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1738 Brazilian Journal of Physics, vol. 34, no. 4B, December, 2004 Surface Charges and External Electric Field in a Toroid Carrying a Steady Current J. A. Hernandes and A. K. T. Assis Instituto de F´ ısica ‘Gleb Wataghin’, Universidade Estadual de Campinas – Unicamp 13083-970 Campinas, S ˜ ao Paulo, SP, Brasil Received on 2 February, 2004; revised version received on 7 May, 2004 We solve the problem of a resistive toroid carrying a steady azimuthal current. We use standard toroidal coor- dinates, in which case Laplace’s equation is R-separable. We obtain the electric potential inside and outside the toroid, in two separate cases: 1) the toroid is solid; 2) the toroid is hollow (a toroidal shell). Considering these two cases, there is a difference in the potential inside the hollow and solid toroids. We also present the electric field and the surface charge distribution in the conductor due to this steady current. These surface charges gene- rate not only the electric field that maintains the current flowing, but generate also the electric field outside the conductor. The problem of a toroid is interesting because it is a problem with finite geometry, with the whole system (including the battery) contained within a finite region of space. The problem is solved in an exact analytical form. We compare our theoretical results with an experimental figure demonstrating the existence of the electric field outside the conductor carrying steady current. 1 Toroidal Ring The electric field outside conductors with steady currents has been studied in a number of cases. These cases, howe- ver, consider infinitely long conductors: coaxial cable, [1, pp. 125–130], [2, pp. 318 and 509–511], [3], [4, pp. 336- 337] and [5]; solenoid with azimuthal current, [2, p. 318] and [6]; transmission line, [7, p. 262] and [8]; straight wire, [9]; and conductor plates, [10]. The only cases solved in the literature where the geometry of the conductor is finite are those of a finite coaxial cable considered by Jackson, [11], and that of a toroidal conducting ring, [12]. Here we consider the case of the conducting toroid with a steady current. Our goal is to find the electric potential inside and outside the toroid, and from the potential we can find the electric field and surface charges. More details about this problem and its analytical solution can be found in [12]. Consider a toroidal conductor with uniform resistivity. It has greater radius R 0 and smaller radius r 0 and carries a steady current I in the azimuthal direction, flowing along the circular loop. The toroid has rotational symmetry around the z-axis and is centered in the plane z =0. The battery that maintains the current is located at ϕ = π rad, see Fig. 1. Air or vacuum surrounds the conductor. The electric potential φ can be calculated using toroidal coordinates (η,ξ,ϕ) [13, p. 112], defined by: x = a sinh η cos ϕ cosh η - cos ξ , y = a sinh η sin ϕ cosh η - cos ξ , z = a sin ξ cosh η - cos ξ . (1) Figure 1. A toroidal ohmic conductor with symmetry axis z, smal- ler radius r 0 (m) and greater radius R (m). A thin battery is loca- ted at ϕ = π rad maintaining constant potentials (represented as the “+” and “-” signs) in its extremities. A steady current flows azimuthally in this circuit loop in the clockwise direction, from ϕ =+π rad to ϕ = -π rad. Here, a is a constant such that when η →∞ we have the circle x = a cos ϕ, y = a sin ϕ and z =0. The toroi- dal coordinates can have the possible values: 0 η< , -π rad ξ π rad and -π rad ϕ π rad. We take η 0 as a constant that described the toroid surface in toroidal coordinates. The internal (external) region of the toroid is characterized by η>η 0 (η<η 0 ). The potential along the surface of the toroid is linear in ϕ, φ(η 0 ,ξ,ϕ)= A + . This potential can be expanded in Fourier series in ϕ: φ(η 0 ,ξ,ϕ)= A + = A +2B X q=1 (-1) q-1 q sin(). (2)

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  • 1738 Brazilian Journal of Physics, vol. 34, no. 4B, December, 2004

