Electrostatics I. Potential due to Prescribed Charge Distribution, Dielectric Properties, Electric Energy and Force

Transcription

1 Chapter Electrostatics I. Potential ue to Prescribe Charge Distribution, Dielectric Properties, Electric Energy an Force. Introuction In electrostatics, charges are assume to be stationary. Electric charges exert force on other charges through Coulomb s law which is the generic law in electrostatics. For a given charge istribution, the electric el an scalar potential can be calculate by applying the principle of superposition. Dielectric properties of matter can be analyze as a collection of electric ipoles. Atoms having no permanent ipole moment can be polarize if place in an electric el. Most molecules have permanent ipole moments. In the absence of electric el, ipole moments are oriente ranomly through thermal agitation. In an electric el, permanent ipoles ten to align themselves in the irection of the applie el an weaken the el. Some crystals exhibit anisotropic polarization an the permittivity becomes a tensor. The well known ouble i raction phenomenon occurs through eviation of the group velocity from the phase velocity in irection as well as in magnitue.. Coulomb s Law Normally matter is charge-neutral macroscopically. However, charge neutrality can be broken relatively easily by such means as mechanical friction, bombarment of cosmic rays, heat (e.g., canle ame is weakly ionize), etc. The rst systematic stuy of electric force among charge boies was mae by Cavenish an Coulomb in the 8th century. (Cavenish s work precee Coulomb s. However, Coulomb s work was publishe earlier. Lesson: it is important to publish your work as early as possible.) The force to act between two charges, q at r an q at r ; follows

2 the well known Coulomb s law, Figure -: Repelling Coulomb force between like charges. F = const. q q r q q = const. ; (N) (.) jr r j where r = jr r j is the relative istance between the charges. In CGS-ESU (cm-gram-secon electrostatic unit) system, the constant is chosen to be unity. Namely, if two equal charges q = q = q separate by r = cm exert a force of yne = 5 N on each other, the charge is e ne as ESU ' 3 9 C. The electronic charge in ESU is e = 4:8 ESU. In MKS-Ampere (or SI) unit system, Coulomb of electric charge is e ne from Coulomb = Ampere sec, where Ampere of electric current is e ne as follows. If two in nitely long parallel currents of equal amount separate by meter exert a force per unit length of 7 N/m on each other (attracting if the currents are parallel an repelling if antiparallel), the current is e ne to be Ampere. Since the force per unit length to act on two parallel currents I an I separate by a istance is given by F l = B I = I I ; (N/m) (.) the magnetic permeability = 4 7 H/m in MKS-Ampere unit system is an assigne constant introuce to e ne Ampere current. (The permittivity " is a measure constant that shoul be etermine experimentally.) In MKS-Ampere unit system, the measure proportional constant in Coulomb s law is approximately 9: 9 N m C : It is customary to write the constant in the form const. = 4" ;

3 Figure -: In MKS unit system, I = Ampere current is e ne if the force per unit length between in nite parallel currents m apart is 7 N/m. The magnetic permeability = 4 7 H/m is an assigne constant to e ne Ampere current. where is the vacuum permittivity. Since " = 8:85 C N m c = " ; m s ; = Fara ; m the permittivity " can be euce from the spee of light in vacuum that can be measure experimentally as well..3 Coulomb Electric Fiel an Scalar Potential The Coulomb s law may be interprete as a force to act on a charge place in an electric el prouce by other charges, since the Coulomb force can be rewritten as where q (r r ) F = 4" jr r j 3 q = q E q ; (N) (.3) E q = 4" q (r r ) jr r j 3 ; N C = V m is the electric el prouce by the charge q at a istance r r : The el is associate with the charge q regarless of the presence or absence of the secon charge q : (The factor 4 is the soli (.4) 3

4 angle pertinent to spherical coorinates. If it is ignore as in the CGS-ESU system, it pops up in planar coorinates. Corresponing Maxwell s equation is r E = " ; in MKS-Ampere, or r E = 4; in CGS-ESU. Which unit system to choose is a matter of conveneince an it is nonsense to argue one is superior to other.) Since two charges q an q exert the Coulomb force on each other, assembling a two-charge system requires work. A i erential work neee to move the charge q against the Coulomb force is Noting W = F r = 4" q q (r r ) jr r j 3 r : r r jr r j = + r r jr r j 3 ; integration from r = to r can be reaily carrie out, W = q q ; (J) (.5) 4" jr r j This is the potential energy of two-charge system. Note that it can be either positive or negative. If W is positive, the charge system has store a potential energy that can be release if the system is isassemble. Energy release through nuclear ssion process is essentially of electrostatic nature associate with a system of protons closely packe in a small volume. On the other han, if W is negative, an energy jw j must be given to isassemble the system. For example, to ionize a hyrogen atom at the groun state, an energy of 3:6 ev is require. In a hyrogen atom, the total system energy is given by W = e mv = 4" r B e 8" r B < ; where r B = 5:3 m is the Bohr raius an mv = e 8" r B = 3:6 ev, is the electron kinetic energy. The electric potential energy is e 4" r B = 7: ev. 4

5 The potential energy of two-charge system may be written as W = q 4" jr r j q ; (J) which e nes the potential associate with a point charge q at a istance r; an the electric el is relate to the potential through = 4" q r ; (J C = V); (.6) E = r; (V m ): (.7) Both potential an electric el E are create by a charge q regarless of the presence or absence of a secon charge..4 Maxwell s Equations in Electrostatics In electrostatics, electric charges are stationary being hel by forces other than of electric origin such as molecular bining force. magnetic els are present, B = : Since charges are stationary, no electric currents an thus no For a istribute charge ensity (r), the electric el can be calculate from E(r) = 4" V (r )(r r ) jr r j 3 V = r 4" (r ) jr r j V ; (.8) since the i erential electric el ue to a point charge q = (r )V locate at r is E = 4" (r )V jr r j 3 (r r ): Note that r jr r j = r jr r j = r r jr r j 3 ; where r is the graient with respect to r. Eq. (.8) allows one to calculate the electric el for a given charge ensity istribution (r): In orer to n a i erential equation to be satis e by the electric le, we rst note that the surface integral of the electric el I E S; can be converte to a volume integral of the ivergence of the el (Gauss mathematical theorem, 5

