Lecture Notes on Electromagnetism

Transcription

1 Prepared for submission to JCAP Lecture Notes on Electromagnetism Salah Nasri a United Arab Emirates University, Al-Ain, UAE snasri@uaeu.ac.ae Abstract. These are my notes on electromagnetism for the course of PHSY 530 (Electrodynamics I) that I have taught at UAEU during the Spring of 2014 and Spring Most of the figures have been produced by the author (using "Jaxo-Draw").

2 Contents 1 Introduction to Electrostatics Coulomb s Law The Electric Field Gauss s Law Curl of Electric Field and the Electrostatic Potential An Application to Surface Charge Distribution The Electrostatic Potential Energy of a Discrete Charge Distribution The Electrostatic Potential Energy of a Continuous Distribution Poisson and Laplace Equations Green s Theorem Appearance of Boundary Condition in Poisson Equation The Uniqueness of the Solution in Dirichlet and Neumann B.Cs Solution of the Electrostatic Boundary-Value Problem with Green Functions 29 2 The Conductors The Electric Field Screening Surface Charge and Force on a Conductor Capacitance 35 3 Boundary-Value Problem in Electrostatic The Method of Images The Green s Function Method for the Sphere 48 4 Solving the Laplace Equation The Orthogonal Functions Separation of Variables: Laplace Equation in Rectangular Coordinates Separation of Variables: Laplace Equation in Spherical Coordinates Expansion of point-like potential in spherical harmonics Expansion of Green s Functions in Spherical Coordinates Fields and Potentials Around Conical Hole 65 5 Multipole Expansion Dipole Moment of pair of charges +q and q Linear Quadrupole Multipole Expansion of a Charge Distribution in Rectangular Coordinates Multipole Expansion of a Charge Distribution in Spherical Coordinates Energy of a Charge Distribution in an External Filed Force and Torque on a Charge Distribution in External Electric Field 79 i

3 6 Electrostatic in Dielectric Media Introduction Macroscopic description of dielectric Boundary Conditions for E and D at medium interfaces Boundary-value problems in dielectrics Clausius-Mossotti Relation Electrostatic energy in dielectric media Application of the method of images to dielectric media 95 7 Exercises on Electrostatics 96 8 Answers to the Exercises on Electrostatics Magnetostatics Introduction Current and Continuity Equation Magnetic Force Biot-Savart law Magnetic Force between two currents The divergence and the curl of B Vector potential Magnetic field of a circular current loop Magnetic moment of a localized current distribution Force on a current distribution in an external magnetic field Gyromagnetic ratio Magnetostatics in Matter Introduction Magnetically permeable media Boundary conditions for B and H at mediums interface Boundary-value problem in magnetostatics for J = Magnetic shielding Exercices on Magnetostatics Answers to the Exercices on Magnetostatics Time-Dependent Phenomena and Maxwell s Equations Faraday s Law of Induction Displacement current and Maxwell s equations Appendices Some Physical Quantities in SI and CGS system of units Useful Formulas in Vector Caculus Cylindrical Coordinates 161 ii

4 14.4 Spherical Coordinates Helmholtz theorem (1855) Legendre Polynomials Spherical Harmonics Addition Theorem The constant α in Faraday s law of induction Who is Who in Electromagnetism Charles Augustin de Coulomb ( ; French) Andre Marie Ampere ( ; French) Johann Carl Friedrich Gauss ( ; German) Jean-Baptiste Biot ( ; French) and Felix Savart ( ; French) Simeon Denis Poisson ( ; French) Michael Faraday ( ; English) 176 1

