hernandez electrostatics of infinite cylinder

Journal of Electrostatics 63 (2005) 1115–1131 Electric potential due to an infinite conducting cylinder with internal or external point charge J.A. Hernandes/C3, A.K.T. Assis Instituto de Fı ´sica ‘Gleb Wataghin’, Universidade Estadual de Campinas—Unicamp, 13083-970 Campinas, Sa˜o Paulo, Brasil Received 16 September 2004; accepted 19 February 2005 Available online 6 April 2005 Abstract We utilize the Green’s function method in order to calculate the electric potential due to an infinite conducting cylinder held at zero potential and a point charge inside and outside it.We calculate and plot the net force upon the point charge as a function of its distance to the axis of the cylinder.We show that this force goes to zero when the radius of the cylinder goes to zero,no matter the distance of the external point charge to the conducting line.r2005 Published by Elsevier B.V. Keywords: Electric potential; Electric induction; Surface charges; Green’s function method 1. Introduction The goal of this work is to calculate the electrostatic force between an infinite conducting cylinder of radius aheld at zero potential and an external point charge q. To our knowledge this has never been done before.To this end we consider theGreen’s function method, [1, Chapters 1–3] .We begin reviewing a known solution of the potential inside a grounded, closed, hollow and finite cylindrical box with a point charge inside it [1, p.143] .We analyse the limit of an infinite cylinder and explore theARTICLE IN PRESS www.elsevier.com/locate/elstat 0304-3886/$ - see front matter r2005 Published by Elsevier B.V. doi:10.1016/j.elstat.2005.02.005/C3Corresponding author. E-mail addresses: julioher@ifi.unicamp.br (J.A. Hernandes), assis@ifi.unicamp.br (A.K.T. Assis). URL: http://www.ifi.unicamp.br/ /C24assis. force exerted upon the point charge.We then perform a similar analysis for the case of an external point charge.We consider in detail the particular situation of a thinwire, that is, with the point charge many radii away from the axis of the cylinder. 2. Finite conducting cylinder with internal point charge: solution to Poisson’s equation Consider a finite conducting cylindrical box of radius aand length Lba,w i t h z being its axis of symmetry, see Fig.1 .With cylindrical coordinates ðr;f;zÞthe center of the box is supposed to be at ðr;zÞ¼ð0;L=2Þ.We consider also a point charge q located at ~x 0¼ðr0oa;f0;z0Þinside the box.We wish to calculate the electric potential of the system, the electric field, the surface charge distribution induced by q and the net force between the cylinder and q. The electrostatic potential Fobeys Poisson’s equation: r2 xF¼/C04pxr, where x¼1=4pe0in SI units ( e0is the electric permittivity of vacuum) or x¼1i n Gaussian units.In this work, we will suppose a vacuum inside and outside thecylinder.The derivation and results might also be useful with a homogeneousdielectric or insulating liquid inside and outside the cylinder, by utilizing thestandard approach described in most textbooks dealing with