the spherical bowl. Thomson suggests that we always choose the plane ABOP to contain the “south pole” Sof the bowl, i.e., the intersection of the axis of the bowl with its surface, as shown in Fig. 4. We introduce angles αandβas shown in Fig. 4 so the the lengths of lines AS,PS,AP andPBare related by AS=2dsinα, (49) PS=2dsinβ, (50) AP=2dsin(α−β), (51) PB=2dsin(α+β). (52) Using the identity sin( α−β)s in )α+β)=s i n 2α−sin2β, we find that AP·PB=(AS)2−(PS)2=2a2(cosθ0−cosθ), (53) noting also that ( AS)2=2a2(1 + cos θ0),etc.By a similar construction that emphasizes point Orather than point P,w ea l s oh a v et h a t OA·OB=(OS)2−(AS)2=2a2(cosθ/prime−cosθ0). (54) Forms (53) and (54) are convenient in that their lefthand sides appear to depend on the azimuthal coordinates of points A,B,OandP, while the righthand sides depend only on the polar coordinates. 12 Figure 4: The plane ABOP may be chosen to contain the “south pole” Sof the spherical bowl without changing the result of eq. (46). This plane does not, in general, contain the center of the spherical bowl, but its intersectionwith the bowl is an arc APSB of a circle with radius d≤a.T h e a r c AS subtends angles 2 αwith respect to the center of arc APSB ,a r cPSsubtends angle 2 β.T h e na r c APsubtends angle 2( α−β), and arc PSB subtends angle 2(α+β). Thus, the only remaining azimuthal dependence of the intergrand of eq. (48) is that due to length OP, which can be expressed as (OP) 2=2a2(1−cosγ)=2a2(1−cosθcosθ/prime−sinθsinθ/primecosϕ/prime), (55) where γis the angle subtended by arc OPwith respect to the center of the spherical bowl. Combining eqs. (48) and (53)-(55), we have σ−(θ)=V0 16π3a/integraldisplay1 cosθ0dcosθ/prime/radicalBigg cosθ/prime−cosθ0 cosθ0−cosθ/integraldisplay2π 0dϕ/prime 1 1−cosθcosθ/prime−sinθsinθ/primecosϕ/prime =V0 16π3a/integraldisplay1 cosθ0dcosθ/prime/radicalBigg cosθ/prime−cosθ0 cosθ0−cosθ2π /radicalBig (1−cosθcosθ/prime)2−sin2θsin2θ/prime =V0 8π2a√ cosθ0−cosθ/integraldisplay1 cosθ0dcosθ/prime/radicalBig cosθ/prime−cosθ0 cosθ/prime−cosθ =V0 8π2a√ cosθ0−cosθ/integraldisplay√ 1−cosθ0 02x2dx x2+c o sθ0−cosθ =V0 4π2a√ cosθ0−cosθ/bracketleftBigg x−/radicalBig cosθ0−cosθtan−1 x √ cosθ0−cosθ/bracketrightBigg√ 1−cosθ0 0 =V0 4π2a⎛ ⎝√ 1−cosθ0 √ cosθ0−cosθ−tan−1/radicalBigg 1−cosθ0 cosθ0−cosθ⎞ ⎠, (56) using Dwight 858.536 to go from the first line to the second, and Dwight 122.1 to go from the 4rth line to the 5th. 13 Thomson [6] also gave an extension of eq. (46) in which the unit charge is not necessarily on the cap of the spherical bowl. 2.5 Solution in Toroidal Coordinates The problem of a charged, conducting spherical bowl can also be solved in toroidal coordi- nates [10]. 3 Appendix: Charge Distribution on a Conducting El- lipsoid and on a Conducting Circular Disk The charge distribution (38) on a thin, conducting disk can be deduced in a variety of ways. We record here a highly geometric derivation following Thomson (pp. 7 and 178-179 of [6]). The starting point is the “elementary” result that the electric field is zero in the interior of a spherical shell of any thickness that has a uniform volume charge density between theinner and outer surfaces of the shell. A well-known geometric argument (due to Newton) forthis is illustrated in Fig. 5. Figure 5: For any point r0in the interior of a uniformly charged shell of charge, the axis of a narrow bicone intercepts the inner surface of the shell at points r1 andr2. The corresponding areas on the inner surface of the shell intercepted by the bicone are A1andA2. In the limit of small areas, A1/R2 01=A2/R2 02. The electric field at point r0in the interior of the shell due to a lamina of thickness δ and area A1centered on point