6 not to be confuse with Gauss law), I E S = S = r EV V V r (r ) 4" V jr r j V : V However, the function = jr r j satis es a singular Poisson s equation, Therefore, I S E S = " r jr r j = 4(r r ): (.9) V V (r r )(r )V = V " (r)v: (.) This is known as Gauss law for the electric el. Note that Gauss law is a consequence of the Coulomb s inverse square law. From r EV = " (r)v; (.) it also follows that r E = " : (.) This is the i erential form of the Gauss law for the longituinal electric el an constitutes one of Maxwell s equations. For the electric el ue to a static charge istribution, E(r) = 4" V (r )(r r ) jr r j 3 V = r 4" V (r ) jr r j V ; its curl ientically vanishes because r E = r r 4" (r ) jr r j V : (Note that for any scalar function F; r rf :) Therefore, the secon Maxwell s equation in electrostatics is r E = : (.3) This is a special case of the more general Maxwell s equation r ; (.4) which etermines the transverse component of the electric el. Eviently, if all charges an els are stationary, there can be no magnetic el. It shoul be remarke that a vector el can be uniquely etermine only if both its ivergence an curl are speci e. (This is known as Helmholtz s 6

7 theorem.) In electroynamics, two vector els, the electric el E an magnetic el B, are to be foun for given charge an current istributions (r; t) an J(r; t): Therefore it is not acciental that four Maxwell s equations emerge specifying the four functions r E; r E; r B; an r B which etermine the longituinal an transverse components of E an B. Digression: A vector A can be ecompose into the longituinal an transverse components A l an A t which are, by e nition, characterize by r A l = ; r A t = : The longituinal component can be calculate from since The transverse component is given by A l (r) = r 4 r A (r ) jr r V ; j r jr r j = 4 r r : A t (r) = A (r) A l (r) = A (r 4 r r ) jr r j V ; where the ientity r r A =rr A r A; is exploite. It is known that for a vector to be uniquely e ne, its ivergence an curl have to be speci e. Since r A =r A l ; ivergence of a vector spci es the ivergence of the longituinal component. Likewise, r A =r A t specifes the curl of the tarnverse component. As a concrete example, let us consier the electric el. Its ivergence is r E = ; " an the solution to this i erential equation is E l (r) = r 4" V (r ) jr r j V : The tranverse component is speci e by the magnetic el as r : The vanishing curl of the static electric el allows us to write the electric el in terms of a 7

8 graient of a scalar potential, E = r; (.5) for the curl of a graient of a scalar function ientically vanishes, r r : For a given charge istribution, the potential has alreay been formulate in Eq. (.8), (r) = 4" V In terms of the scalar function (r); Eq. (.) can be rewritten as (r ) jr r j V : (.6) r = " ; (.7) which is known as Poisson s equation. In general, solving the scalar i erential equation for is easier than solving vector i erential equations for E. The physical meaning of the scalar potential is the amount of work require to move aiabatically a unit charge from one position to another. Consier a charge q place in an electric el E. The force to act on the charge is F = qe an if the charge moves over a istance l; the amount of energy gaine (or lost, epening on the sign of qe l) by the charge is qe l = qr l: Therefore, the work to be one by an external agent against the electric force to move the charge from position r to r is W = r r q E l = q r l = q [(r ) (r )] ; (.8) r r where (r ) (r ) is the potential i erence between the positions r an r : The work is inepenent of the choice of the path of integration. The potential is a relative scalar quantity an a constant potential can be ae or subtracte without a ecting the electric el. In a conuctor, an electric el rives a current ow accoring to the Ohm s law, J = E; (A m ) (.9) where (S m ) is the electrical conuctivity. In static conitions (no time variation an no ow of charges), the electric el in a conuctor must therefore vanish. A steay current ow an constant electric el can exist in a conuctor if the conuctor is a part of close electric circuit. However, such a circuit is not static because of the presence ow of charge. For the same reason, a charge given to an isolate conuctor must entirely resie on the outer surface of the conuctor. Since E (static) = in a conuctor, the volume charge ensity must also vanish accoring to 8

9 = " r E = : A charge given to a conuctor can only appear as a surface charge (C m ) on the outer surface. The time scale for a conuctor to establish such electrostatic state may be estimate from the charge conservation + r J = : The current ensity can be estimate from the equation of motion for = ee m ev; (.) where c (s ) is the electron collision frequency with the lattice ions. Noting J = the ensity of conuction electrons, we obtain nev + c J = ne E: (.) m Therefore, the equation for the excess charge ensity in a + c +! p = ;! p = ne " m ; (.3) which escribes a ampe plasma oscillation e t i!pt with an exponential amping factor = c =: The typical electron collision frequency in metals is of orer c ' 3 = sec an the time constant to establish electrostatic state is inee very short. (If the low frequency Ohm s law is use, an unrealistically short transient time emerges, where = = ' + = ; = e t= ; The principle that a charge given to a conuctor must resie entirely on its outer surface is also a consequence of Coulomb s =r law. For a conuctor of an arbitrary shape, the surface charge istribution is so arrange that the electric el insie the conuctor vanishes everywhere. In Fig. -3, one conucting spherical shell surrouns a smaller one. The spheres are connecte through a thin wire. A charge originally given to the inner sphere will en up at the outer surface of the larger sphere an the electric el insie shoul vanish if Coulomb s law is correct. Experiments have been conucte to measure resiual electric els insie an it has been establishe that the power in the Coulomb s law F _ =r is inee very close to within one part in 6 : is probably exactly equal to.. If not, there woul be a grave consequence that a photon shoul have a nite mass. This is because if a photon has a mass m p ; corresponing Compton wavenumber k C = m p c=~ will moify the wave equation for all potentials, incluing the Coulomb potential, in the form r k = : (.4) " 9

10 Figure -3: Charge initially given to the inner conuctor will all en up at the outer surface of the outer conuctor. The absence of the electric el insie the sphere is a consequence of Coulomb s law. In static = ; this reuces to r kc = " ; (.5) an for a point charge (r) = q(r); yiels a Debye or Yukawa type potential, an electric el (r) = 4" q r exp ( k Cr) ; (.6) E r = q 4" r + k C e kcr : (.7) r The power in the Coulomb s law E r _ q=r a experimentally establishe is = O( 6 ); an this correspons to an upper limit of photon mass of m p < 5 kg. The lower limit of photon Compton wavelength is C > 4 5 km at which istance a signi cant eviation from the Coulomb s law, if any, is expecte. (Incientally, in a plasma, the static scalar potential oes have a Debye form, where (r) = 4" q r exp ( k Dr) ; (.8) k D = s ne " T ; (.9) is the inverse Debye shieling istance, n is the plasma ensity an T (in Joules) is the plasma