electromagnetism. By the standard Green’s function method, the solution of Poisson’s equation for this case with Dirichlet boundary conditions (potential specified on a closed surface)is given by Fð~xÞ¼xZZZ Vrð~x00ÞGð~x;~x00Þd~x00/C01 4paSFð~x00ÞqG qn00da00, (1) where Vis the volume of the cylindrical box, Sits closed surface and q=qn00is the normal derivative at the surface Sof the box directed outwards.Here, Gð~x;~x00Þis Green’s function satisfying the equation r2 x00Gð~x;~x00Þ¼/C0 4pdð~x/C0~x00Þ.(2) As the surface of the cylinder in electrostatic equilibrium is at a constant potential F0 we imposed that Gð~x;~x00Þ¼0 at this surface.ARTICLE IN PRESS Fig.1. Finite conducting cylinder of length Land radius acentered at ðr;zÞ¼ð0;L=2Þ, with zbeing its axis of symmetry.There is a point charge qlocated at ðr;f;zÞ¼ðr0oa;f0;0oz0oLÞ.J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1116 We can expand the Dirac delta function in cylindrical coordinates as given by dð~x/C0~x00Þ¼dðr/C0r00Þdðf/C0f00Þ rdðz/C0z00Þ.(3) The delta functions for fand zcan be written in terms of orthonormal functions: dðz/C0z00Þ¼2 LX1 n¼1sinnpz Lsinnpz00 L"# , (4) dðf/C0f00Þ¼1 2pX1 m¼/C01eimðf/C0f00Þ"# .(5) Note our particular choice of expansion for z, Eq.(4).This choice satisfies the condition Gð~x;~x00Þ¼0 in the upper and lower covers of the cylindrical box.The Green function can be expanded in a similar fashion: Gð~x;~x00Þ¼1 pLX1 m¼/C01eimðf/C0f00ÞX1 n¼1sinnpz Lsinnpz00 Lgmðk;r;r00Þ"# () , (6) where k¼np=Land gmðk;r;r00Þis the radial Green function to be determined. Substituting this expression into Eq.(2) and using (3)–(5) we obtain 1 rd drrdgm dr/C18/C19 /C0k2þm2 r2/C18/C19 gm¼/C04p rdðr/C0r00Þ.(7) Forrar00the right-hand side of Eq.(7) is equal to zero.This means that gmis a linear combination of modified Bessel functions, ImðkrÞand KmðkrÞ.Suppose that c1ðkrÞsatisfies the boundary conditions for ror00and that c2ðkrÞsatisfies the boundary conditions for r4r00: c1ðkroÞ¼AImðkroÞþBK mðkroÞ, (8) c2ðkr4Þ¼CImðkr4ÞþDK mðkr4Þ.(9) Here, A,B,Cand Dare coefficients to be determined.The symmetry of the Green function in randr00requires that gmðk;r;r00Þ¼c1ðkroÞc2ðkr4Þ, (10) where r4androare, respectively, the larger and the smaller of randr00.The potential must not diverge for r!0, so we must have B¼0.The Green function must vanish at r¼a, that is, c2ðkaÞ¼0.This yields C¼/C0DK mðkaÞ=ImðkaÞ.TheARTICLE IN PRESS J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1117 function gmcan then be written as gmðk;r;r00Þ¼HImðkroÞKmðkr4Þ/C0Imðkr4ÞKmðkaÞ ImðkaÞ/C20/C21 .(11) The normalization coefficient H¼ACis determined by the discontinuity implied by the delta function in Eq.(7): dgm dr/C12/C12/C12/C12 þ/C0dgm dr/C12/C12/C12/C12 /C0¼/C04p r00¼kW½c1;c2/C138, (12) where the /C6signs mean evaluation at r¼r00/C6/C15and taking the limit /C15!0.In the last equality, W½c1;c2/C138is the Wronskian of c1andc2.Substituting gminto Eq.