Chapter. Electrostatic II Notes: Most of the material presented in this chapter is taken from Jackson, Chap.,, and 4, and Di Bartolo, Chap... Mathematical Considerations.. The Fourier series and the Fourier transform Any periodic signal g( x), of period a, can always be expressed with the so-called Fourier series, where with G n the Fourier coefficient G n e g x, (.) n π nx i a G n π nx a i a g( x)e dx a. (.) a Alternatively, the Fourier series can be expressed using sine and cosine functions instead of an exponential function. Equations (.) and (.) are then replaced by g( x) G 0 + A n cos πnx a + B n sin πnx a n A n a g( x)cos πnx a a dx a B n a g( x)sin πnx a a dx. a (.) Note that G 0 A 0, if n 0 is allowed in the second of equations (.). If we set δ ( ma) g x, for x < a m (.4) then it seen from equations (.) and (.) that This is the closure relation. δ ( ) a π n( x x ) e i a, for x < a n. (.5) 9
If we let a, we have the following transformations πn a n k dn a π dk (.6) G n π a G( k). The pair of equations (.) and (.) defining the Fourier series are replaced by a corresponding set G k g x G( k) e ikx dk g( x)e ikx dx π. (.7) These equations are usually rendered symmetric by, respectively, dividing and multiplying them by π. We then get the Fourier transform pair π g x π G k G( k)e ikx dk g( x)e ikx dx. (.8) We can evaluate the closure relation corresponding to the Fourier transform by setting g x x δ ( x ) in equations (.8), we thus obtain δ ( ) ik e ( x x ) π dk. (.9) The generalization of both the Fourier series and Fourier transform to functions of higher dimensions is straightforward... Spherical harmonics One possible solution to the Laplace equation is a potential function of polynomials of rank l. If we use Cartesian coordinates we can write this potential as Φ l ( x) C n n n x n yn zn. (.0) n +n +n l Effecting a change of coordinates from Cartesian to spherical with 0
x r sin( θ)cos ϕ y r sin( θ)sin( ϕ) z r cos( θ), (.) and further x + iy r sin( θ)e iϕ x iy r sin( θ)e iϕ, (.) we can write the potential as Φ l l ( r,θ,ϕ) C lm r l Y lm ( θ,ϕ), (.) where the functions Y lm ( θ,ϕ), with l < m < l, are the spherical harmonics. The Laplacian operator in spherical coordinates is given by m l r r r + r sin θ r r r + Ω r, θ sin( θ) θ + sin θ ϕ (.4) with Ω sin θ θ sin( θ) θ + sin θ ϕ. (.5) Applying the Laplace equation to the term ( l, m) of the potential using the last of equations (.4) yields A lm r l + B lm Φ r,θ,ϕ l r l+ Y lm ( θ,ϕ). (.6) l0 m l The spherical harmonics also satisfy the following relations: a) The orthonormality relation π π * dϕ Y lm ( θ,ϕ)y l m ( θ,ϕ)sin( θ)dθ δ l 0 0 l δ m m. (.7)
b) If we write Y lm r ( θ,ϕ) Y lm r Y lm ( n), (.8) where n has the orientation π θ and π + ϕ, then the following symmetry relation applies Y lm ( n) l Y lm n. (.9) c) If we substitute m m Y l, m ( θ,ϕ) m Y lm * ( θ,ϕ). (.0) d) The closure relation l * Y lm ( θ, ϕ )Y lm ( θ,ϕ) δ ϕ ϕ l 0 m l δ ( cos( θ) cos ( θ )). (.) Since the spherical harmonics form a complete set of orthonormal functions, any arbitrary function can be expanded as a series of spherical harmonics with A lm Y lm ( θ,ϕ) g θ,ϕ l l 0 m l π π * A lm dϕ g( θ,ϕ)y lm ( θ,ϕ)sin( θ)dθ. 0 0 (.) Here are some examples of spherical harmonics
Y 00 4π Y 0 Y,± Y 0 4π cos θ 8π sin( θ)e±iϕ 5 6π cos θ Y,± 5 8π sin( θ)cos( θ)e±iϕ Y,± 5 ( π sin θ)e ±iϕ. (.) The Legendre polynomials are the solution to the generalized Legendre differential equation (i.e., equation Error! Reference source not found.) when m 0. They can be defined using the spherical harmonics as P l cos( θ) 4π l + Y l 0 ( θ). (.4) The Legendre polynomials also form a complete set of orthogonal functions with an orthogonality relation and a closure relation ( x)p l ( x) dx P l l + δ (.5) l l where x cos( θ). Any arbitrary function of x cos θ series of Legendre Polynomials * P l ( x )P l ( x) l + δ ( ), (.6) l 0 can, therefore, be expanded in a A l P l ( x) g x l 0 A l l + g( x)p l ( x)dx. (.7) The first few Legendre polynomials are