11 temperature. Likewise, in a superconuctor, static magnetic el obeys r k L B = ; (.3) where k L =! p =c is the Lonon skin epth with! p = p ne =" m e the electron plasma frequency.) where The potential energy of a two-charge system is W = q q 4" jr r j = q q 4" jr r j ; (.3) q q 4" jr r j ; (.3) is the potential energy associate with each charge. For many charge system, this can be generalize in the form where W = X j q j = X q i q j 4" jr i r j j ; (.33) j i6=j j = X q i 4" jr i r j j ; (.34) i6=j is the potential at the location of charge q j ue to all other charges. For istribute charge with a local charge ensity (r) (C/m 3 ); the potential energy of a i erential charge q = V, which can be regare as a point charge if V is su ciently small, is W = q = V: Since = " r E; W = " r EV: (.35) Integrating over the entire volume, we n W = " = " = I " V V S r EV [r (E) (r )E] V ES + " E V: (.36) V where use is mae of Gauss theorem, V I r FV = S FS; (.37) for an arbitrary well e ne vector el F. The surface integral vanishes because at in nity, both potential an electric el vanish. Therefore, the potential energy associate with a istribute

12 charge system is which is positive e nite. W = " E V; (.38) In the expression for the potential energy of iscrete charge system, the potential energy ue to a point charge itself is exclue while in the integral form, spatial istribution of charge is assume even for point charges an the so-calle self energy is inclue. The quantity u e = " E ; (J/m 3 ) (.39) is the electric energy ensity associate with an electric el. This expression for the energy ensity hols regarless of the origin of the electric el, an can be use for els inuce by time varying magnetic el as well. The self-energy of an ieal point charge eviently iverges because the electric el proportional to =r oes in the limit of r!. However, if the electron is assume to have a nite raius r e ; the self-energy turns out to be of the orer of W e ' 4" e r e : (.4) This shoul not excee the rest energy of the electron m e c : Equating these two, the following estimate for the electron raius emerges, r e ' 4" e mc = :85 5 m. (.4) Although this result shoul not be taken seriously, the scattering cross section of free electron place in an electromagnetic wave (the process known as Thomson scattering) oes turn out to be = 8 3 r e; an the concept of electron raius is not totally absur. For proton whose raius is also of orer 5 m, the analogy obviously breaks own. The Poisson s equation in Eq. (.7) is the Euler s equation to make the energy-like integral stationary U = Inee, the variation of this integral E V = (r) V: (.4) U = = ( r r ) V r + V;

13 becomes stationary (U = ) when the Poisson s equation hols, r + " = : In other wors, electrostatic els are realize in such a way that the total electric energy becomes minimum. In general, electric force act so as to reuce the energy in a close (isolate) system. Macroscopically, electric force tens to increase the capacitance as we will see in Chapter. If the charge ensity istribution is known, the potential (r) can be reaily foun as a solution of the Poisson s equation. However, if the charge ensity is unknown a priori, as in most potential bounary value problems, the potential an electric el must be foun rst. Then the charge ensity is to be foun from = " r E; or in the case of surface charge on a conuctor surface, = " E n ; (C m ); where E n is the electric el normal to the conuctor surface..5 Formal Solution to the Poisson s Equation As shown in the preceing section, if the spatial istribution of the charge ensity (r) is given, the solution for the Poisson s equation r (r) = (r) ; (.43) can be written own as (r) = 4 V (r ) jr r j V : (.44) This is unerstanable because the i erential potential ue to a point charge q = V is = 4" (r )V jr r j ; an the solution in Eq. (.44) is a result of superposition. It is note that the Poisson s equation for the scalar potential still hols even for time varying charge ensity, r C (r; t) = (r; t) " ; (.45) if one employs the Coulomb gauge characterize by the absence of the longituinal vector potential r A = : (.46) 3

14 Figure -4: Di erential potential ue to a point charge q = V. In the Coulomb gauge, the transverse vector potential satis es the wave equation, A t = J t ; (.47) where J t is the transverse current. The solution for the scalar potential in the Coulomb gauge is non-retare, C (r; t) = 4 V an so is the resultant longituinal electric el, E l (r; t) = r C = 4" V (r ; t) jr r j V ; (.48) r r jr r j 3 (r ; t)v : (.49) Such non-retare (instantaneous) propagation of electromagnetic isturbance is clearly unphysical an shoul not exist. In fact, the non-retare Coulomb electric el is exactly cancelle by a term containe in the retare transverse electric el E t (r; t) Jt (t ) jr r j V ; = jr r j c ; as will be shown in Chapter 6. The potential in Eq. (.44) is in the form of convolution between the function G(r; r ) = G(r r ) = 4jr r j ; (.5) 4

15 an the source function (r )= : The function G(r; r ) is calle the Green s function an introuce as a solution to the following singular Poisson s equation r G = (r r ); (.5) subject to the bounary conition that G = at r = : (G subject to this bounary conition is calle the Green s function in free space. It is a particular solution to the singular Poisson s equation. Later, we will generalize the Green s function so that it vanishes on a given close surface by aing general solutions satisfying r G = :) It is evient that the Green s function is essentially the potential ue to a point charge, for the charge ensity of an ieal point charge (without spatial extension) q locate at r can be written as c = q(r r ): (.5) Here (r r ) is an abbreviation for the three imensional elta function. For example, in the cartesian coorinates, in the spherical coorinates (r; ; ), an in the cylinrical coorinates (; ; z), The singular Poisson s equation (r r ) = (x x )(y y )(z z ); (.53) (r r ) = (r r )[r( )][r sin ( )] (.54) = (r r ) rr sin ( )( ) (.55) = (r r ) rr (cos cos )( ); (.56) (r r ) = ( )[( )](z z ) (.57) = ( ) ( )(z z ): (.58) can be solve formally as follows. Let the Fourier transform of G(r g(k) = r G = (r r ); (.59) r ) be g(k); namely, G(r r )e ik(r r) r: (.6) Its inverse transform is G(r r ) = () 3 Noting that the operator r is Fourier transforme as ik an (r g(k)e ik(r r) 3 k: (.6) r ) as unity, we reaily n from 5