(12), and using that W½Imðkr00Þ;Kmðkr00Þ/C138 ¼ /C0 1=ðkr00Þ, we find H¼4p.The Green function for the problem of a finite conducting cylinder with a charge inside it can befinally written as Gð~x;~x 00Þ¼4 LX1 m¼/C01eimðf/C0f00ÞX1 n¼1sinðkzÞsinðkz00ÞImðkroÞKmðkr4Þ/C2( ( /C0Imðkr4ÞKmðkaÞ ImðkaÞ/C21/C27/C27 . ð13Þ 2.1. Cylinder held at zero potential Consider the cylinder to be held at zero potential, namely, Fð~x00Þ¼0: Fða;f;0pzpLÞ¼Fðrpa;f;LÞ¼Fðrpa;f;0Þ¼0.(14) Substituting Eqs.(13) and (14) into Eq.(1) yields the potential inside the cylinder as (withrð~x00Þ¼qdð~x0/C0~x00Þ) Fð~x;~x0Þ¼4xq LX1 m¼/C01X1 n¼1eimðf/C0f0Þsinnpz L/C16/C17 sinnpz0 L/C18/C19 Imnpro L/C16/C17( ( /C2Kmnpr4 L/C16/C17 /C0Imnpr4 L/C16/C17Kmnpa L/C0/C1 Imnpa L/C0/C1" #)) . ð15Þ Here, r4(ro) is the larger (smaller) of randr0. 3. Infinite conducting cylinder with internal point charge The solution for an infinite cylinder differs from the solution of the finite cylinder by changing essentially the expansion of the delta function in Eq.(4).In the infiniteARTICLE IN PRESS J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1118 cylinder, there is no restriction on the choice of n(ork): dðz/C0z00Þ¼1 2pZ1 /C01eikðz/C0z00Þdk¼1 pZ1 0cos½kðz/C0z00Þ/C138dk.(16) The Green function can be written as Gð~x;~x00Þ¼2 pX1 m¼/C01eimðf/C0f00ÞZ1 0cos½kðz/C0z00Þ/C138ImðkroÞ( /C2Kmðkr4Þ/C0Imðkr4ÞKmðkaÞ ImðkaÞ/C20/C21 dk) . ð17Þ Note that we can pass from Eq.(4) to Eq.(16) by transforming the Fourier series into the Fourier transform.That is, by letting L!1 , setting np=L¼k,dk¼p=L, z!zþL=2,z00!z00þL=2 and by replacing the infinite sum by the integral over k. 3.1. Cylinder held at zero potential Consider the cylinder to be held at zero potential, namely, Fða;f;zÞ¼0. Substituting Eq.(17) into Eq.(1), the potential inside the cylinder can be writtenas (with rð~x 00Þ¼qdð~x0/C0~x00Þ) Fð~x;~x0Þ¼2xq pX1 m¼/C01eimðf/C0f0ÞZ1 0cos½kðz/C0z0Þ/C138ImðkroÞ( /C2Kmðkr4Þ/C0Imðkr4ÞKmðkaÞ ImðkaÞ/C20/C21 dk) . ð18Þ Once more r4(ro) is the larger (smaller) of randr0. The electric field is given by ~E¼/C0 rF, with components Erðror0Þ¼ /C0qF qr¼/C02xq pX1 m¼/C01eimðf/C0f0ÞZ1 0kcos½kðz/C0z0Þ/C138I0 mðkrÞ( /C2Kmðkr0Þ/C0Imðkr0ÞKmðkaÞ ImðkaÞ/C20/C21 dk) , ð19Þ Erðr4r0Þ¼ /C0qF qr¼/C02xq pX1 m¼/C01eimðf/C0f0ÞZ1 0kcos½kðz/C0z0Þ/C138Imðkr0Þ( /C2K0 mðkrÞ/C0I0 mðkrÞKmðkaÞ ImðkaÞ/C20/C21 dk) , ð20ÞARTICLE IN PRESS J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1119 Ef¼/C01 rqF qf¼4xq prX1 m¼1msin½mðf/C0f0Þ/C138Z1 0cos½kðz/C0z0Þ/C138ImðkroÞ( /C2Kmðkr4Þ/C0Imðkr4ÞKmðkaÞ ImðkaÞ/C20/C21 dk) , ð21Þ Ez¼/C0qF qz¼2xq pX1 m¼/C01eimðf/C0f0ÞZ1 0ksin½kðz/C0z0Þ/C138ImðkroÞ( /C2Kmðkr4Þ/C0Imðkr4ÞKmðkaÞ ImðkaÞ/C20/C21 dk) . ð22Þ The force ~F¼q~Eð~x0Þacting upon the charge qis given by Eq.(19) at ~x¼~x0 without the first term between brackets (it is the field generated by the charge q itself).There is only a radial component when the cylinder has an infinite length ~Fð~x0Þ¼2xq2 pX1 m¼/C01Z1 0kImðkr0ÞI0 mðkr0ÞKmðkaÞ ImðkaÞdk() ^r ¼/C0xq2 pr02X1 m¼/C01Z1 0I2 mðxÞd dxxKmðxa=r0Þ Imðxa=r0Þ/C20/C21 dx() ^r. ð23Þ In the last equation we integrated by parts.We plotted in Fig.2 the force of Eq.