r1that lies within a narrow cone whose vertex is point 0 is given by E1=ρdVol 1 R2 01ˆR01, (57) where ρis the volume charge density, dVol 1=A1δ,R01=r0−r1, and the center of the sphere is taken to be at the origin. Likewise, the electric field from a lamina of area A2centered 14 on point r2defined by the intercept with the shell of the same narrow cone extended in the opposite direction (forming a bicone) is given by E2=ρdVol 2 R2 02ˆR02, (58) In the limit of bicones with small half angle, the two parts of the bicone as truncated by the shell are similar, so that A1 R2 01=A2 R2 02,dVol 1 R2 01=dVol 2 R2 02, (59) and, of course, ˆR02=−ˆR01. Hence E1+E2= 0. Since this construction can be applied to all points in the material of the spherical shell, and for all pairs of surface elements subtended by (narrow) bicones,the total electric field in the interior of the shell is zero. We now reconsider the above argument after arbitrary scale transformations have been applied to the rectangular coordinate axes, x→k1x, y →k2y, z →k3z. (60) A spherical shell of radius sis thereby transformed into an ellipsoid, x2 s2+y2 s2+z2 s2=1 →x2 s2/k2 1+y2 s2/k2 2+z2 s2/k2 3=1. (61) As parameter sis varied, one obtains a set of similar ellipsoids, centered on the origin. A small volume element obeys the transformation dVol = dxdydz →k1k2k3dxdydz =k1k2k3dVol. (62) The three points 0, 1, and 2 in Fig. 5 lie along a line, so that R01=r0−r1=CR02=C(r0−r2), (63) where Cis a (negative) constant. This relation is invariant under the scale transformation (60), so that together with eq. (62) the relation dVol 1 R2 01=dVol 2 R2 02, (64) is also invariant. Hence, if the ellipsoidal shell, which is the transform of the spherical shell of Fig. 5, contains a uniform volume charge density, the relation E1+E2= 0 remains true at the vertex of any bicone in the interior of the shell, which implies that the total electricfield is zero there. This proof is based on the premise that the ellipsoidal shell is bo unded by two similar ellipsoids, and that the volume charge density in the shell is uniform. If we let the outer ellipsoid of the shell approach the inner one, always remaining similar to the latter, we reach a configuration that is equivalent to a thin, conducting e llipsoid, since in both cases the electric field is zero in the interior. Hence, the surface charge distribution 15 on a thin, conducting e llipsoid must the same as the projection onto its surface of a uniform charge distribution between that surface and a similar, but slightly larger ellipsoidal surface. The charge σper unit area on the surface of a thin, conducting e llipsoid is therefore proportional to the thickness, which we write as δd, of the ellispodal shell formed by that surface and a similar, but slightly larger ellipsoid: σ=ρδd, (65) where constant ρis to be determined from a knowledge of the total charge Qon the con- ducting ellipsoid. The thickness δdof a thin ellipsoidal shell at some point on its inner surface is the distance between the plane that is tangent to the inner surface at the specified point, and the planethat is tangent to the outer surface at the point similar to the specifiedl point. These planes are parallel since the ellipsoids are parallel. In particular if the semimajor axes of the inner ellipsoid are called a,b,a n dc, then those of the outer ellipsoid can be written a+δa,b+δb andc+δc. Let the (perpendicular) distance from the plane tangent to the specified point on the inner ellipsoid to its center be called d, and the corresponding distance from the outer tangent plane be d+δd,s ot h a t δdis the desired thickness of the shell at the specified point. Then, the condition of similarity