P 0 P P P ( x) ( x) x ( x) x ( x) 5x x (.8) P 4 ( x) ( 8 5x4 0x + ). There exists a useful relation between the Legendre polynomials and the spherical harmonics. The so-called addition theorem for spherical harmonics is expressed as follows P l cos( γ ) 4π l * Y lm ( θ, l + ϕ )Y lm ( θ,ϕ) (.9) m l and where γ is the angle made between two vectors x and x of coordinates r,θ,ϕ ( r, θ, ϕ ), respectively. Finally, another useful equation is that for the expansion of the potential of point charge, in a volume excluding the charge (i.e., x x ), as a function of Legendre polynomials l x x r < l + r P l cos( γ ), (.0) > l 0 where r < and r > are, respectively, the smaller and larger of x and x, and γ is the angle between x and x (see equation Error! Reference source not found.). Upon substituting the addition theorem of equation (.9) for the Legendre polynomials in equation (.0), we find l l x x 4π r < l + l + r Y * lm ( θ, ϕ )Y lm ( θ,ϕ). (.) > l 0 m l. Solution of Electrostatic Boundary-value Problems As we saw earlier, the expression for the potential obtained through Green s theorem is a solution of the Poisson equation Φ ρ ε 0. (.) 4
It is always possible to write Φ as the superposition of two potentials: the particular solution Φ p, and the characteristic solution Φ c Φ Φ p + Φ c. (.) The characteristic potential is actually the solution to the homogeneous Laplace equation Φ 0 (.4) due to the boundary conditions on the surface S that delimitates the volume V where the potential is evaluated (i.e., Φ c is the result of the surface integral in equation (.06) after the desired type of boundary conditions are determined). Consequently, Φ p is the response to the presence of the charges present within V. More precisely, Φ p ( x) ρ x R d x. (.5) V This suggests that we can always break down a boundary-value problem into two parts. To the particular solution that is formally evaluated with equation (.5), we add the characteristic potential, which is often more easily solved using the Laplace equation (instead of the aforementioned surface integral)... Separation of variables in Cartesian coordinates The Laplace equation in Cartesian coordinates is given by Φ x + Φ y + Φ 0, (.6) z where we dropped the subscript c for the potential as it is understood that we are solving for the characteristic potential. A general form for the solution to this equation can be attempted by assuming that the potential can be expressed as the product of three functions, each one being dependent of one variable only. More precisely, Φ( x, y, z) X ( x)y ( y)z ( z). (.7) Inserting this relation into equation (.6), and dividing the result by equation (.7) we get d X ( x) X ( x) + dx Y y d Y ( y) + dy Z z d Z ( z) 0. (.8) dz If this equation is to hold for any values of x, y, and z, then each term must be constant. We, therefore, write 5
d X ( x) α X ( x) dx d Y y Y ( y) d Z z Z ( z) dy β dz γ, (.9) with α + β γ. (.40) The solution to equations (.9), when α and β are both positive, is of the type Example Φ( x, y,z) e ±iα x e ±iβy e ± α +β z. (.4) Let s consider a hollow rectangular box of dimension a,b and c in the x, y, and z directions, respectively, with five of the six sides kept at zero potential (grounded) and the top side at a voltage V ( x, y). The box has one of its corners located at the origin (see Figure.). We want to evaluate the potential everywhere inside the box. Since we have Φ 0 for x 0, y 0, and z 0, it must be that Figure. A hollow, rectangular box with five sides at zero potential. The topside has the specified voltage Φ V x, y. X sin αx Y sin βy Z sinh( α + β z). (.4) 6