(23), normalized by F0/C17xq2=a2, as a function of r0=a.This force goes to zero when r0=a¼0 and diverges when r0!a, as expected.ARTICLE IN PRESS Fig.2. Force Fbetween an infinite grounded conducting cylinder of radius acentered on the zaxis and a point charge qat a distance r0from the zaxis, normalized by F0¼xq2=a2.J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1120 The surface charges can be calculated using Gauss’ law, yielding sða;f;zÞ¼Erða;f;zÞ 4px ¼/C0q 2p2aX1 m¼/C01eimðf/C0f0ÞZ1 0cos½kðz/C0z0Þ/C138Imðkr0Þ ImðkaÞdk"# . ð24Þ The surface charges per unit of length lðzÞis given by lða;zÞ¼Z2p 0sða;f;zÞadf¼/C0q pZ1 0cos½kðz/C0z0Þ/C138I0ðkr0Þ I0ðkaÞdk.(25) The total charge induced in the cylinder supposing z0¼0 can be obtained integrating Eq.(25) from z¼/C0 1 to1.Utilizing dðkÞ¼1 2pZ1 /C01cosðkzÞdz¼1 pZ1 0cosðkzÞdz, (26) this yields Q¼Z1 /C01lða;zÞdz¼/C0q.(27) 4. Infinite conducting cylinder with external point charge Suppose that the point charge qis located at ~x0¼ðr0;f0;z0Þ,w i t h r04a.Green’s function can be written analogously in this case as Gð~x;~x00Þ¼1 2p2X1 m¼/C01eimðf/C0f00ÞZ1 0cos½kðz/C0z00Þ/C138gmðk;r;r00Þdk"# , (28) where gmcan be written as the product of c0 1ðror00Þc0 2ðr4r00Þ.The functions c0 1and c0 2satisfy the modified Bessel equation.They can be written as a linear combination of the possible solutions c0 1ðkroÞ¼A0ImðkroÞþB0KmðkroÞ, (29) c0 2ðkr4Þ¼C0Imðkr4ÞþD0Kmðkr4Þ.(30) Forr!1 Green’s function must remain finite.This means that C0¼0. Additionally, Green’s function must be zero at the boundary surface.That is, G¼0 at the surface of the cylinder r¼a.This yields c0 1ðaÞ¼A0ImðkaÞþB0KmðkaÞ¼0! B0¼/C0A0ImðkaÞ KmðkaÞ.(31)ARTICLE IN PRESS J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1121 In order to obtain the function gmwe still have to find the constant H0: gmðk;r;r00Þ¼H0ImðkroÞ/C0KmðkroÞImðkaÞ KmðkaÞ/C20/C21 Kmðkr4Þ, (32) where r4(ro) is the larger (smaller) of randr00. From Eq.(12) we have that H0¼4p: gmðk;r;r00Þ¼4pImðkroÞ/C0KmðkroÞImðkaÞ KmðkaÞ/C20/C21 Kmðkr4Þ.(33) The Green function is then given by Gð~x;~x00Þ¼2 pX1 m¼/C01eimðf/C0f00ÞZ1 0cos½kðz/C0z00Þ/C138( /C2ImðkroÞ/C0KmðkroÞImðkaÞ KmðkaÞ/C20/C21 Kmðkr4Þdk) . ð34Þ 4.1. Cylinder held at zero potential Suppose that the surface of the cylinder is held at zero potential, namely Fða;f;zÞ¼0.(35) Applying Eqs.(34) and (35) in Eq.(1) with rð~x00Þ¼qdð~x0/C0~x00Þyield Fð~x;~x0Þ¼2xq pX1 m¼/C01eimðf/C0f0ÞZ1 0cos½kðz/C0z0Þ/C138( /C2ImðkroÞ/C0KmðkroÞImðkaÞ KmðkaÞ/C20/C21 Kmðkr4Þdk) . ð36Þ Here, r4(ro) is the larger (smaller) of randr0. Far from the origin, ris much larger than r0, hence we can express Eq.(36) in approximate form.The first term that appears between brackets is given by ImðkroÞKmðkr4Þ, with ro¼r0andr4¼r.Note the presence of the term KmðkrÞ, withrbr0, which decays rapidly for increasing k.This implies that the main contribution of the integrand is in the region 0 oko1=r.Then we can approximate Imðkr0Þfor small arguments, that is, for kr051, yielding Imðkr0Þ/C25ð kr0=2Þm=m!. From this we can see that the most relevant term is the first one, m¼0.The integral of the first term between brackets in Eq.