5 1.1 Coulomb s Law 1 Introduction to Electrostatics Experimentally it was observed that certain bodies long range forces on each other that are not gravitational; they are not proportional to the mass, and they can be repulsive as well attractive. The source of this new force was defined to be the electric charge of the object, which can be positive or negative. It is believed that the electric charge is quantised in unit of e, which is the magnitude of the charge of the electron. This means that the net charge of in a any material is Q = (N p N e )e where N p and N e represent the number of protons and electrons in the material, respectively. In this section, we will discuss in a quantitative way the form of the electrostatic force between charges. In 1785, Charles Augustin de Coulomb, using a very sensitive torsional balance was able to empirically infer that for two static bodies one of charge q 1 at r 1 and the second one of charge q 2 at r 2, and if their separation is much greater than the size of either body, then the force, say on q 1 due to q 2, is given by 1 q F 12 = k 1 q 2 e ˆr r (1.1) where r 12 = ( r 1 r 2 ), ˆr 12 is the unit vector directed from q 2 to q 1, and k e is a positive universal constant. The force exerted on q 2 due to q 1 has the same magnitude as the one exerted on q 2 due to q 1, but opposite in direction, i.e. F 21 = k e q 2 q 1 r 2 21 q 2 q 1 ˆr 21 = k e ˆr r = F 12 (1.2) Thus, the magnitude of the electrostatic force is inversely proportional to the square of the distance between the two charges. Since, we only know about charges by measuring the Coulomb force, we are, in principle, free to choose k e to be any thing we want. So, our choice of k e then determines the unit that charge is measured in. In the SI system of units, i.e., the MKS system (also called the practical units), charge is measured in the historical unit, the Coulomb, then k e has the value 1 k (SI) e = 10 7 ( c 0 m/s ) 2 N.m 2 C N.m2 C 2 (1.3) 1 Actually Henri Cavendish ( ) was the first to discover and understand the Physics aspects of Coulomb s laws, but he was lazy to publish it. 1 The choice for the value of k e in the MKS system of units given above, was made official in

6 Z F 21 = F 12 q 2 r 12 r 2 q 1 F 12 r 1 Y X Figure 1. The electrostatic force between two charges (here of the same sign). where c m/s is the speed of light in vacuum, and "C" denotes the unit of the electric charge, the Coulomb 2 Note that the electrostatic force is parametrically similar to the gravitational force. Thus, one can compare their strength by considering their ratio. As example, consider two protons (m p kg), then F elec = k SI e 2 [C 2 ] e F grav G[m 3.s 2.kg 1 ]m 2 p[kg 2 ] (1.4) where G = m 3.s 2.kg 1 is the universal constant of gravity. So we see that one can neglect the gravitational force as compared to the electrostatic force. It is customary to write k (SI) e in the form 2 In the SI system of units the electron charge is e C 3

7 k (SI) e := 1 4πɛ 0 (1.5) where ɛ [C 2.N 1.m 2 ] which is called the permittivity of the vacuum 3. In the cgs system of unit 4, also called the Gaussian units, where c stands for centimeter, g for grams, and s for seconds, one has k (cgs) e = 1 (1.6) Thus, in this system of units the electric charge is not an independent unit, instead it is a derived quantity with dimension [q (CGS) ] = M 1/2 L 3/2 T 1 (1.7) and it is measured in unit of stat Coulomb, also called the electro-static-unit, denoted by esu, which is defined by the statement that the force between two charges each of 1 edu, and 1 cm apart is 1 dyne 5, i.e. 1esu = ( dyne.cm 2) 1/2 = [ g 1/2.cm 3/2.s 1] (1.8) To find the conversion factor from the unit esu to Coulomb we write q (cgs) = k e (SI) q (SI) = /2 N.m 2 q(si) C = dyne.cm 2 q(si) C Taking q [SI] to be 1 Coulomb, the above equality reads (1.9) (1.10) q (cgs) = esu (1.11) Hence, we obtain 3 The reasons why k (SI) e was written as 1/4πɛ 0 are: The factor of 4π was introduced in the definition of k e so that it cancels the 4π factor that arises in the calculation of the electric flux (see later). The constant ɛ 0, was put in the denominator in order to make the electrostatic theory in 1800 s look more like the theory of fluid mechanics. 4 Formally introduced by the British Association of Advancement in Science (BAAS) in dyne is the force that would give a mass of 1 gram an acceleration of 1 cm/cm 2. In terms of Newton, 1 dyne = 10 5 N. 4

8 1C esu (1.12) In the appendix we list some physical quantities in both SI and cgs unit systems and give their corresponding conversion factors. The main advantage of the Gaussian units is that the various formulas of electromagnetism are simpler and easier to remember. Also, the synthesis of electricity and magnetism into electromagnetism is more symmetrical. A simple rule for translating formulas of electrostatic from SI to Gaussian units and vice versa is a s follows: Equations of electrostatics without the displacement electric field vector D (see the section on the dielectrics in these notes) can be converted from SI to cgs units by the replacement of 1 4πɛ 0 [SI] 1[cgs] (1.13) Gaussian electrostatic equations without D containing the square of electron electric charge, e 2, can be converted to SI units by the replacement of e 2 [cgs] e2 4πɛ 0 [SI] (1.14) It is useful to remember that 1 stat volt[cgs] = 300 volts[si] (1.15) Therefore, there are no ɛ 0 s in Gaussian units. There are no µ 0 s, called the permeability of the vacuum (see the section on magnetostatics on these notes), either, since these can be expressed in terms of the speed of light in vacuum by the relation ɛ 0 µ 0 = 1 c 2 0 (1.16) So, in cgs unit system, instead of ɛ 0 s and µ 0 s one sees factors of c 0. It is important to note that ɛ 0 and µ 0 are not physical properties of the vacuum, but rather artifacts 5