16 Eq. (.59), g(k) = k : (.6) Substitution into Eq. (.6) yiels G(r r ) = = = = () 3 () 3 () r) 3 k k eik(r k e ikjr (jr r j cos sin r j cos ) sin 4jr r j ; (.63) where the polar angle in the k-space is measure from the irection of the relative position vector r r ; an 3 k = k k sin ; (.64) e ikx k = (x); (.65) (ax) = (x); (.66) jaj are note. (The same technique will be use in ning a Green s function for the less trivial wave equation, in Chapter 6.) G(r; r ; t; t ) = (r r )(t t ); (.67) If the charge ensity istribution is spatially con ne within a small volume such that r r ; the inverse istance function =jr an so is the potential (r) = 4 V jr r j r j may be Taylor expane as follows: = r r + r r! rr : r r r = r + r r r 3 + X r 5 (3r i r j r ij )rir j + (.68) i;j (r ) jr r j V = (r )V + r 4 r V r 3 r (r )V + 3rr r V r 5 : Q + ; (.69) where is the unit yaic. The quantity q = R (r )V is the total charge containe in the (small) volume V an the corresponing lowest orer monopole potential is monopole = 4 q r : (.7) 6

17 The vector quantity p = r(r)v; (.7) is the ipole moment an the corresponing ipole potential is ipole = 4 p r r 3 : (.7) The tensor Q = Q ij = r i r j (r)v; (.73) e nes the quarupole moment an the corresponing quarupole potential is quarupole = 3rr r 4 r 5 : Q = X 4 r 5 (3r i r j r ij )Q ij : (.74) i;j Example Electric Fiel Lines of a Dipole The potential ue to a ipole moment irecte in z irection p = pe z at the origin is given by (r; ) = p cos : 4 r The equipotential surfaces are escribe by cos r = const. The electric el can be foun from E = r = 4 p cos The equation to escribe an electric el line is, by e nition, r 3 e r + p sin r 3 e : (.75) r = r ; (.76) E r E which gives an thus r r = cot ; r = const. sin : (.77) Eviently, the electric el lines are normal to the equipotential surfaces. In Fig.-5, one equipotential surface an three electric el lines are shown. Example Linear quarupole 7

18 Figure -5: Equipotential surface an electric el lines of an electric ipole p z : The z-axis is vertical. A linear quarupole consists of charges q at z = a an q at z = as shown in Fig.-6. The charge ensity is (r) = q[(z a) (z) + (z + a)](x)(y): Therefore, only Q zz is nonvanishing, Q zz = z (r)v = a q; an the quarupole potential at r a is given by where (r) = 4 r 5 (3z r )Q zz = qa 3 cos 4 r 3 = qa 4 r 3 P (cos ); is the Legenre function of orer l =. P (cos ) = 3 cos ; Alternatively, the potential can be foun from the irect superposition of the potentials prouce by each charge, (r; ) = q p 4 r + a ar cos r + p : r + a + ar cos For r > a; the functions = p r + a ar cos can be expane in terms of the Legenre functions 8

19 Figure -6: Linear quarupole. P l (cos ) as p r + a ar cos = r X a l () l P l (cos ); r > a: r l= Retaining terms up to l = (quarupole), we recover (r; ) ' 4 qa r 3 P (cos ); r a: At r a; the leaing orer potential is of quarupole. The total charge is zero, thus no monopole potential at r a. Also, the charge system consists of two ipoles of equal magnitue oriente in opposite irections. Therefore, the ipole potential vanishes at r a as well. Equipotential pro le is shown in Fig.??..6 Potential ue to a Ring Charge: Several Methos A given potential problem can be solve in i erent coorinates systems. Of course, the solution is unique, an answers foun by i erent methos shoul all agree. For the purpose of becoming familiar with several coorinates systems an some useful mathematical techniques, here we n the potential ue to a ring charge having raius a an total charge q uniformly istribute using several inepenent methos. Metho : Direct integration The potential is symmetric about the z-axis because of the uniform charge istribution. Therefore, we may evaluate the potential at arbitrary azimuthal angle an here we choose = =, that is, observing point in the y z plane. The i erential potential ue to a point charge 9

20 Figure -7: Equipotential surface of a linear quarupole. The z axis ( = ) is in the vertical irection. The thick line is at + unit potential, an the thin line is at : q = q = locate at angle is = 4 q jr r j = 4 q p r + a ar cos ; (.78) where is the angle between the vectors r = (r; ; = =) an r = (a; = =; ): cos reuces to Integrating over from to ; we n cos = cos cos + sin sin cos( ) = sin sin : (.79) (r; ) = 4 q Changing the variable from to through the integral can be reuce to p r + a ar sin sin : (.8) = + ; 4 = p p r + a + ar sin k sin = 4 p r + a + ar sin K(k ); (.8)

21 Figure -8: Uniform ring charge. where k = 4ar sin r + a + ar sin ; (.8)

22 is the argument of the complete elliptic integral of the rst kin e ne by The nal form of the potential is (r; ) = K(k ) = p : (.83) k sin q p r + a + ar sin K(k ): (.84) f (x) = = p x sin Figure -9: K(x) the complete elliptic integral of the rst kin. It iverges logarithmically at x. ; K(x) ' ln 4= p x : The function K(k ) is shown in -9. It iverges as x = k approaches unity, that is, near the ring itself, as expecte. However, the ivergence is only logarithmic, lim k! 4p = ln : Example 3 Capacitance of a Thin Conuctor Ring Let us assume a thin conucting ring with ring raius a an wire raius ( a) shown in Fig.-. The potential on the ring surface can be foun by letting r = a ; = =: In this limit,

23 the argument k approaches unity, k = an the ring potential becomes 4a(a ) (a ) + a + a(a ) ' ; a ring ' Then the self-capacitance of the ring can be foun from C = q ring = q 8a 4 a ln : (.85) 4 a ; a : (.86) 8a ln This formula is fairly accurate even for a not-so-thin ring because of the mere logarithmic epenence on the aspect ratio, a=: For example, when a= = 5 (which is a fat torus rather than a thin ring); Eq. (.86) still gives a capacitance within % of the correct value. An exact formula for the capacitance of a conucting torus will be worke out later in terms of the toroial coorinates. Figure -: A thin conuctor ring with major raius a an minor raius ; a : Metho : Multipole Expansion in the Spherical Coorinates As the secon metho, we irectly solve the Poisson s equation in the spherical coorinates, r = (r) = q (r a)(cos ): (.87) a Except at the ring itself, the charge ensity vanishes. Therefore, in most part of space, the potential satis es Laplace equation, r = ; r 6= a; 6= ; an we seek a potential in terms of elementary solutions to the Laplace equation. = ; 3