(36) is then given by F1ðrbr0Þ/C252xq pZ1 0cos½kðz/C0z0Þ/C138K0ðkrÞdk¼xq r, (37) where we have used in the last equation the identity: ð2=pÞR1 0cosðxtÞK0ðytÞdt¼1=ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2þy2p ,[2, Problem 11.5.11] .The second term that appears between brackets in Eq.(36) can be treated in a similar way.The main contribution of the integrand is in the region 0 oko1=r.Again, the most relevantARTICLE IN PRESS J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1122 term is the first one.Accordingly, we approximate the function K0ðkr0Þfor small arguments: K0ðkr0Þ/C25/C0 lnðkr0Þ.This yields F2ðrbr0Þ/C25/C02xq pZ1 0cos½kðz/C0z0Þ/C138lnðkr0Þ lnðkaÞK0ðkrÞdk.(38) From Eq.(36) the electric field is given by ~E¼/C0 rF, with components Erðror0Þ¼ /C02xq pX1 m¼/C01eimðf/C0f0ÞZ1 0kcos½kðz/C0z0Þ/C138( /C2I0 mðkrÞ/C0K0 mðkrÞImðkaÞ KmðkaÞ/C20/C21 Kmðkr0Þdk) , ð39Þ Erðr4r0Þ¼ /C02xq pX1 m¼/C01eimðf/C0f0ÞZ1 0kcos½kðz/C0z0Þ/C138( /C2Imðkr0Þ/C0Kmðkr0ÞImðkaÞ KmðkaÞ/C20/C21 K0 mðkrÞdk) , ð40Þ Ef¼4xq prX1 m¼1msin½mðf/C0f0Þ/C138Z1 0cos½kðz/C0z0Þ/C138( /C2ImðkroÞ/C0KmðkroÞImðkaÞ KmðkaÞ/C20/C21 Kmðkr4Þdk) , ð41Þ Ez¼2xq pX1 m¼/C01eimðf/C0f0ÞZ1 0ksin½kðz/C0z0Þ/C138( /C2ImðkroÞ/C0KmðkroÞImðkaÞ KmðkaÞ/C20/C21 Kmðkr4Þdk) . ð42Þ The force ~F¼q~Eð~x0Þacting upon the charge qis given by Eq.(39) at ~x¼~x0 without the first term between brackets (it is the field generated by the charge q itself).There is only a radial component ~Fð~x0Þ¼2xq2 pX1 m¼/C01Z1 0kK mðkr0ÞK0 mðkr0ÞImðkaÞ KmðkaÞdk"# ^r ¼/C0xq2 pr02X1 m¼/C01Z1 0K2 mðxÞd dxxImðax=r0Þ Kmðax=r0Þ/C20/C21 dx() ^r, ð43Þ In the last equation we integrated by parts.We plot the force of Eq.(43) in Fig.2 , normalized by F0/C17xq2=a2, as a function of r0=a.This force goes to zero when r0=a!1 and diverges when r0!a, as expected.ARTICLE IN PRESS J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1123 The surface charges can be calculated using Gauss’ law, yielding sða;f;zÞ¼Erða;f;zÞ 4px ¼/C0q 2p2aX1 m¼/C01eimðf/C0f0ÞZ1 0cos½kðz/C0z0Þ/C138Kmðkr0Þ KmðkaÞdk"# . ð44Þ The surface charge per unit of length lðzÞis given by lða;zÞ¼Z2p 0sða;f;zÞadf¼/C0q pZ1 0cos½kðz/C0z0Þ/C138K0ðkr0Þ K0ðkaÞdk.(45) It is interesting to obtain the behaviour of lfor a thin wire, far from z0 (jz/C0z0jbr0ba).Utilizing Eq.(3. 150) of [1]we obtain l/C25/C0q 2l nðjzj=aÞ1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r02þz2p .(46) The total charge induced in the cylinder can be obtained integrating Eq.(45) from z¼/C0 1 to1.Utilizing Eq.(26) this yields Q¼Z1 /C01lða;zÞdz¼/C0q.(47) A plot of lða;zÞas a function of z,w i t h z0¼0 and normalized by q=r0, is given in Fig.3 .The maximum value of lða;zÞis given at z¼z0, as expected.In Fig.4 we plot lmax as a function of r0=a, normalized by q=r0.From this figure we can see thatARTICLE IN PRESS Fig.3. Induced linear charge density lon the conducting cylinder with a point charge outside it, Eq.