9 of the SI system of units, which disappear in Gaussian system of units. Another reasonable choice of units would be to measure charge in integer multiples of the electron charge. This would yield different value of k e. An important property of Coulomb s force is that it satisfies the superposition principle. It states that the electrostatic force exerted on a stationary charge q 1 at position r 1 by a system of stationary point charges {q i, i = 1, 2,.., N} at { r i }, is given by the vector sum of all Coulomb s forces exerted on q 1. That is F( r 1 ) = j 1 F 1j = k e q 1 j 1 where r 1j = r 1 r j, and ˆr 1j is the unit vector directed from q j to q The Electric Field q j r 2 1j ˆr 1j (1.17) We define the electric field, E( r), at point r as the limiting force per unit charge exerted on a test charge δq at that point as δq 0: E( r) := lim δq 0 F( r) δq (1.18) The limit δq 0 is introduced so that the test charge does not influence the charge sources that produce the electric field. Thus, E is an entity independent of the probe (i.e. the test charge) placed at that point. Moreover, the notion of field becomes almost unavoidable for the description of the time-dependent phenomena, such as the electromagnetic waves. As a consequence of the principle of superposition, the electric field at a fixed point r due to a system of stationary discrete charges q i located at r i, with i = 1, 2,.., N, is given by E( r) = k e N i=1 q i ( r r i) r r i 3 (1.19) If many charges {q i } within some volume are located so closely we can approximate them with a continuous charge distribution, such as the charge contained in a very small volume d 3 r is 6 dq( r ) = ρ( r )d 3 r (1.20) 6 For areal charge density, σ, and linear density λ, we have dq = σd 2 r and dq = σdx, respectively. 6

10 where ρ( r ) is the volume charge density. To find the electric field generated by all charges, we rewrite the sum (1.20) as E( r) = k e dq( r ) ( r r ) (1.21) r r 3 d 3 r which can be expressed as an integral over the whole volume containing the charges, i.e. 7 E( r) = k e d 3 r ρ( r ) ( r r ) r r 3 (1.22) It is possible to treat the discrete and continuous charge on the same footing by introducing the Dirac δ-function 8, which, in one dimension, is defined as: (i) δ(x a) = 0, if x a; (ii) { x 2 1, if x1 < a < x x 1 δ(x a)dx = 2 ; 0, if a is outside [x 1, x 2 ]. (1.23) This implies that we can write 9 x2 x 1 f(x)δ(x a)dx = f(a); a [x 1, x 2 ] (1.35) 7 Note that for a smooth charge distribution function ρ, the integral (1.22) does not diverges at ( r r ) 0, because in this limit the fraction under the integral increases as r r 2, i.e. slower than the decrease of the elementary volume d 3 r, proportional to r r 3. 8 To be more precise, δ(x) is not a function because it is not integrable in the conventional sense of Riemann s notion of integral. Instead, it is a mathematical entity called distribution, which is well defined only when it appears under an integral sign. 9 Another useful property of δ-distribution is δ [f(x)] = n j=1 1 f (x j ) δ(x x j) (1.24) where x j is the j th root of the equation f(x) = 0, and f (x j ) denotes the derivative of f at x j. This can be shown as follows. Cconsider the following integral: + g(x)δ [f(x)] dx (1.25) where g(x) is any function. Near the j th root of the function f(x), we can consider the integrand as g(x)δ [f (x j ) (x x j )], and for a small positive number, we have 7

11 δ( x a) a x Figure 2. The Dirac distribution δ(x a). xj+ x j g(x)δ [f(x)] dx = = xj+ x j 1 f (x j ) g(x j )δ [f (x) (x x j )] dx (1.26) x j + f x j f g(x j )δ(y)dy where in the second step above we made the change of variable y = f (x j )(x x j ). We have two cases : 1. If f > 0, then the lower limit is smaller than the upper one, and thus the integrand is g(x j )/f (x j ). 2. If f < 0, then the lower limit is greater than the upper one, and thus the integrand is g(x j )/f (x j ). So, the integral in (1.27) reads + g(x)δ [f(x)] dx = j 1 g(x j ) f (x j ) (1.27) 8