24 the Laplace equation r = : Assuming that the potential is separable in the form (r; ) = R(r)F (), we n r R R r + r R + r F sin sin F = : (.89) Since the rst term is a function of r only an the secon term is a function of only, each term must be constant cancelling each other. Introucing a separation constant l(l + ); we obtain two orinary i erential equations for R an F; which can be written as sin R r + r R r sin F l(l + ) r R = ; (.9) + l(l + )F = ; (.9) ( ) F + l(l + )F = ; (.9) where = cos : Eq. (.9) is known as the Legenre s i erential equation. The solutions for R(r) are an the solutions for F () are the familiar Legenre functions, R(r) = r l an ; (.93) rl+ F () = P l (cos ) an Q l (cos ); (.94) where Q l (cos ) is the Legenre function of the secon kin. Some low orer Legenre functions for real x (jxj ) are liste below. P (x) = ; P (x) = x; P (x) = 3x Q l (x) = P l(x) ln + x x ; P 3 (x) = 5x3 3x ; (.95) W l (x); W (x) = ; W (x) = ; W (x) = 3 x; (.96) For a general complex variable z; the e nition for Q l (z) is moi e as follows: Q l (z) = P l(z) ln z + z W l (z); W (z) = ; W (z) = ; W (z) = 3 z; (.97) In the present problem, the potential shoul remain nite everywhere except at the ring. Since Q l (cos ) iverges at = an ; it shoul be iscare an we assume the following series solutions 4

25 for the potential separately for interior (r < a) an exterior (r > a); 8 >< (r; ) = >: P l A l r a l Pl (cos ); r < a P l B l a r l+ Pl (cos ); r > a (.98) where A l an B l are constants to be etermine. The potential shoul be continuous at r = a because there are no ouble layers to create a potential jump. Therefore, A l = B l : To etermine A l ; we substitute the assume potential into the Poisson r = a (r a)(cos ); to obtain X l r + r r l(l + ) r A l R l (r)p l (cos ) = q (r a)(cos ); (.99) a where 8 >< R l (r) = >: r l ; r < a a (.) a l+ ; r > a r If we multiply Eq. (.99) by P l (cos ) sin an integrate over from = to ; the summation over l isappears because of the orthogonality of the Legenre functions, P l (cos )P l (cos ) sin = P l ()P l () = l + ll: (.) We thus obtain A l r + r r l(l + ) r R l (r) = q + (r a)l P a l (): (.) The LHS shoul contain a elta function (r a) to be compatible with that in the RHS. This can be seen as follows. The raial function R l (r) has a iscontinuity in its erivative at r = a; R l r = l + r=a+ a ; R l r = + l r=a a : (.3) Therefore, the secon orer erivative R l =r yiels a elta function, R l r = l + (r a): (.4) a 5

26 We thus nally n the expansion coe cient A l ; A l = q 4 a P l() = q ( ) l= l! 4 a l ; (l = ; ; 4; 6; ); (.5) [(l=)!] an the potential, (r; ) = q 4 a X l even ( ) l= l! l [(l=)!] R l(r)p l (cos ): (.6) Figure -: The erivative of the raial function R(r) is iscontinuous at r = r an thus its secon orer erivative yiels a elta function. The isappearance of the o harmonics is unerstanable because of the up-own symmetry of the problem. At r a; the leaing orer term is the monopole potential, followe by the quarupole potential, l= (r) = q a 4 a r ; l= (r; ) = q a 3 P (cos ); 4 a r an so on. The ipole potential (l = ) vanishes because of the symmetric charge istribution, ( r) = (r); p = r(r)v = : For problems with axial symmetry (@=@ = ) as in this example, knowing the potential along the axis (z) is su cient to n the potential at arbitrary point (r; ) in the spherical coorinates. This is because for all of the Legenre functions, P l () = ; P l ( ) = ( ) l : In the case of the ring 6

27 charge, the axial potential can be reaily foun, (z) = q p 4 z + a : (.7) For jzj > a; this can be expane as (z) = q 4 jzj a 3 a 4 + : (.8) z 8 z Therefore, at arbitrary point (r > a; ); the potential is (r; ) = q 4 r For interior region (r < a); the potential becomes (r; ) = q 4 a They agree with the potential given in Eq. (.6). a P (cos ) + 3 a 4 P4 (cos ) ; r > a: (.9) r 8 r r P (cos ) + a 3 r 4 P4 (cos ) ; r < a: (.) 8 a 7

28 Metho 3: Cylinrical Coorinates In the cylinrical coorinates (; ; z); the Poisson s equation for the potential ue to a ring (; z) = Since the z-coorinate extens from q ( a) (z): (.) a to ; we seek a solution in the form of Fourier transform, (; z) = (; k)e ikz k; (.) where (; k) is the one-imensional Fourier transform of the potential having imensions of Vm. Since is Fourier transforme (; ik(; k); an (z) as unity, the equation for (; k) becomes an orinary i erential equation, + k (; k) = q ( a) : (.3) a Elementary solutions of the i erential equation at 6= a; + k (; k) = ; (.4) are the zero-th orer moi e Bessel functions, shown in Fig.-. (; k) = I (k); K (k); (.5) The Fourier potential (; k) which is continuous at = a an remains boune everywhere may be constructe in the following form, 8 >< AI (k)k (ka); (; k) = >: AI (ka)k (k); < a > a (.6) The coe cient A can be etermine reaily from the iscontinuity in the erivative of (; k) at = a; = AkI (ka)k(ka); =a+ = AkI(ka)K (ka); (.7) =a 8

29 x.5 Figure -: Moi e Bessel functions I (x) (starting at ) an K (x) (iverging at x = ): from which it follows that = Ak I (ka)k(ka) =a I(ka)K (ka) ( a) = A ( a a); (.8) where the Wronskian of the moi e Bessel functions I (ka)k (ka) I (ka)k (ka) = ak ; (.9) has been exploite. From = A =a a ( a) = q ( a) ; (.) a we n an the nal form of the potential is (; z) = q 4 8 >< >: A = q ; (.) I (k)k (ka) I (ka)k (k) 9 >= >; eikz k; < a > a (.) Noting that the moi e Bessel functions are even with respect to the argument, this may be further rewritten as (; z) = q 8 >< I (k)k (ka) >: I (ka)k (k) 9 >= >; cos(kz)k; < a > a (.3) 9