(45), as a function of z=a.We utilized z0¼0,r0=a¼2 and normalized by q=r0.J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1124 lmax!0 when r0=a!1 , that is, for a conducting cylinder of zero thickness, a simple conducting straight line. 4.2. Thin wire held at zero potential Consider that the grounded conducting cylinder is very thin, i.e., a5r0.The modified Bessel functions can be approximated for small argument by [3, Section 8.44] Imðy51Þ/C251 m!ym 2m, (48) Kmðy51Þ/C25ðm/C01Þ!2m/C01 ym;m40, (49) K0ðy51Þ/C25/C0 lny 2/C0g.(50) Hereg¼0:577 is the Euler–Mascheroni constant. The term between brackets in Eq.(43) for m¼0 and for m40 can be approximated by, respectively, d dxx1 /C0lnðax=2r0Þ/C0g/C20/C21 /C25/C01 lnða=r0Þ, d dxx1 m!ðax=r0Þm 2mðax=r0Þm ðm/C01Þ!2m/C01/C20/C21 /C25ð2mþ1Þx2mða=r0Þ2m m!ðm/C01Þ!22m/C01.ARTICLE IN PRESS Fig.4.Maximum induced linear charge density lmaxðz¼z0Þon the conducting cylinder with a point charge outside it, Eq.(45), as a function of r0=a.We normalized the plot by q=r0.J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1125 The most relevant term for r0bais therefore m¼0.Using the identityR1 0K2 0ðxÞdx¼p2=4 we have the force acting upon the charge qas given by ~Fðr0baÞ/C25 /C0xq2 pr0Z1 0K2 0ðxÞlnð2r0=xaÞ/C0gþ1 ½g/C0lnð2r0=xaÞ/C1382dx^r /C25/C0xq2 pr02lnðr0=aÞZ1 0K2 0ðxÞdx^r¼/C0xq2p 4r02lnðr0=aÞ^r. ð51Þ Alternatively, another expression for the force can be found by integrating the force exerted by the linear charge density of a thin cylinder, lða;zÞof Eq.(45), acting upon the point charge q.Utilizing that Z1 /C01r0cos½kðz/C0z0Þ/C138 ½r02þðz/C0z0Þ2/C1383=2dz¼2kK1ðkr0Þ we obtain ~Fð~x0Þ¼/C0xqZ1 /C01r0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r02þz2plða;zÞ r02þz2dz^r¼/C02xq2 pr02Z1 0xK0ðxÞK1ðxÞ K0ðxa=r0Þdx^r. (52) To compare Eqs.(51) and (52) we can expand the latter using the approximation r0ba.Using that K1ðxÞ¼/C0 dK0=dx, integrating by parts, and K0ðxa=r0Þ/C25 /C0lnðxa=2r0Þ/C0g/C25lnðr0=aÞwe obtain ~F¼2xq2 pr02Z1 0xK0ðxÞðdK0=dxÞ K0ðxa=r0Þdx^r/C25/C0xq2 pr0Z1 0K2 0ðxÞlnð2r0=xaÞ/C0gþ1 ½g/C0lnð2r0=xaÞ/C1382dx^r /C25/C0xq2p 4r02lnðr0=aÞ^r, ð53Þ which is exactly Eq.(51). 4.3. Infinite cylinder held at constant potential Suppose that the conducting cylinder is held at a constant potential, Fða;f;zÞ¼F0.From Eq.(34) we obtain (with n00¼roandr4¼r) qG qn00/C12/C12/C12/C12 r00¼a¼/C02 pX1 m¼/C01eimðf/C0f00ÞZ1 0kcos½kðz/C0z00Þ/C138KmðkrÞ KmðkaÞ( /C2½I0 mðkr00ÞKmðkaÞ/C0ImðkaÞK0 mðkr00Þ/C138dk) r00¼a ¼2 paX1 m¼/C01eimðf/C0f00ÞZ1 0cos½kðz/C0z00Þ/C138KmðkrÞ KmðkaÞdk() . ð54Þ In the last equality, we used the Wronskian relation W½Imðkr00Þ;Kmðkr00Þ/C138 ¼/C01=ðkr00Þ.ARTICLE IN PRESS J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1126 The second term given by Eq.(1) can be written as Fþ¼/C01 4paSFð~x00ÞqG qn00da00¼F0 2p2aZ1 /C01adz00Z2p 0df00 /C2X1 m¼/C01eimðf/C0f0ÞZ1 0cos½kðz/C0z0Þ/C138KmðkrÞ KmðkaÞdk() ¼F0 pZ1 /C01dz00Z1 0cos½kðz/C0z0Þ/C138K0ðkrÞ K0ðkaÞdk ¼2F0 pZ1 0dz00Z1 0cos½kðz/C0z0Þ/C138K0ðkrÞ K0ðkaÞdk. ð55Þ In the last equality, we changed the limits of the integral over z00. In order to calculate the last integral, we utilize Eq.(16).Changing variables, we have Z1 0cos½kðz/C0z0Þ/C138dz0¼pdðkÞ.