is that δa a=δb b=δc c=δd d. (66) Since the volume of an ellipsoid with semimajor axes a,b,a n d cis 4πabc/3, the volume of the ellispoidal shell is 4 π(a+δa)(b+δb)(c+δc)/3−4πabc/ 3=4πabc(δd/d), using eq. (66). Since the constant ρhas an interpretation as the uniform charge density within the material of the ellipsoidal shell, we find that the total charge Qon the conducting e llipsoid is related by Q=ρVol shell=4πabc dρδd, (67) and hence, σ=ρδd=Qd 4πabc. (68) It remains to find an expression for the distance dto the tangent plane. If we write the equation for the ellipsoid in the form f(x,y,z )=x2 a2+y2 b2+z2 c2−1=0, (69) then the gradient of fis perpendicular to the tangent plane. Thus, the vector dfrom the center of the ellipsoid to the tangent plane is proportional to ∇f.T h a ti s , d∝∇f=2/parenleftbiggx a2,y b2,z c2/parenrightbigg . (70) The unit vector ˆdis therefore ˆd=/parenleftBig x a2,y b2,z c2/parenrightBig /radicalBig x2 a4+y2 b4+z2 c4. (71) 16 The magnitude dof the vector dis related to the vector r=(x,y,z ) of the specified point on the ellipse by d=r·ˆd=x2 a2+y2 b2+z2 c2 /radicalBig x2 a4+y2 b4+z2 c4=1 /radicalBig x2 a4+y2 b4+z2 c4. (72) At length, we have found the charge density on the surface of a conducting e llipsoid to be σellipsoid =Q 4πabc/radicalBig x2 a4+y2 b4+z2 c4, (73) where Qis the total charge. The case of a thin, conducting e lliptical disk in the x-yplane can be obtained from eq. (73) by letting cgo to zero. For this we note that eq. (69) for a general ellipsoid permits us to write c/radicalBigg x2 a4+y2 b4+z2 c4=/radicaltp/radicalvertex/radicalvertex/radicalbt c2/parenleftBiggx2 a4+y2 b4/parenrightBigg +1−x2 a2−y2 b2→/radicalBigg 1−x2 a2−y2 b2. (74) The charge density on each side of a conducting e lliptical disk is therefore σelliptical disk =Q 4πab/radicalBig 1−x2 a2+y2 b2. (75) The charge density on each side of a conducting circular disk of radius bfollows immedi- ately as σcircular disk =Q 4πb√ b2−r2, (76) where r2=x2+y2. Such a disk has potential V0, which can be found by calculating the potential at the center of the disk according to V0=V(r=0,z=0 )=/integraldisplayb 02σ(r) r2πr dr =Q b/integraldisplayb 0dr √ b2−r2=πQ 2b. (77) Hence, a conducting disk of radius bat potential V0has charge density σcircular disk =V0 2π2√ b2−r2(78) on each side, which is the result quoted in eq. (38). References [1] G. Green, Mathematical Papers , (Chelsea Publishing Co., Bronx, NY, 1970), pp. 50-55, 57-61. [2] W.R. Smythe, Static and Dynamic Electricity , 3rd ed. (McGraw-Hill, New York, 1968), problems 46 and 47, p. 228. 17 [3] J.D. Jackson, Classical Electrodynamics , 3rd ed. (Wiley, New York, 1999). [4] Green’s derivation of eq. (29) is considered in prob. 5 of Ph501 Set 2, http://puhep1.princeton.edu/~mcdonald/examples/ph501set2.pdf [5] See, for example, secs. 5.09-5.102 of [2].[6] W. Thomson (Lord Kelvin), Determination of the Distribution of Electricity on a Cir- cular Segment of Plane or Spherical Conducting Surface, Under Any Given Influence , inPapers on Electricity and Magnetism , 2nd ed. (Macmillan, London, 1884), pp. 178- 191, dated 1869, published in the 1st ed. of 1872. See also pp. 153-154 for the originalstatement of the result in a letter (in French) to Liouville. [7] J.C. Maxwell, A Treatise on Electricity and Magnetism , 3rd ed. (Dover Publications, New York, 1954), Vol. 1, pp. 276-280. [8] J. Jeans, The Mathematical Theory of Electricity and Magnetism , 5th ed. (Cambridge U. Press, 1951), pp. 250-251. [9] See, for example, sec. 4.21 of [2]. [10] See problem. 501, p. 239 of N.M. Lebedev, I.P. Skalskaya and Y.S. Ufyland, Worked Problems in Applied Mathematics (Dover Publications, New York, 1979). 18