    Surface Charges and External Electric Field in aToroid Carrying a Steady Current

    J. A. Hernandes and A. K. T. AssisInstituto de Fsica Gleb Wataghin, Universidade Estadual de Campinas Unicamp

    13083-970 Campinas, Sao Paulo, SP, Brasil

    Received on 2 February, 2004; revised version received on 7 May, 2004

    We solve the problem of a resistive toroid carrying a steady azimuthal current. We use standard toroidal coor-dinates, in which case Laplaces equation is R-separable. We obtain the electric potential inside and outside thetoroid, in two separate cases: 1) the toroid is solid; 2) the toroid is hollow (a toroidal shell). Considering thesetwo cases, there is a difference in the potential inside the hollow and solid toroids. We also present the electricfield and the surface charge distribution in the conductor due to this steady current. These surface charges gene-rate not only the electric field that maintains the current flowing, but generate also the electric field outside theconductor. The problem of a toroid is interesting because it is a problem with finite geometry, with the wholesystem (including the battery) contained within a finite region of space. The problem is solved in an exactanalytical form. We compare our theoretical results with an experimental figure demonstrating the existence ofthe electric field outside the conductor carrying steady current.

    1 Toroidal RingThe electric field outside conductors with steady currentshas been studied in a number of cases. These cases, howe-ver, consider infinitely long conductors: coaxial cable, [1,pp. 125130], [2, pp. 318 and 509511], [3], [4, pp. 336-337] and [5]; solenoid with azimuthal current, [2, p. 318]and [6]; transmission line, [7, p. 262] and [8]; straight wire,[9]; and conductor plates, [10]. The only cases solved in theliterature where the geometry of the conductor is finite arethose of a finite coaxial cable considered by Jackson, [11],and that of a toroidal conducting ring, [12].

    Here we consider the case of the conducting toroid witha steady current. Our goal is to find the electric potentialinside and outside the toroid, and from the potential wecan find the electric field and surface charges. More detailsabout this problem and its analytical solution can be foundin [12].

    Consider a toroidal conductor with uniform resistivity.It has greater radius R0 and smaller radius r0 and carriesa steady current I in the azimuthal direction, flowing alongthe circular loop. The toroid has rotational symmetry aroundthe z-axis and is centered in the plane z = 0. The batterythat maintains the current is located at = pi rad, see Fig. 1.Air or vacuum surrounds the conductor.

    The electric potential can be calculated using toroidalcoordinates (, , ) [13, p. 112], defined by:

    x = asinh coscosh cos , y = a

    sinh sincosh cos ,

    z = asin

    cosh cos . (1)

    Figure 1. A toroidal ohmic conductor with symmetry axis z, smal-ler radius r0 (m) and greater radius R (m). A thin battery is loca-ted at = pi rad maintaining constant potentials (represented asthe + and - signs) in its extremities. A steady current flowsazimuthally in this circuit loop in the clockwise direction, from = +pi rad to = pi rad.

    Here, a is a constant such that when we have thecircle x = a cos, y = a sin and z = 0. The toroi-dal coordinates can have the possible values: 0 < ,pi rad pi rad and pi rad pi rad. We take0 as a constant that described the toroid surface in toroidalcoordinates. The internal (external) region of the toroid ischaracterized by > 0 ( < 0).

    The potential along the surface of the toroid is linear in, (0, , ) = A + B. This potential can be expandedin Fourier series in :

    (0, , ) = A+B = A+ 2Bq=1

    (1)q1q

    sin(q).

    (2)

  • J. A. Hernandes and Andre K. T. Assis 1739

    Figure 2. Equipotentials for a resistive full solid toroidal conductorin the plane z = 0. The bold circles represent the borders of the to-roid. The current runs in the azimuthal direction, from = +pi radto = pi rad. The thin battery is on the left ( = pi rad). Wehave used 0 = 2.187.

    Figure 3. Equipotentials in the plane x = 0 for a resistive full solidtoroidal conductor carrying a steady azimuthal current, Eq. (7) withA = 0 and B = 0/2pi. The bold circles represent the conductorsurface. We have used Eq. (22) with 0 = 2.187.

    Eq. (2) can be used as the boundary condition for our parti-cular solution to this problem.