(56) The approximation for small argument of K0ðyÞ, namely, K0ðyÞ/C25/C0 lny,i si n this case inappropriate, because the term lim k!0K0ðkrÞ=K0ðkaÞ!1 for any r. This is true for an infinite cylinder, but gives no physical insight into the behaviour of the potential as a function of r.We should use instead k51=ro1=a, yielding Fþ/C25F0lnðkrÞ lnðkaÞfork51=ro1=a.(57) The potential outside an infinite conducting cylinder with an external charge, held at a constant potential F0, is then given by the summation of Eqs.(36) and (57). We can find the potential of a cylinder held at a constant potential F0by a different method.Suppose we have a long straight line of length ‘along the zaxis, uniformly charged with a linear charge density l.The potential at a distance rfrom thez-axis, for ‘br, is given by Fline/C252xlln‘ r.(58) At a distance r¼afrom the z-axis, we have a constant potential F0¼2llnð‘=aÞ, which is the same boundary condition as before.This implies that the solution is thesame.Substituting l, we obtain the potential as given by F line¼F0lnð‘=rÞ lnð‘=aÞ.(59) Note that Eq.(57) with k51=ro1=aand Eq.(59) with ‘ba4rare essentially the same.Henceforth, we utilize Eq.(59) as the solution for a long cylinder held at a constant potential.ARTICLE IN PRESS J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1127 The final potential of the problem of a long conducting cylinder held at a constant potential F0with an external charge qis given by Fð~x;~x0Þ¼2xq pX1 m¼/C01eimðf/C0f0ÞZ1 0cos½kðz/C0z0Þ/C138Kmðkr4Þ( /C2ImðkroÞ/C0ImðkaÞ KmðkaÞKmðkroÞ/C20/C21 dk) þF0lnð‘=rÞ lnð‘=aÞ. ð60Þ The electric field, the force exerted on q, the surface charge density and the linear charge density are given by, respectively, Erðror0Þ¼ /C02xq pX1 m¼/C01eimðf/C0f0ÞZ1 0kcos½kðz/C0z0Þ/C138( /C2I0 mðkrÞ/C0ImðkaÞ KmðkaÞK0 mðkrÞ/C20/C21 Kmðkr0Þdk) þF0 rlnð‘=aÞ, ð61Þ Erðr4r0Þ¼ /C02xq pX1 m¼/C01eimðf/C0f0ÞZ1 0kcos½kðz/C0z0Þ/C138( /C2Imðkr0Þ/C0ImðkaÞ KmðkaÞKmðkr0Þ/C20/C21 K0 mðkrÞdk) þF0 rlnð‘=aÞ, ð62Þ Ef¼4xq prX1 m¼1msin½mðf/C0f0Þ/C138Z1 0cos½kðz/C0z0Þ/C138( /C2½ImðkroÞ/C0ImðkaÞ KmðkaÞKmðkroÞ/C138Kmðkr4Þdk) , ð63Þ Ez¼2xq pX1 m¼/C01eimðf/C0f0ÞZ1 0ksin½kðz/C0z0Þ/C138( /C2ImðkroÞ/C0ImðkaÞ KmðkaÞKmðkroÞ/C20/C21 Kmðkr4Þdk) , ð64Þ ~Fð~x0Þ¼/C0xq2 pr02X1 m¼/C01Z1 0K2 mðxÞd dxxImðax=r0Þ Kmðax=r0Þ/C20/C21 dx() ^rþqF0 r0lnð‘=aÞ^r, (65) sða;f;zÞ¼ /C0q 2p2X1 m¼/C01eimðf/C0f0ÞZ1 0kcos½kðz/C0z0Þ/C138Kmðkr0Þ KmðkaÞdk() þF0 4pxalnð‘=aÞ, ð66ÞARTICLE IN PRESS J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1128 lða;zÞ¼/C0q pZ1 0cos½kðz/C0z0Þ/C138K0ðkr0Þ K0ðkaÞdkþF0 2xlnð‘=aÞ.(67) From Eq.(67) we can calculate the total charge on the cylinder: Q¼Z1 /C01lða;zÞdz¼/C0q pZ1 /C01dzZ1 0cos½kðz/C0z0Þ/C138K0ðkr0Þ K0ðkaÞdkþ‘F0 2xlnð‘=aÞ ¼/C0q pr02lim k!0lnðkr0Þ lnðkaÞþ‘F0 2l nð‘=aÞ¼/C0qþ‘F0 2xlnð‘=aÞ. ð68Þ For a neutral charged cylinder, i.e., Q¼0, we can relate the constant potential F0 with the charge qby F0¼2xqlnð‘=aÞ ‘.(69) 5. Discussion We can express the force exerted by the grounded conducting infinite cylinder of radius aupon the external point charge qat a distance r0from the axis of the cylinder as given by ~F¼/C0aLxq2 r02^r, (70) where aLis a dimensionless parameter.In this work, we obtained three different expressions for this force, namely, Eqs.