12 This suggests that f(x)δ(x a) can be interpreted intuitively as a strongly concentrating function at point a. Therefore, collection of discrete charges q i, with i = 1, 2,,..N, at positions r i can be represented by the charge density ρ( r) = N q i δ( r r i ) (1.36) i=1 or, equivalently, δ [f(x)] = j 1 f (x j ) δ(x x j) (1.28) The δ-distribution can be represented also as a limit of functions. For example, a very useful representation is given by lim x 0 1 πσ e x2 /σ = δ(x) (1.29) The above definitions of the δ-function can be generalised to distribution defined on a space M (d) of d dimensions as such that δ(x A) = δ(x 1 a 1 )δ(x 2 a 2 )...δ(x d a d ) (1.30) (i) δ(x A) = 0, if x A; (ii) { 1, if A V; V δ(x A)d3 X = 0, otherwise. (1.31) where X = (x 1, x 2,.., x d ), and A = (a 1, a 2,..a d ) are vectors in M (d) written in a rectangular system of coordinates. Note that δ(x a) has dimension of 1/L d, where L is some length scale. In a curvilinear coordinates, X = (u 1, u 2,..., u d ), the volume element has the form dv (d) = h u1 h u2...h ud du 1 du 2...du d (1.32) where h u1 h u2...h ud is the Jacobian of the coordinate transformation from rectangular to curvilinear. In this system of coordinates, the δ-distribution is given by δ(x X 0 ) = 1 h u1 h u2...h ud δ(u 1 u (0) 1 )δ(u 2 u (0) 2 )...δ(x d u (0) d ) (1.33) with (u (0) 1, u(0) 2,..., u(0) d ) are the coordinates of A in the curvilinear system. For example, in the spherical coordinates in d = 3, (r, θ, φ), we have h r = r 2.h θ = sin θ, h φ = 1, hence δ( r r 0 ) = 1 r 2 sin θ δ(r r 0)δ(θ θ 0 )δ(φ φ 0 ) (1.34) 9

13 By inserting it in (1.22) yields the expression of the electric field of N charges given in (1.19). Remarque The introduction of the concept of field is necessary so that causality does not get violated. To see this, consider two static charges q 1 and q 2 at a distance r apart. Imagine that we suddenly displace one of the charges, say, q, by a distance r. Then, according to Coulomb s law, the force felt by q 2 is q 1 q 2 F = k e 2 ˆr (1.37) (r + r) which means that action at a distance has occurred. Does it happen instantaneously? According to the theory of special relativity, it takes time for the information about the change to reach q 2. In fact, the information propagates with a speed of light in vacuum, which means that it could take t = r/c. For Coulomb s law to be consistent with causality, Faraday and Maxwell introduced the concept of electric field such that F 12 = q 1 E 2 ( r 1 ) or F 21 = q 2 E 1 ( r 1 ) (1.38) with E i being the field due to charge q i and r j is the position of q j. Then a charge interacts only with the field at its own position; that is the interaction is local. So when we change the position of the charge, the electric field takes time to respond. As we will show later, Maxwell s equations correctly predict how the electric (and magnetic) field will respond when we move the charge. 1.3 Gauss s Law Let us consider a single charge q inside a smooth, closed surface S. We would like to calculate the flux, Φ E, of the electric field E through a closed surface S, i.e. Φ E = E.dS = E( r).ˆnds (1.39) where ds is an infinitesimal element of the surface area. If θ is the angle between E and the unit vector ˆn normal to the element of surface, then using the Coulomb s law, we get 10