Moreover, since Φ 0 at x a, and y b, we must further have that α n nπ a β m mπ b (.4) γ nm π n a + m b, where the subscripts n and m were added to identify the different modes allowed to exist in the box. To each mode nm corresponds a potential Φ nm sin( α n x)sin( β m y)sinh( γ nm z). (.44) The total potential will consists of a linear superposition of the potentials belonging to the different modes A nm sin α n x Φ x, y, z sin( β m y)sinh( γ nm z), (.45) n,m where the still undetermined coefficient A nm is the amplitude of the potential associated with mode nm. Finally, we must also take into account the value of the potential at z c with A nm sin α n x V x, y sin( β m y)sinh( γ nm c). (.46) n,m We see from equation (.46) that A nm sinh( γ nm c) is just a (two-dimensional) Fourier coefficient, and is given by A nm sinh( γ nm c) 4 ab a b dx dy V ( x, y)sin( α n x)sin( β m y). (.47) 0 0 If, for example, the potential is kept constant at V over the topside, then the coefficient is given by A nm sinh( γ nm c) V 4 π nm n m. (.48) We see that no even modes are allowed, and that the amplitude of a given mode is inversely proportional to its order. 7
. Multipole Expansion Given a charge distribution ρ ( x ) contained within a sphere of radius R, we want to evaluate the potential Φ( x) at any point exterior to the sphere. Since the potential is evaluated at points where there are no charges, it must satisfy the Laplace equation and, therefore, can be expanded as a series of spherical harmonics Φ( x) l Y c lm θ,ϕ lm, (.49) r l + l 0 m l where the coefficients c lm are to be determined. The radial functional form chosen for the potential in equation (.49) is the only physical possibility available from the general solution to the Laplace equation (see equation (.6)), as the potential must be finite when r. In order to evaluate the coefficients c lm, we use the well-known volume integral for the potential Φ( x) ρ x d x, (.50) V with equation (.) for the expansion of the denominator. We, then, find (with r < r x and r > r x, since x > x ) l ε 0 Φ x * Y lm θ, l + ( ϕ ) r l ρ ( x )d x Y lm θ,ϕ. (.5) r l + l 0 m l Comparing equations (.49) and (.5), we determine the coefficients of the former equation to be c lm 4π l + Y * lm ( θ, ϕ ) r l ρ ( x )d x. (.5) The multipole moments, denoted by q lm ( c lm ( l + ) 4π ), are given by * q lm Y lm ( θ, ϕ ) r l ρ ( x )d x (.5) Using some of equations (.), with r sin ( θ )e ±iϕ x ± i y r cos θ z, (.54) we can calculate some of the moments, with for example 8
q 00 4π ρ ( x )d x q 4π q 0 q,± 4π cos( θ) r ρ ( x ) d x 4π p z 8π sin( θ)e iϕ r ρ ( x ) d x ( 8π p x ip y ) z ρ 4π ( x ) d x x i 8π ( y )ρ( x ) d x (.55) for l, and the following for l q 0 5 cos ( θ) 6π r ρ ( x )d x 5 6π Q 5 6π q,± 5 8π sin( θ)cos( θ)e iϕ r ρ ( x )d x 5 8π q,± 5 ( 8π Q iq ) 5 π ( sin θ)e iϕ r ρ ( x )d x 5 ( π Q iq Q ). 5 π z r ρ ( x )d x z ( x i y )ρ( x )d x (.56) ( x i y ) ρ ( x )d x In equations (.55) and (.56), q is the total charge or monopole moment, p is the electric dipole moment p x ρ ( x )d x, (.57) and Q ij is a component of the quadrupole moment tensor (here, no summation is implied when i j ) Q ij ( x i x r δ )ρ( x )d x. (.58) j ij 9
Using a Taylor expansion (see equation (.84)) for x x, we can also express the potential in rectangular coordinate (the proof will be found with the solution to the first problem list) with Φ( x) q r + p x + r Q ij x i x j + r 5. (.59) The electric field corresponding to a given multipole is given by and from equations (.49), (.5), and (.5) we find E Φ lm, (.60) q lm E r Φ r l + ε 0 l + E θ Φ r θ ε 0 l + E ϕ Φ r sin( θ) q lm ( θ,ϕ) Y lm r l + r l + q lm ϕ ε 0 l + θ Y lm θ,ϕ r l + im sin( θ) Y lm ( θ,ϕ). (.6) For example, for the electric monopole term ( l 0 ) we find E r q r and E E θ ϕ 0, (.6) which is equivalent to the field generated by a point charge q, as expected. For the electric dipole term ( l ), we have E r ε 0 r q, Y, + q 0 Y 0 + q Y ( p ε 0 r x + ip y )sin( θ)e iϕ + p z cos θ 8π r sin( θ)e iϕ + p x ip y p x sin( θ)cos( ϕ) + p y sin( θ)sin( ϕ) + p z cos( θ) E θ Y q, ε 0 r, θ + q Y 0 0 θ + q Y θ ( p ε 0 r x + ip y )cos( θ)e iϕ p z sin θ 8π r cos( θ)e iϕ + p x ip y p x cos( θ)cos( ϕ) + p y cos( θ)sin( ϕ) p z sin( θ), (.6) 40