(43), (51) and (52).The parameter aLfor these three cases is given by, respectively, aL¼1 pX1 m¼/C01Z1 0K2 mðxÞd dxxImðax=r0Þ Kmðax=r0Þ/C20/C21 dx() , (71) aL/C251 pZ1 0K2 0ðxÞlnð2r0=xaÞ/C0gþ1 ½g/C0lnð2r0=xaÞ/C1382dx/C25p 4l nðr0=aÞ, (72) aL¼2 pZ1 0xK0ðxÞK1ðxÞ K0ðxa=r0Þdx.(73) We plot these three values of aLas functions of a=r0inFigs.5 and6.We can see that these three values of aLconverge to one another as a=r0!0.This was expected because Eq.(43) is valid for a cylinder of finite thickness with arbitrary value of a=r0, while Eqs.(51) and (52) are valid only for a thin cylinder, that is, for a=r0!0. From Eq.(72) we can see that when a=r051, the parameter aLbehaves as p=½4l nðr0=aÞ/C138.That is, it goes to zero when a=r0!0.According to these calculations we conclude that there is no force between a point charge and an idealized groundedconducting line (of zero thickness).One of the authors (AKTA) had expected0oa Lo1[4], not specifically for a grounded conducting line, but for a conducting line with zero total charge.In particular, he expected that 0 :1oaLo0:9, by guessingARTICLE IN PRESS J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1129 the result based on dimensional analysis and in analogy with the case of a point charge qat a distance r0from an infinite conducting plane.In this last case, the net force upon the test charge is given by aPxq2=r02, with aP¼1 4.The results of the calculations presented here, on the other hand, indicate that aL¼0 when a=r0¼0 (in the case of a grounded infinite line).This is an interesting result indicating that the existence of a force upon the external test charge requires not only that it is at aARTICLE IN PRESS Fig.6. Dimensionless parameter aLgiven by Eq.(70) as a function of a=r0, for the region a=r051.The continuous line represents the parameter from Eq.(71); the tight-dashed line that of Eq.(72); and the light-dashed line that of Eq.(73). Fig.5. Dimensionless parameter aLgiven by Eq.(70) as a function of a=r0.The continuous line represents the parameter from Eq.(71); the tight-dashed line that of Eq.(72); and the light-dashed line that of Eq. (73).J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1130 finite distance to the cylinder, but also the existence of a surface area different from zero in the conductor with which it is interacting. Acknowledgements The authors thank Prof.J. D.Jackson for important suggestions related to this work and Faep/Unicamp for financial support.JAH thanks CNPq (Brazil) forfinancial support. References [1] J.D. Jackson, Classical Electrodynamics, third ed., Wiley, New York, 1999. [2] G.B. Arfken, H.J. Weber, Mathematical Methods for Physicists, fourth ed., Academic Press, San Diego, 1995. [3] I.S. Gradshteyn, I.M. Ryzhik, Table of Integrals, Series, and Products, fourth ed., Academic Press, San Diego, 1963. [4] A.K.T. Assis, W.A. Rodrigues Jr., A.J. Mania, The electric field outside a stationary resistive wire carrying a constant current, Found.Phys.29 (1999) 729–753.ARTICLE IN PRESS J.A. Hernandes, A.K.T. Assis / Journal of Electrostatics 63 (2005) 1115–1131 1131