30 The convergence of the integral is rather poor since the function I K ecreases with k only algebraically. An alternative solution for the potential can be foun in terms of Laplace transform rather than Fourier transform, (; z) = (; k)e kjzj k: (.4) The Laplace s equation thus reuces to + + k (; k) = ; (.5) whose solution is the orinary Bessel function J (k): We thus assume for (; k) (; k) = A(k)J (k); (.6) where A(k) may still be a function of k: The Poisson s A(k)J (k)e kjzj k = q ( a) (z): (.7) a Noting z e kjzj = k e kjzj k(z) (.8) we thus n ka(k)j (k)k = q ( a) : (.9) 4 a However, the Bessel function forms an orthogonal set accoring to kj (ka)j (k)k = ( a) ; (.3) a which uniquely etermines the function A(k); A(k) = Therefore, the nal form of the potential is q 4 J (ka): (.3) (; z) = q 4 J (ka)j (k)e kjzj k: (.3) The convergence of this solution is much faster than the solution in Eq. (.3) an thus more suitable for numerical evaluation. 3

31 Metho 4: Toroial Coorinates The toroial coorinates (; ; ) are relate to the cartesian coorinates (x; y; z) through the following transformation, 8>< x = R sinh cos cosh cos ; y = R sinh sin cosh cos ; (.33) >: z = R sin cosh cos : This coorinate system is one example in which the Laplace equation is not separable, that is, the potential cannot be written as a prouct of inepenent functions, (; ; ) 6= F ()F ()F 3 (): However, the potential is partially separable an can be sought in the form (; ; ) = p cosh cos F ()F ()F 3 (): (.34) Figure -3: Cross-section of the toroial coorinates (; ; ):! correspons to a thin ring of raius R: In the toroial coorinates, the = constant surfaces are torois having a major raius R coth an minor raius R= sinh :! egenerates to a thin ring of raius R: In the other limit,! escribes a thin ro on the z-axis. The = constant surfaces are spherical bowls as illustrate in Fig.-3. Assuming the partial separation for the potential in Eq. (.34) leas to the following orinary 3

32 equations for the functions F (); F () an F 3 (); Comparing Eq. sinh F + sinh 4 + l F = ; l m sinh F = ; (.35) + m F 3 = : (.35) with the stanar form of the i erential equation for the associate Legenre function Pl m (cos ); Q m l (cos ); we see that solutions for F () are sin + l(l + ) sin m sin (Pl m ; Q m l ) = ; (.36) The functions F an F 3 are elementary, F () = P m (cosh ); Q m l l (cosh ): (.37) F () = e il ; F 3 () = e im ; (.38) an the general solution for the potential may be written as (; ; ) = p cosh cos X l;m A lm P l (cosh ) + B lm Q l (cosh ) e il+im : (.39) Of course, the potential is a real function. The solution above is an abbreviate form of a more cumbersome expression, (; ; ) = p cosh cos X l= A l P l X (E l cos m + F l sin m) : m= (cosh ) + B l Q l (cosh ) (C l cos l + D l sin l) We now consier a conucting toroi with a major raius a an minor raius b: Its surface is escribe by = const. where a = R coth ; b = R sinh : (.4) Because of axial symmetry, only the m = term is present. If the toroi is at a potential V; the potential o the toroi can be written as (; ) = p cosh cos X l= A l P l (cosh ) cos l; (.4) 3

33 where the Legenre function of the secon kin Q l (cosh ) has been iscare because it iverges at! which correspons to the z axis. (The potential along the z axis shoul be boune.) The expansion coe cients A l can be etermine from the bounary conition, = V at = ; V = p cosh cos X l= A l P l (cosh ) cos l; (.4) or X l= A l P l (cosh ) cos l = V p cosh cos : Multiplying both sies by cos l an integrating over from to ; we obtain A l = V Q l P l (cosh ) (cosh ) p l; (.43) where = ; l = (l ); an the following integral representation has been exploite, Then, the nal form of the potential is cos l p = p Q x cos l (x): (.44) (; ) = p V p cosh cos X l= Q l (cosh ) (cosh ) P l P l (cosh ) cos(l) l : (.45) The capacitance of a conucting torus can be foun from the behavior of the potential at a large istance from the torus which shoul be in the form of monopole potential, (r! ) = At a large istance from the torus r R; both an approach zero, q 4 r : (.46) r = x + y + z = R (sinh + sin ) (cosh cos )! 4R + ; (.47) p cosh cos ' p + p ' p R r ; (.48) P l (cosh ) ' : (.49) 33

34 Therefore, the asymptotic potential is (r R) = V R r X l= Q l (cosh ) (cosh ) = P l q 4 r ; (.5) from which the capacitance of the torus can be foun, C q V = 8 R X l= Q l (cosh ) P l (cosh ) l = 8 a tanh X l= Q l (cosh ) P l (cosh ) l; (.5) where a is the major raius of the torus. The function Q l (cosh ) can be evaluate from Eq. (.44) an P l (cosh ) from P l (cosh ) = Example 4 Capacitance of a Fat Torus : (.5) (cosh + sinh cos ) l+ We wish to n the capacitance of a conucting torus having a major raius of a = 5 cm an minor raius of b = cm. The torus is e ne by cosh = 5: an R = a tanh = 49 cm. For cosh = 5:; the Legenre functions numerically evaluate are: l Q l (5) P l (5) Q l (5) (5) P l :8 :74575 :3434 :563 :3557 :487 :384 3:384 :9 The ratio Q l (5)=P l (5) rapily converges as l increases an it su ces to truncate the series at l = : The capacitance is C = 8" R(:343 + :49 + :3 + ) ' 8 R :393 = 8 a :36: The approximate formula worke out earlier gives C ' 4 a = 8 a :34; 8a ln b which is in reasonable agreement with the numerical result. 34

35 .7 Spherical Multipole Expansion of the Scalar Potential The potential ue to a prescribe charge istribution, (r) = 4" (r ) jr r j V ; can be expane in terms of the general spherical harmonics as follows. For this purpose, it is su cient to expan the Green s function, which satis es the singular Poisson s equation, The Green s function G satis es Laplace equation except at r = r ; G(r; r ) = 4 jr r j ; (.53) r G = (r r ): (.54) r G = ; r 6= r ; (.55) an we rst seek elementary solutions for Laplace equation. Assuming that the function G(r; ; ) is separable in the form an substituting this into + r sin we obtain three orinary equations, G(r; ; ) = R(r)F ()F (); (.56) r + @ r sin + l(l + ) sin + m G(r; ; ) = ; (.57) l(l + ) r R(r) = ; (.58) m sin F () = ; (.59) F () = ; (.6) where l(l + ) an m are separation constants. Solutions of the raial function R(r) are as before, R(r) = r l ; r l+ : In free space, the Green s function must be perioic with respect to the azimuthal angle : Therefore, F () = e im with m being an integer. 35