    Laplaces equation for the electric potential 2 =0 can be solved in toroidal coordinates with the methodof separation of variables (by a procedure known as R-separation), leading to a solution of the form, [13, p. 112]:

    (, , ) =cosh cos H()X()(), (3)

    where the functions H , X , and satisfy the general equati-ons (where p and q are constants):

    c

    (cosh2 1)H + 2 cosh H [(p2 1/4) + q2/(cosh2 1)]H = 0, (4)X + p2X = 0, (5) + q2 = 0. (6)

    Using the boundary condition (2) and the possible solutions of Eqs. (4) to (6) we obtain the potential as given by, [12]:

    ( 0, , ) =cosh cos

    [ p=0

    Ap cos(p)Pp 12 (cosh ) +q=1

    sin(q)p=0

    Bpq cos(p)Pq

    p 12(cosh )

    ], (7)

    d

    where the coefficients Ap and Bpq are given by, respecti-vely:

    Ap =2A(2 0p)

    pi

    Qp 12 (cosh 0)

    Pp 12 (cosh 0), (8)

    Bpq =22B(1)q1(2 0p)

    qpi

    Qp 12 (cosh 0)

    P qp 12

    (cosh 0), (9)

    where wp is the Kronecker delta, which is zero for w 6= pand one for w = p. The functions P q

    p 12(cosh ) and

    Qqp 12

    (cosh ) are known as toroidal Legendre polynomialsof the first and second kind respectively, [14, p. 173].

    For the region inside the hollow toroid (that is, > 0),the potential is given by:

    ( > 0, , ) = A+cosh cos

    q=1

    sin(q)p=0

    Bpq cos(p)Qq

    p 12(cosh ), (10)

    where the coefficients Bpq are defined by:

    Bpq =22B(1)q1(2 0p)

    qpi

    Qp 12 (cosh 0)

    Qqp 12

    (cosh 0). (11)

    We plotted the equipotentials of a full solid toroid on theplane z = 0 in Fig. 2 with A = 0 and B = 0/2pi. Fig. 3shows a plot of the equipotentials of the full solid toroid inthe plane x = 0 (perpendicular to the current).

  • 1740 Brazilian Journal of Physics, vol. 34, no. 4B, December, 2004

    2 Electric Field and Surface ChargesThe electric field can be calculated by ~E = in toroidalcoordinates, as given by: c

    E = sinh cosh cos a

    p=0

    cos(p){Ap

    [12Pp 12 (cosh ) + (cosh cos )Pp 12

    (cosh )]

    +q=1

    sin(q)Bpq

    [12P qp 12

    (cosh ) + (cosh cos )P qp 12

    (cosh )]}

    , (12)

    E = cosh cos

    a

    p=0

    [sin cos(p)

    2 p(cosh cos ) sin(p)

    ]

    [ApPp 12 (cosh ) +

    q=1

    sin(q)BpqPq

    p 12(cosh )

    ], (13)

    E = (cosh cos )3/2

    a sinh

    q=1

    q cos(q)p=0

    Bpq cos(p)Pq

    p 12(cosh ), (14)

    where P qp 12

    (cosh ) are the derivatives of the P qp 12

    (cosh ) relative to cosh . The electric field inside the full solid toroid( > 0) is given simply by:

    E = 0, E = 0, E = cosh cos a sinh

    B = Bx2 + y2

    . (15)

    For the full solid toroid, the surface charge distribution that creates the electric field inside (and outside of) the conductor,keeping the current flowing, can be obtained with Gauss law (by choosing a Gaussian surface involving a small portion of theconductor surface):

    (0, , ) = 0[~E( < 0) () + ~E( > 0)

    ]0=

    0 sinh 0a

    {A+B2

    + (cosh 0 cos )3/2

    p=0

    cos(p)[ApPp 12

    (cosh 0) +q=1

    sin(q)BpqPq

    p 12(cosh 0)

    ]}. (16)

    d

    3 Thin Toroid Approximation

    Here we treat the case of a thin toroid, such that the ou-ter radius R0 = a cosh 0/ sinh 0 a and the inner ra-dius r0 = a/ sinh 0 are related by r0 R0, see Fig. 1.In this case we have cosh 0 0 1. The functionQp 12 (cosh 0) that appears in Eqs. (8) and (9) for the coef-

    ficients Ap and Bpq can be approximated by, [14, p. 164]:

    Qp 12 (cosh 0) pi(p+ 12

    )2p+

    12 p! coshp+

    12 0

    , (17)

    where is the gamma function, [15, p. 591].Because Eq. (17) has a factor of coshp 12 0 1, we

    can neglect all terms with p > 0 in Eq. (7) compared withthe term with p = 0. The potential outside the thin toroid(0 1):

    c

    ( 0, , ) =cosh cos

    cosh 0

    [AP 12 (cosh )

    P 12 (cosh 0)+2B

    q=1

    (1)q1q

    sin(q)P q 12

    (cosh )

    P q 12(cosh 0)

    ]. (18)

  • J. A. Hernandes and Andre K. T. Assis 1741

    We are especially interested in the expressions for the potential and electric field outside but in the vicinity of the conductor,0 > 1. A series expansion of the functions P q 12 () and P

    q

    12() around gives as the most relevant terms [14,

    p. 173]:

    P q 12()

    2/pi

    (12 q

    ) ln(2) ( 12 q)

    , P q 12()

    2/pi

    (12 q

    ) 13/2

    [1 ln(2)

    (12 q

    ) 2

    ], (19)

    where (z) = (z)/(z) is the digamma function, and 0.577216 is the Euler gamma. The potential just outside the thintoroid, Eq. (7), can then be written in this approximation as:

    (0 1, , ) = A ln(8 cosh )ln(8 cosh 0) + 2Bq=1

    (1)q1q

    sin(q)ln(2 cosh ) ( 12 q) ln(2 cosh 0)

    (12 q

    ) . (20)This is a new result not presented in [12].

    Far from the battery (that is, for || pi rad) the potential (20) can be fitted numerically by trial and error by the followingsimpler expression (valid for 0 > 103):

    (0 1, , ) = A ln(8 cosh )ln(8 cosh 0) +Bln(1.67 cosh )ln(1.67 cosh 0)

    . (21)

    This is a correction from Eq. (35) of [12].The surface charge distribution in this thin toroid approximation is given by:

    (0 1, , ) = 0 sinh 0a

    [A

    ln(8 cosh 0)+ 2B

    q=1

    (1)q1q

    sin(q)ln(2 cosh 0)

    (12 q

    ) ]. (22)

    This is another new result not presented in [12]. In Fig. 4 we plotted the density of surface charges as a function of theazimuthal angle . We can see that is linear with only close to = 0 rad. Close to the battery diverges to infinity (thatis, when pi rad).

    Far from the battery Eq. (22) can be fitted numerically by a linear function on , namely:

    (0 1, , ) = 0 sinh 0a

    [A

    ln(8 cosh 0)+

    B

    ln(1.67 cosh 0)

    ]=

    0r0

    [A

    ln(8a/r0)+

    B

    ln(1.67a/r0)

    ] A + B. (23)

    This is a correction from Eq. (37) of [12]. We defined A and B by this last equality.We can calculate the total charge qA of the thin toroid as a function of the constant electric potential A. For this, we

    integrate the surface charge density , Eq. (22), in and (in the approximation cosh 0 1):

    qA = pipi

    hd

    pipi

    hd(, ) =4pi20Aa

    ln(8 cosh 0)=

    4pi20Aaln(8a/r0)

    , (24)

    d

    Figure 4. Density of surface charges as a function of the azi-muthal angle in the case of a thin resistive toroid carrying a ste-ady current. We used 0 = 10.

    where h = h = a/(cosh cos ) and h =

    a sinh /(cosh cos ) are the scale factors in toroidalcoordinates, [16]. Notice that from Eq. (24) we can obtainthe capacitance of the thin toroid, [17, p. 127]:

    C =qAA

    =4pi20a

    ln(8 cosh 0)=

    4pi20aln(8a/r0)

    . (25)

    It is useful to define a new coordinate system:

    = a, =

    (x2 + y2 a

    )2+ z2. (26)