14 E.dS = E cos θds = k e q r 2 ds (1.40) where ds = ds cos θ is the projection of ds onto the direction of the vector r connecting the charge q with the point on ds. But the ratio ds /r 2 is the elementary solid angle dω, thus Φ E = k e q dω = 4πk e q (1.41) S since the full solid angle equals 4π. Eq (1.41) is the Gauss s Law for one point charge. However, it is only valid if the charge is located inside the volume limited by S. For a charge q outside the volume V, for each positive contribution ds /r 2 there is always equal and opposite contribution to the integral. As a result, at the integration over the whole closed surface the positive and negative contributions cancel exactly, and one gets S E.dS = 0; q / V (1.42) In general, for a collection of charges, either discrete or continuous, the superposition principle implies that the flux created by these charges equals the sum of the individual fluxes from each charge, for which either E (1.41) or Eq (1.42) are valid independent on the charge position (in or out of the volume). As a result, the total flux reads or, equivalently, Φ Etot = 4πk e q j = 4πk e j V V ρ( r )d 3 r (1.43) E.dS = 4πke Q enc (1.44) where Q enc is the net charge enclosed in V. This is the full statement of Gauss s Law. It is important to note that the integral form of Gauss Law (1.44) is useful only for highly symmetrical geometries, like the two previous examples we discussed. However, it may be recast into an alternative differential form whose field of application is much wider. Using the divergence theorem, we can write the electric flux as 11

15 S E. ds = V (. E ) d 3 r (1.45) where V is the volume limited by the closed surface S. Combining Eq (1.45) with the Gauss Law (1.44), we obtain V [. E 4πke ρ] d 3 r = 0 (1.46) Since the above equation is valid for any choice of volume, the function under the integral must vanish at each point, i.e. 10. E = 4πk e ρ (1.47) which is Gauss s law in differential form, which shows that in space the distribution of charges i are the source of the electric field. Example 1.1: Electric field of a charged sphere Consider a spherically symmetric charge distribution with density ρ(r). Due to the spherical symmetry, the electric field should be perpendicular to the sphere at each its point, and its magnitude is the same at all points. As a result, the flux of the electric field is E.ˆnd 2 r = 4πr 2 E(r) (1.48) Now, applying Gauss s Law (1.44), we get 4πr 2 E(r) = 4πk e so that we get r <r ρ(r )4πr 2 dr (1.49) E(r) = k e Q(r) r 2 (1.50) 10 In SI and Gaussian unit system, the differential form of Gauss Law reads 4πρ,. E = ρ ɛ 0, in Gaussian units; in MKS units. 12

16 where Q(r) is the net charge inside the sphere of radius r: Q(r) = 4π r 0 ρ(r )r 2 dr (1.51) This implies that the electric field outside the sphere of radius R, where the whole charge is contained, is given by E out (r) = k e Q(R) r 2 k e Q tot r 2 (1.52) which shows that outside the sphere, the electric field is exactly the same as if all the total charge Q was concentrated in the sphere s center 11. For the field inside the sphere, it requires an explicit knowledge of the profile of the density. For instance, If the charge Q is uniformly distributed inside the sphere of radius R, then and so Eq (1.50) yields ρ(r ) = Q V = 3 Q (1.53) 4π R 3 E in (r) = k e Q R 3 r (1.54) Thus, for a spherically-symmetric charge distribution with uniform density, the electric field is given by r, r < R; R 3 E(r) = k e Q (1.55) 1, r > R. r 2 Note that the above results of the electric field for r < R and r > R give the same value k e Q/R 2 at the charged sphere s surface, so that the electric field is continuous. 13

17 Example 1.2: Electric field of a charged surface Consider an infinite thin sheet (see the figure below), with a uniform surface charge density σ. Let us choose a Gauss s surface in the form of a planar "pillbox" of thickness 2z. Because the sheet is infinitely large, this system has a planar symmetry, and so the electric field should be: directed along the z-axis, constant on each of the upper and bottom side of the pillbox, equal and opposite on these sides, parallel to the side surface of the box. As a result the total electric field flux through the pillbox surface is just 2AE(z), where A is the area of the upper and lower side of the pillbox. According to Gauss s Law (1.44), this should be equal to the charge within the pillbox, i.e. 4πk e σa. Hence we obtain E(z) = 2πk e σ (1.56) which in vector notation can be written as 2πk e σẑ, z > 0; E = 2πk e σẑ, z < 0. (1.57) Note that although the magnitude of the electric field E = E is constant, its vertical component E z changes sign at z = 0, experiencing a discontinuity equals to E z = 4πk e σ. This jump disappear if the surface is not charged (i.e. σ = 0). Law Example 1.3: Electric field of charged sphere using local Gauss s As an application of Eq (1.47), let us consider the previous example of the charged sphere. Since E = E(r) ˆn, where ˆn is an outward-normal unit vector, the divergence E in spherical coordinates give. E = 1 d ( r 2 E ) (1.58) r 2 dr As a result, Eq (1.47) yields a linear differential equation for E(r): 1 de(r) 4πk e ρ, r R; = r 2 dr 0, r R. 14