36 where Equation (.59) is known as the moi e Legenre equation an its solutions are F () = P m l (cos ); Q m l (cos ); (.6) Pl m (x) = ( x m= m ) x m P l(x); (.6) Q m l (x) = ( x m= m ) x m Q l(x): (.63) That Pl m (cos ) given in Eq. (.6) satis es the moi e Legenre equation can be seen as follows. The orinary Legenre function P l (cos ) = P l () satis es Di erentiating m times yiels ( ) P l() + l(l + )P l () = : ( ) m+ P l m+ (m + ) m+ P l + [l(l + ) m+ m(m + )] m P l m = : Let us assume F () = ( ) m= f(); an substitute this into Eq. (.59). f() satis es the same equation as m P l m ; ( ) f (m + ) f + [l(l + ) m(m + )] f = : Therefore, f() = m P l m : Since the Legenre function P l (x) is a polynomial of orer l; the azimuthal moe number m is limite in the range m l: For later use, we combine the functions F () an F () an introuce the spherical harmonic function e ne by where the constant is chosen from the following normalization, Y lm (; ) = const. P m l (cos )e im ; (.64) Y lm (; )Ylm (; ) = : (.65) Noting we n [Pl m (cos )] sin = l + const. = s l + (l m)! 4 (l + m)! ; (l + m)! (l m)! ; (.66) 36

37 Y lm = For historical reasons, it is customary to write Y lm (; ) in the form s Y lm (; ) = ( ) m s l + (l m)! 4 (l + m)! P l m (cos )e im : (.67) l + (l m)! 4 (l + m)! P l m (cos )e im ; for m l; Y l; m (; ) = ( ) m Ylm (; ) ; for l m ; or for arbitrary m; positive or negative, Y lm (; ) = ( Some low orer forms of Y lm (; ) are: ) (m+jmj)= s l = m = Y = p 4 l + (l jmj)! 4 (l + jmj)! P jmj l (cos )e im : (.68) l = m = Y = r 3 4 cos m = r 3 Y ; = 8 sin ei l = m = Y = r 5 3 cos 4 m = r 5 Y ; = sin cos ei 8 m = Y ; = 4 r 5 sin e i Having foun the general solutions of the Green s function, we now assume the following expansion for G; G(r; r ) = X lm A lm g l (r; r )Y lm (; ); where the raial function g l (r; r ) is 8 >< g l (r; r ) = >: r l r l+ ; r < r ; r l r l+ ; r > r : 37

38 In this form, the function g l (r; r ) remains boune everywhere. The expansion coe cient A lm can be etermine by multiplying the both sies of r G = (r r ) = (r r ) rr (cos cos )( ); (.69) by Y l m (; ) an integrating the result over the entire soli angle, A lm r g l(r; r ) = (r r ) rr Ylm ( ; ); where orthogonality of Y lm (; ); Y lm (; )Yl m (; ) = ll mm; (.7) is exploite to remove summation over l an m: Since the raial function g l (r; r ) has a iscontinuity in its erivative at r = r ; it follows that Therefore, r g l(r; r ) = (l + ) (r r ) r : (.7) A lm = l + Y lm ( ; ); (.7) an the esire spherical harmonic expansion of the Green s function is given by G(r; r ) = 4 jr r j = X lm l + g l(r; r )Y lm (; )Y lm ( ; ): (.73) For a given charge istribution (r); this allows evaluation of the potential in the form of spherical harmonics, (r) = = = (r ) 4" jr r j V X 4 4" l + r l+ Y lm(; ) r l Ylm ( ; )(r ; ; )V lm X 4 4" l + r l+ Y lm(; )q lm ; in the region r > r ; (.74) lm where q lm r l Y lm ( ; )(r ; ; )V ; (.75) is the electric multipole moment of orer (l; m): Example 5 Planar Quarupole A planar quarupole consists of charges +q an q alternatively place at each corner of a 38

39 square of sie a as shown in Fig.-4. The charge ensity may be written as (r) = q (r) (r a) a (cos ) () + (r a p a) (cos ) 4 (r a) a (cos )! : The potential in the far el region r a is of quarupole nature an given by (r; ; ) = " 5r 3 (Y ;q ; + Y ; q ; ) ; where q ; = The potential reuces to r Y;(; )(r)v = r 5 a qe i= : (r; ; ) ' 3 8" a r 3 sin sin(); r a: The reaer shoul check that this is consistent with the irect sum of four potentials in the limit r a: Figure -4: Planar quarupole in the x y plane..8 Collection of Dipoles, Dielectric Properties Dielectric properties of material meia originate from ipole moments carrie by atoms an molecules. Some molecules carry permanent ipole moments. For example, the water molecule has a ipole moment of about 6 3 Cm ue to eviation of the center of electron clou from the center of proton charges. When water is place in an external electric el, the ipoles ten to be 39

40 aligne in the irection of the electric el an a resultant electric el in water becomes smaller than the unperturbe external el by a factor "=" ; where " is the permittivity of water. Even a material compose of molecules having no permanent ipole moment exhibits ielectric property. For example, a ipole moment is inuce in a hyrogen atom place in an electric el through perturbation in the electron clou istribution which is spatially symmetric in the absence of external electric el. In this section, the potential an electric el ue to a collection of ipole moments will be analyze. The potential ue to a single ipole p locate at r is (r) = (r r ) p 4 jr r j 3 : (.76) Consier a continuous istribution of many ipoles. It is convenient to introuce a ipole moment ensity P =np (Cm/m 3 = C/m ) where n is the number ensity of ipoles. potential ue to an incremental point ipole p = P(r )V is Integration of this yiels Since An incremental = 4 (r r ) P(r ) jr r j 3 V : (.77) (r) = (r r ) P(r ) 4 jr r j 3 V : (.78) r r jr r j 3 = r jr r ; (.79) j where r means i erentiation with respect to r ; the integral in Eq. (.78) can be rewritten as (r r ) P(r ) jr r j 3 V = P(r r ) r jr r j V P (r) jr r V : j The rst term in the right can be converte into a surface integral through the Gauss theorem, V P(r r I ) jr r V P(r ) = j S jr r j S ; which vanishes on a close surface with in nite extent on which all sources shoul be absent. Therefore, the potential ue to istribute ipole moments is ipole (r) = r P(r ) 4 jr r V : (.8) j Comparing with the stanar form of the potential, (r) = 4" (r ) jr r j V ; 4