and E ϕ i ε 0 r sin θ i ε 0 r sin θ r q, Y, + q Y 8π ( p + ip x y )sin( θ)e iϕ + ( p x ip y )sin θ p x sin( ϕ) + p y cos( ϕ). e iϕ (.64) Because the unit basis vectors of the spherical and Cartesian coordinates are related by the following relations e r sin( θ)cos( ϕ)e x + sin( θ)sin( ϕ)e y + cos( θ)e z e θ cos( θ)cos( ϕ)e x + cos( θ)sin( ϕ)e y sin( θ)e z e ϕ sin( ϕ)e x + cos( ϕ)e y, (.65) then we can write the dipole electric field vector as E r ( p r e r p). (.66) If the dipole is located at x 0 (which we now consider to be the origin of the coordinate system), with r x x 0 and n the unit vector linking x 0 to x (we could use e r instead of n ), we can write the dipole electric field vector as follows n( n p) p E x x x 0. (.67) It is important to note that in general the multipole moments q lm depend on the choice of the origin. Equation (.67) can be augmented to account for the fact that the charge distribution ρ( x) may possess multipoles of higher order than the dipole. It can be shown that the result is E( x) n( n p) p x x 0 4π pδ x x 0. (.68) Equation (.68) does not change the value of the electric field as calculated with equation (.67) when x x 0. 4
.4 Electrostatic Potential Energy We know from the calculations that led to equation (.60), on page, that the product of the potential and the charge can be interpreted as the potential energy of the charge as it is brought from infinity (where the potential is assumed to vanish) to its final position. More precisely, we define the potential energy W i of a charge q i with W i q i Φ( x i ). (.69) If the potential is due to an ensemble of n charges q j, then the potential energy of the charge q i becomes and the total potential energy W is W i q i W n q j, (.70) j x i x j n q i q j. (.7) i j <i x i x j Alternatively, we can write a more symmetric equation for the total energy by summing over all charges and dividing by two W 8πε 0 n q i q j. (.7) i j i x i x j Equation (.7) can easily be generalized to continuous charge distributions with W 8πε 0 or, alternatively, if we use equation (.57) for the potential ρ( x)ρ ( x ) d x d x (.7) W ρ( x)φ( x)d x (.74) Furthermore, we can express the potential energy using the electric field instead of the charge distribution and the potential. To so, we replace the charge distribution with the Poisson equation in equation (.74) and proceed as follows 4
W ε 0 ε 0 ε 0 Φ Φd x ( Φ) Φ Φ d x ( Φ Φ) nda E d x S, (.75) where we used equation (.6) and the divergence theorem for the second and last lines, respectively. Because the integration is done over all of space, the surface in integral will vanish since lim R ( Φ Φ) nda S R S lim 4π lim R R 0. R R R dω (.76) The potential energy then becomes W ε 0 E d x (.77) It follows from this result that the potential energy density is defined as w ε 0 E. (.78) It is important to note that the energy calculated with equation (.77) will be greater than that evaluated using equation (.7) as it contains (that is, equation (.77)) the selfenergy of the charge distribution. We are now interested in expressing the potential energy of a charge distribution subjected to an external field as consisting of the contributions from the different terms in the multipole expansion. In this case, we know from equation (.69) that W ρ( x)φ( x) d x. (.79) We start by expanding the potential function with a Taylor series (see equation (.84)) around some predefined origin 4
Φ( x) Φ( 0) + x Φ x0 + x Φ ix j + x i x j x0 Φ( 0) x E( 0) x E j ix j +, x i x0 (.80) where we used E Φ, and summations over repeated indices were implied. Since E 0 for the external field, we can subtract the following relation from the last term of equation (.80) without effectively changing anything 6 r E x0 6 r δ ij E j x i x0 (.8) to get Φ( x) Φ( 0) x E( 0) ( 6 x ix j r δ ij ) E j + (.8) x i x0 If we insert equation (.8) into equation (.74) for the potential energy, and using equation (.58) for the components of the quadrupole moment tensor, we get W qφ( 0) p E( 0) 6 Q E j ij + (.8) x i x0 Using equation (.68) for the electric field generated by a dipole, and equation (.8), we can calculate the energy of interaction between two dipoles p and p located at x and x, respectively, when x x. Thus, W p E x p E ( x ) ( p n) p p p n, x x (.84) where n ( x x ) x x. The dipoles are attracted to each other when the energy is negative, and vice-versa. For example, when the dipoles are parallel in their orientation, and to the line joining them, then, W < 0 and they will attract each other..5 Electrostatic fields in Matter So far in dealing with the equations of electrostatic, we were concerned with, and derived, the microscopic equations of electrostatic. That is, we considered problems involving mainly charge distributions without the presence of any ponderable media. When analyzing electrostatic fields in matter, we need to make averages over 44