18 which for a uniform charge density, may be readily integrated on each of the segments, and we find E(r) = 4πk ρr 3 + C e 3 1, r R; (1.59) r 2 C 2, r R. where C 1 and C 2 are the integration constants. Using the boundary conditions E(0) = 0, (1.60) E(R 0) = E(R + 0),. and hence we get E(r) = k e Q r, r < R; R 3 1, r > R. r 2 which is identical to the expression we obtained in example 1.1. (1.61) Although it might seem that in this example the differential form of Gauss Law is not simpler than its integral form, it is actually more convenient for asymmetric charge distributions. Moreover, it is invaluable in cases where the charge distribution is not known a priori and has to be found in a self-consistent way (see more later). 1.4 Curl of Electric Field and the Electrostatic Potential In general, Eq (1.47) is not enough to determine the electric field. However, if in addition one specifies the curl of E, and boundary conditions (here E 0 as r ), then E( r) is uniquely determined 12. To obtain the equation of E, we use the fact that ( ) ( r r ) 1 r r = 3 r r (1.62) where the gradient is with respect to r. Hence, E( r) = so the electric field can be written as d 3 r k e ρ( r ) r r (1.63) 12 This is known as the Helmholtz theorem. See Appendix for more details. 15

19 where the function Φ( r) is given by E = Φ( r) (1.64) Φ( r) := k e d 3 r ρ( r ) r r (1.65) So, it might be easier to find the field Φ first, and then calculating E by taking the gradient. Moreover, as a consequence of (1.4) we have E = 0 (1.66) The work done to move a point charge q from position r A to r B along some curve C AB in electric field E is rb rb W AB = F. dl = q E. d l (1.67) r A r A It represents the energy of the system ( the electric field E and the charged particles) which increases if work is done on the system, say by moving the charge q against E, but decreases when work is done by the system, say when E moves the charge. We can rewrite (1.67) as rb rb W AB = q Φ.d l = q r A r A Φ. dl (1.68) l where in the second equality we used the definition of the directional derivative. Integrating over dl yields where W AB = q r B r A dφ = U(r B ) U(r A ) (1.69) U( r) = qφ( r) (1.70) represents the electrostatic potential energy of the charge q. Hence, the field Φ has the interpretation of "potential energy" per unit charge, and so it is called the scalar potential. Note that Φ( r) and Φ( r) + Constant give rise to the same electric field E (since.constant = 0) and W AB (where the constant cancels out). 16

20 1.5 An Application to Surface Charge Distribution Consider a surface S of surface charge density σ( r). Let E 2 and E 1 be the electric fields at points just above and just below r on two sides of the surface. To find the relation between E 2 and E 1, we construct a "pillbox" through the surface as shown in the figure below. Applying Gauss law to the pillbox gives E 2. S top + E 1. S bottom + sides = 4πk e σ S (1.71) The contribution from the sides vanishes in the limit where h 0 and S top = S ˆn and S bottom = S ˆn, and so, we obtain (E 2 E 1 ). ˆn = 4πk e σ (1.72) Thus, the surface charge density creates a discontinuity in the normal component of the electric field. To see how the tangential component of E behaves, consider a small rectangular loop through the surface (see figure below). Using Eq (1.4) and Stokes theorem, we have E. dl = l (E 2. ˆt) l (E 2. ˆt) + E. d l = 0 (1.73) where ˆt is a unit vector along the elementary line vector l in the loop 13. Taking h 0, the contribution from the sides vanishes, and we get sides (E 2 E 1 ). ˆt = 0 (1.74) So, the tangential component of the electric field remains continuous across the surface and it is not affected by the charge density σ. Now, by integrating Eq (1.4), we find Φ( r 2 ) Φ( r 1 ) = r2 r 1 E. d l (1.75) where r 1 is a point just below the surface, and r 2 just above the surface. Taking the limit δ l 0, we obtain (since E is finite) Hence, the potential is continuous at surface charge layer. Φ (top) = Φ (bottom) (1.76) 13 Note that the orientation of the loop is not fixed, and so ˆt could be any tangent to the surface. 17