41 we see that e = r P; can be regare as an e ective charge ensity. To istinguish it from the free charge ensity, e e ne above is calle boun charge ensity. In general, e cannot be controlle by external means. In a ielectric boy, the boun charge appears as a surface charge ensity. Aing the monopole potential ue to a free charge ensity free (r); we n the total potential Noting we n In this form too, it is evient that (.8) can be rearrange as (r) = free (r ) 4 jr r j V 4 r jr r j = 4(r r ); r E = r = free Introucing a new vector D (isplacement vector) by r P(r ) jr r V : (.8) j r P : (.8) r P can be regare as an e ective charge ensity. Eq. r ( E + P) = free : (.83) D = E + P; (.84) we write Eq. (.83) as r D = free ; (.85) which is equivalent to the original Maxwell s equation r E = all ; (.86) provie that the total charge ensity all consists of free charges an ipole charges an that higher orer charge istributions (quarupole, octupole, etc.) are ignorable. The vector D was name the isplacement vector by Maxwell. Its funamental importance in electroynamics will be appreciate later in time varying els because it was with this isplacement vector that Maxwell was able to preict propagation of electromagnetic waves in vacuum. As brie y iscusse in Section, Maxwell s equations woul not be consistent with the charge conservation principle if the isplacement current, @t ; In usual linear insulators, both D an P are proportional to the local electric el E which 4

42 allows us to introuce an e ective permittivity "; D = " E + P ="E: (.87) In some soli an liqui crystals, the permittivity takes a tensor form, D = E or D i = ij E j ; (.88) because of anisotropic polarizability. Double refraction phenomenon in optics alreay known in the 7th century is ue to the tensorial nature of permittivity in some crystals as we will stuy in Chapter 5. Figure -5: In an electric el, the hyrogen atom exhibits a ipole moment ue to the isplacement of the center of electron clou from the proton. The origin of el inuce ipole moment may be seen qualitatively for the case of hyrogen atom as follows. In an electric el, the istribution of the electron clou aroun the proton becomes asymmetric because of the electric force acting on the electron. The center of electron clou is isplace from the proton by x given by m! x = ee; where! is the frequency of boun harmonic motion of the electron an m is the electron mass. (~! is of the orer of the ionization potential energy.) Then, the ipole moment inuce by the electric el is p = ex = e m! If the atomic number ensity is n; the ipole moment ensity P is E: P = np = ne m! E; (.89) 4

43 an the isplacement vector is given by The permittivity can therefore be e ne by D = " E + P = " + ne " m! E = "E: (.9) " = " +! p!! ; (.9) where s ne! p = " m ; (.9) is an e ective plasma frequency. (The plasma frequency pertains to free electrons in a plasma.! p introuce here is so calle only for imensional convenience.) In the incremental contribution to the permittivity normalize by ",! p! = n e m"! ; the quantity e 4" m! ; is of the orer of r 3 where r is the Bohr raius. This can be seen from the force balance for the electron, which inee gives The quantity e mr! = 4" r ; e 4" m! = 4" r 3 = = r 3 : e m! ; (.93) is calle the atomic polarizability. A more rigorous quantum mechanical calculation yiels for this quantity = 9 4" r; 3 (.94) inicating a substantial correction (by a factor 4.5) to the classical picture base on the assumption of rigi shift of the electron clou. For molecualr hyrogen H ; the polarizability at T = 73 K is approximately given by ' 5:4 4" r: 3 43

44 Satisfactory quantum mechanical calculation was performe by Kolos an Wolniewicz relatively recently (967). This is in a fair agreement with the permittivity experimentally measure in the stanar coniition C, atmospheric pressure, " ' + :7 4 " : The atomic polarizability an the macroscopic relative permittivity " r are relate through a simple relationship known as the Clausius-Mossoti equation, n = 3 " r " " r + ; (.95) where n is the number ensity of atoms. This may qualitatively be seen as follows. A ipole moment inuce by an atom, prouces an electric el p = ex = E = e m! E = E ext ; (.96) p 4" r 3 ; (.97) at a istance r: If the number ensity of atoms is n; the average inter-atom istance R can be foun from Then, an the total electric el is E = E = = 4 3 R3 = n : (.98) 4" R 3 E ext; (.99) 4" R 3 n 3" E ext (.) E ext : (.) Therefore, using this total el in calculation of the polarization el P; we n P = n n=" n E = n " E; (.) 3" 3" which e nes the relative permittivity " r as " r = n=" n : (.3) 3" Solving for n=" ; we obtain Eq.(.95). Note that the relative permittivity is an easily measurable quantity an it re ects, somewhat surprisingly, microscopic atomic polarizability through a simple 44

45 relationship. In an oscillating electric el, the isplacement x is to be foun from the equation of +! x = ee e i!t ; (.4) which yiels A resultant permittivity is x(t) = ee e i!t!! : (.5) "(!) = "! p!!! : (.6) This exhibits a resonance at the frequency! : The resonance plays an important role in calculation of energy loss of charge particles moving in a ielectric meium as will be shown in Chapter 8. Most molecules have permanent electric ipole moments. In the absence of electric els, ipoles are ranomly oriente ue to thermal agitation. When an electric el is present, each ipole acquires a potential energy, U = p E = pe cos ; (.7) where is the angle between the ipole moment p an electric el E: Dipoles are ranomly oriente but shoul obey the Boltzmann istribution in thermal equilibrium, n(cos ) = n exp U = n exp k B T pe cos k B T ; (.8) where n(cos )=n is the probability of ipole orientation in irection, n is the number ensity of the ipoles an k B = :38 3 J/ K is the Boltzmann constant. The average ipole number ensity oriente in the irection of the electric el can be calculate from n = R n(cos ) cos R : (.9) exp pe cos k B T In practice, the potential energy is much smaller than the thermal energy pe k B T: Therefore, the exponential function can be approximate by pe cos exp ' + k B T pe cos k B T ; (.) an we obtain n ' n p E: (.) 3k B T 45