    We can interpret as a distance along the toroid surfacein the direction, and as the shortest distance from thecircle x2 + y2 = a2 located in the plane z = 0. Consider a

  • 1742 Brazilian Journal of Physics, vol. 34, no. 4B, December, 2004

    certain piece of the toroid between the angles 0 and 0,with potentials in these extremities given by R = A+B0and L = A B0, respectively. This piece has a length

    of ` = 2a0. When 0 > 1 (that is, r0 < a) thepotential can be written as:

    c

    = Aln(`/) ln(`/8a)ln(`/r0) ln(`/8a) +

    2B0`

    ln(`/) ln(`/1.67a)ln(`/r0) ln(`/1.67a)

    (R + L

    2+R L

    `)ln(`/)ln(`/r0)

    , (27)

    where in the last approximation we neglected the term ln(`/a) utilizing the approximation r0 < a (so that `/r0 >`/ `/1.67a > `/8a). The electric field can be expressed in this approximation as:

    ~E = (R + L

    2+R L

    `)

    ln(`/r0) R L

    `

    ln(`/)ln(`/r0)

    . (28)

    d

    Eqs. (27) and (28) can be compared to Eqs. (12) and (13)of Assis, Rodrigues and Mania, [9], respectively. They havestudied the case of a long straight cylindrical conductor ofradius r0 carrying a constant current, in cylindrical coordi-nates (, , z) (note that the conversions from toroidal tocylindrical coordinates in this approximation are and z). In their case, the cylinder has a length ` andradius r0 `, with potentials L and R in the extremitiesof the conductor, and RI = L R. Our result of the

    potential in the region close to the thin toroid coincides withthe cylindrical solution, as expected.

    4 Charged Toroid Without CurrentConsider a toroid described by 0, without current but char-ged to a constant potential 0. Using A = 0 and B = 0 inEqs. (7) we have the potential inside and outside the toroid,respectively:

    c

    ( 0, , ) = A = 0, (29)

    ( 0, , ) =cosh cos

    p=0

    Ap cos(p)Pp 12 (cosh ), (30)

    where the coefficients Ap are given by Eq. (8). This solution is already known in the literature, [18, p. 239], [19, p. 1304].It is also possible to obtain the capacitance of the toroid, by comparing the electrostatic potential at a distance r far from

    the origin with the potential given by a point charge q, (r a) q/4pi0r:

    (r a, , ) a2

    r

    p=0

    20(2 0p)

    pi

    Qp 12 (cosh 0)

    Pp 12 (cosh 0)=

    q

    4pi0r. (31)

    d

    The capacitance of the toroid with its surface at a cons-tant potential 0 can be written as C = q/0. From Eq. (31)this yields, [18, p. 239], [20, p. 5-13], [21, p. 9], [22, p. 375]:

    C = 80ap=0

    (2 0p)Qp 12 (cosh 0)

    Pp 12 (cosh 0). (32)

    Utilizing the thin toroid approximation, 0 1, one can

    obtain the capacitance of a circular ring, Eq. (25).Another case of interest is that of a charged circular line

    discussed below, which is the particular case of a toroid withr0 0. With 0 1 and cosh 0 1 we have R a.Keeping only the term with p = 0 in Eqs. (8) and (30),expressed in toroidal and spherical coordinates (r, , ) res-pectively, the potential for the thin toroid becomes:

    c

    (r, , ) =qA

    4pi20a

    cosh cos P 12 (cosh ) (33)

  • J. A. Hernandes and Andre K. T. Assis 1743

    =qA4pi0

    1[(r2 a2)2 + 4a2r2 cos2 ]1/4P 12

    (r2 + a2

    (r2 a2)2 + 4a2r2 cos2

    ). (34)

    We can expand Eq. (34) on r, where r< (r>) is the lesser (greater) between a and r =x2 + y2 + z2. We present

    the first three terms:

    (r, , ) qA4pi0

    {1r>

    1 + 3 cos(2)8

    r2

    +3512

    [9 + 20 cos(2) + 35 cos(4)

    ]r4

    }. (35)

    d

    Eqs. (33) to (35) can be compared with the solution gi-ven by Jackson, [23, p. 93]. Jackson gives the exact elec-trostatic solution of the problem of a charged circular wire(that is, a toroid with radius r0 = 0), in spherical coordinates(r, , ):

    (r, , ) =qA4pi0

    n=0

    r2n

    (1)n(2n 1)!!2nn!