21 1.6 The Electrostatic Potential Energy of a Discrete Charge Distribution The potential energy of a collection of N point charges {q i ; i = 1, 2,...N} placed at positions r i is the work done to assemble the collection by bringing in charges from infinity to the positions r i. So, let us compute the work done by assembling this collection of charges one by one 14 : N = 2: W (2) = k e q 1 q 2 r 1 r 2 (1.77) N = 3: W (3) q 1 q 2 = k e r 1 r 2 + k q 1 q 3 e r 1 r 3 + k q 2 q 3 e r 2 r 3 (1.78) N = 4: W (4) q 1 q 2 = k e r 1 r 2 + k q 1 q 3 e r 1 r 3 + k q 1 q 4 e r 1 r 4 q 2 q 3 +k e r 2 r 3 + k q 2 q 4 e r 2 r 4 + k q 3 q 4 e r 3 r 4 (1.79) Thus, for N charges, we obtain or, equivalently W (N) = k e N i=2 = k e N q 1 q i r 1 r i + k e N i=1 j=i+1 q i q j r i r j N i=3 q 2 q i r 2 r i + k e N i=2 q 3 q i +... (1.80) r 3 r i W (N) = k e 2 i,j i q i q j r i r j (1.81) can be written as W (N) = 1 q i Φ i ( r i ) (1.82) 2 i where Φ i ( r i ) is the electrostatic potential produced at point r i by all the charges other than q i, that is q j Φ i ( r i ) = k e (1.83) r i r j j i 14 We bring them one at a time from infinity and with a velocity v << c, in order to avoid magnetic field effects. 18

22 1.7 The Electrostatic Potential Energy of a Continuous Distribution In analogy with the case of discrete charge distribution in equation (1.81), the potential for a continuous charge distribution, ρ( r), is given by W [ρ] = k e ρ( r )ρ( r ) 2 r r d3 r d 3 r (1.84) or, equivalently W [ρ] = 1 2 ρ( r )Φ( r )d 3 r (1.85) Although the expression above for W [ρ] is the continuum generalization of the discrete case, the difference between the two becomes clear if we consider ρ( r) = N i=1 q iδ( r r i ). Then, substituting in (1.84) gives W [ρ] = k e 2 = k e 2 i q iδ( r r i) j q jδ( r r j ) d 3 rd 3 r (1.86) r r q i q j r i r j i,j Hence, the continuum formula is not valid for a discrete distribution as it leads to a divergent result (when i = j in the sum above). Naively, it may seems that the continuum expression (1.84) also contains self-energy like effect because of the 1/ r i r j factor in the integrand. However, As r i r j 0, the contribution of this factor to the integral is finite. To see this, define new variables Then, υ = r r ; u = r + r (1.87) ρ( r)ρ( r ) W [ρ] υ 2 dυdω υ d 3 u (1.88) υ where we have wrote the volume element d 3 υ in spherical coordinates. Now, the integrand is clearly finite as υ 0 (in fact, it vanishes in this limit) provided that ρ( r) is a continuous function. The discrete limit was problematic since then as υ 0, ρ( r)ρ( r ) δ( r r i )δ( r r j ) which for r i = r j is sharply peaked and not smooth. Alternatively, the continuum result in (1.85) can also be written the form: W = 1.E( r ) Φ( r ) d 3 r (1.89) 8πk e = 1 [ ]. (E Φ) d 3 r E. Φ 8πk d 3 r e 19

23 or, equivalently, W = 1 E Φ d s + 1 E 2 d 3 r (1.90) 8πk e S= V 8πk e V We can choose the integration region V large enough that the fields on its surface is negligibly small and fall off to zero. Then, the surface integral vanishes and we are left with W = 1 8πk e V E 2 d 3 r = 1 8πk e V Φ 2 d 3 r (1.91) From which we can read off the energy density of the electric field as 15 ω = 1 8πk e E 2 (1.92) which is every where positive or zero. The above expression of the energy density is very general and also applies to regions of space where ρ( r) = 0. It gives the energy stored in the electric field E due to the work done in setting up charge distributions that are the sources of E. Remarques We had two different expressions for the potential energy: W = 1 ρ( r)φ( r)d 3 r (1.93) 2 and W = 1 E 2 d 3 r (1.94) 8πk e V If we used the first expression to define an energy density, we would get ω = 1 ρ( r)φ( r) (1.95) 2 15 In the SI and the Gaussian units, the energy density of the electric field reads ɛ 02 E 2, in SI units; ω = 1 8π E 2, in Gaussian units. 20