macroscopically small, but microscopically large, volumes to obtain the macroscopic equations of electrostatics. In the first place, if we think of the macroscopic electric field E at a given point x as some average of the microscopic electric field E µ over some surrounding volume ΔV, we can write E( x) ΔV If we calculate the curl of the macroscopic field, we have E µ ( x + x )d x. (.85) ΔV E( x) ΔV ΔV ΔV E µ ( x + x )d x E µ ( x + x )d x, ΔV (.86) and since from the microscopic equation E µ 0, then E 0 (.87) for the macroscopic field. So the same relation between the electric field and the potential exists when dealing with ponderable media. That is, E( x) Φ. (.88) When a medium made up atoms or molecules is subjected to an external electric field, the charges making up the molecules will react to its presence by individually producing a, or enhancing an already existing, dipole moment. The material as a whole will, therefore, become electrically polarized, the resulting dipole moment being the dominant multipole term. That is, an electric polarization P (dipole moment per unit volume) given by N i x P x p i (.89) i is induced in the medium, where p i is the average dipole moment of the ith type of molecules in the medium (calculated in the same manner as the electric field was in equation (.85)), and N i is the average number density of the same type of molecules at point x. The average charge density ρ x is evaluated using the same process. Similarly, we can use equation (.59) to calculate the contribution to the potential at the position x from the macroscopically small volume element ΔV located at x 45
ΔΦ( x, x ) ρ ( x )ΔV + ( ) P ( x )ΔV, (.90) where ρ ( x )ΔV and P ( x )ΔV are, respectively, the average monopole (or charge) and dipole moments contained in the volume. Taking the limit ΔV d x, and integrate over all space, we get the potential Φ( x) ρ x d x x x + P ( x ), (.9) where we used equation (.55). Transforming the second term, we find P x ( ) d P x x P x d x P ( x ) n d a P ( x ) d x P x d x, (.9) since the first integral on the right-hand side is over an infinite surface. The potential now becomes Φ( x) d x ρ ( x ) P x x ( x ). (.9) Equation (.9) has the same form as equation (.57) for the potential, as long as we define a new effective charge density ( ρ P). By effective charge density, we mean that if the polarization is non-uniform in a given region (i.e., P 0 ), then there will be a net change in the amount of charge within that region. Upon using equation (.88), we can write E( x) Φ ρ ( x ) P ( x ) d x ρ ( ) δ ε x P x 0 d x (.94) P( x) ε ρ x. 0 46
We now define the electric displacement D as and equation (.94) becomes D ε 0 E + P (.95) D ρ (.96) In most media P is proportional to E (i.e., the media are linear and isotropic), and we write P ε 0 χ e E, (.97) where χ e is the electric susceptibility of the medium. In such cases, D is also proportional to E and with the electric permittivity ε defined by D εe, (.98) ε ε 0 ( + χ e ). (.99) The quantity ε ε 0 is called the dielectric constant. If the medium is also homogeneous, then the divergence of the electric field becomes E ρ ε. (.00) Consequently, any electrostatic problem in a linear, isotropic, and homogeneous medium is equivalent to one set in vacuum (solved using the microscopic equations) as long as the electric field is scale by a factor ε 0 ε. Finally, the boundary conditions derived for the microscopic electric field (see equations (.79)) can easily be extended to such medium ( D D ) n σ n 0, E E (.0) where n is a unit vector extending from medium to medium, and normal to the boundary surface, and σ is the surface charge. 47