    P2n(cos ),

    (36)where qA is the total charge of the wire. Eq. (36) expan-ded to n = 2 yields exactly Eq. (35). We have checked thatEqs. (34) and (36) are the same for at least n = 30.

    Eqs. (33) and (36) yield the same result. It is worthwhileto note that in spherical coordinates we have an infinite sum,Eq. (36), while in toroidal coordinates the solution is gi-ven by a single term, Eq. (34). The agreement shows thatEqs. (33) and (36) are the same solution only expressed indifferent forms.

    Figure 5 shows the potential as function of (in cylin-drical coordinates) in the plane z = 0. Eqs. (33) and (36)give the same result.

    Figure 5. Normalized potential as function of (distance from z-axis) on the plane z = 0. Eqs. (33) and (36) give the same result.We utilize 0 = 38 (cosh 0 = 1.6 1016) and a = 1.

    5 Discussion and ConclusionFigure 2 can be compared with the experimental result foundby Jefimenko, [24, Fig. 3], reproduced here in Fig. 6 withFig. 2 overlaid on it. Jefimenko painted a circular conduc-ting strip on a glass plate utilizing a transparent conductingink. A steady current flowed in the strip by connecting its

    extremities with a battery. By spreading grass seeds on theglass plate he was able to map the electric field lines insideand outside the strip (in analogy with iron fillings mappingthe magnetic field lines). The equipotential lines obtainedhere are orthogonal to the electric field lines. There is a veryreasonable agreement between our theoretical result and theexperiment.

    Figure 6. Jefimenkos experiment [24, Fig. 3] in which the linesof electric field were mapped using grass seeds spread over a glassplate. There is a circular conducting strip carrying a steady cur-rent. Fig. 2 has been overlaid on it the equipotential lines areorthogonal to the electric field lines.

    Our solution inside and along the surface of the full so-lid toroid yields only an azimuthal electric field, namely,|E| = /2pi. But even for a steady current we musthave a component of ~E pointing away from the z axis, E,due to the curvature of the wire. Here we are neglecting thiscomponent due to its extremely small order of magnitudecompared with the azimuthal component E. See furtherdiscussion in [12].

    The beautiful experimental result of Jefimenko showingthe electric field outside the conductor is complementedby this present theoretical work, with excellent agreement,Figs. 6. The electric potential and electric field of thethin toroid approximation with a steady current, respecti-vely Eqs. (27) and (28), agree with the known case of along straight cylindrical conductor carrying a steady current,Eqs. (12) and (13) of [9]. The electric potential of the thintoroid approximation without current agrees with the knownresult of a charged wire, [23, p. 93].

  • 1744 Brazilian Journal of Physics, vol. 34, no. 4B, December, 2004

    Here we have obtained a theoretical solution for the po-tential due to a steady azimuthal current flowing in a toroidalresistive conductor which yielded an electric field not onlyinside the toroid but also in the space surrounding it. Oursolution showed a reasonable agreement with Jefimenkosexperiment which proved the existence of this external elec-tric field due to a resistive steady current.

    AcknowledgmentsThe authors wish to thank Fapesp (Brazil) for finan-

    cial support to DRCC/IFGW/Unicamp in the past few ye-ars. They thank Drs. R. A. Clemente and S. Hutcheon forrelevant comments, references and suggestions. One of theauthors (JAH) wishes to thank CNPq (Brazil) for financialsupport.

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    vistic Physics, edited by F. Selleri (Apeiron, Montreal, 1998),pp. 177185.

    [4] D. J. Griffiths, Introduction to Electrodynamics (PrenticeHall, New Jersey, 1999), 3rd ed.

    [5] A. K. T. Assis and J. I. Cisneros, IEEE Trans. Circ. Sys. I 47,63 (2000).

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