24 which can be both positive and negative. That makes it unreasonable to interpret it as energy density. Furthermore, ω vanishes for ρ( r) = 0 and hence it is not consistent with the definition of ω given in (1.92). In fact, ω and ω differ by.(eφ) which is the origin of surface term in W in (1.90). Thus, ω and ω give the same potential energy, however, as energy densities, they are very different. The criterion that chooses for us ω over ω is the gauge invariance, i.e. invariance of the energy density under the transformation Φ Φ + Constant (1.96) which should not affect physical quantities. Clearly, ω is not unchanged under this transformation and can not be physical, where as ω is invariant. Eq (1.92) was derived for the field created only by the charge system under consideration. If the electric field has an external component, E ext, as well, then the total electric field is E tot = E self + E ext, which implies that Φ tot = Φ self + Φ ext (1.97) Here Φ self is the self-interaction scalar potential induced by the charge distribution ρ( r) and is described by Eq (1.65) which do not include, in its righthand side, the charges which create the external field Φ ext (i.e. the external potential Φ ext is independent of the density ρ.). The total energy is [ ] 1 W tot = W self + W ext = ρ( r) 2 Φ self + Φ ext (1.98) Note that the electrostatic energy due to the self interaction is twice less than of the one due to external scalar. This is due to the self-consistent build up of the electric field as the charge of the system is being formed. We can repeat the integration by part that lead us to the last equation in (1.89), assuming that the "internal" charges ρ( r) and the "external" charges (responsible for the field Φ ext ) are well separated in space. In this case, the integration volume V may be selected so that the external charges are outside it, while the field E self at the limiting surface V is negligible. Then the surface integral vanishes again and for the energy we get V ω tot = 1 8πk e ( E 2 self + 2E self.e ext ) (1.99) We can look at this result as follows.t On one hand, we may rewrite it as ω tot = 1 ( ) E 2 8πk self + 2E self.e ext + Eext 2 1 Eext 2 (1.100) e 8πk e = 1 (E self + E ext ) 2 Constant 8πk e 21

25 Besides the last term, which does not depend on ρ( r), this expression coincides with Eq (1.92), where the E is the total electric field. On the other hand, we may consider (1.99) as describing a different potential energy of the system, sometimes called Gibbs energy, whose minimum corresponds to stable equilibrium of the system in a fixed external field. Usually this energy is introduced in thermodynamics and is one of the central notions in statistical physics. Example 1. 4: The energy of a conducting sphere We will calculate the energy of a conducting sphere of radius R carrying total charge Q uniformly distributed on its surface using three methods: (a) Using equation (1.85): W = 1 2 4π 0 Q k e Q Q 2 4πR 2 R R2 dω = k e 2πR (1.101) where we used ρ = Q/4πR 2. (b) Using equation (1.90). Since the charge reside on the surface of the sphere and the electric field inside a conducting material vanishes, we only need to consider the first integral term: W = 1 EΦdS = 1 4π k e Q k e Q 8πk e S= V 8πk e 0 R 2 R R2 dω (1.102) Q 2 = k e 2πR (c) Using eq (1.92): k e W = 8πk e Q 2 = k e 2πR 0 k 2 eq 2 R 4 4πr2 dr (1.103) We see that the three methods give the same result for the electrostatic energy. 1.8 Poisson and Laplace Equations We have seen that electrostatic can be described by the two differential equations for the electric field (1.47) and (1.4), which, according to Helmholtz theorem (see Appendix...), are sufficient to determine E uniquely given appropriate boundary conditions and 22

26 the charge density. However, in general, this is not the best approach to the solution of the problem. Often, it is better to solve for Φ by substituting E = Φ (which results from Eq (1.4)) in the differential form of Gauss s law to get 2 Φ( r) = 4πk e ρ( r) (1.104) which is the Poisson equation. In regions of space where the charge density vanishes, Poisson s equation reduces to Laplace equation: 2 Φ( r) = 0 (1.105) Most problems in electrostatics reduces to solving these equations subject to appropriate boundary conditions. It is worth mentioning that Poisson equation can be obtained by directly computing the Laplacian of the scalar potential due to a charge distribution ρ( r). This will be an opportunity to derive an important expression relating the Laplacian of 1/r potential and the Delta-distribution. So, for a charge distribution ρ( r ), we have 2 Φ( r) = k e 2 ρ( r ) V r r d3 r (1.106) ( ) 1 = k e ρ( r ) 2 d 3 r V r r ( ) 1 = k e ρ( r r) r 2 d 3 r r V where we defined r = r r. Note that r 2 (1/ r) vanishes everywhere except at the origin, i.e. at r = 0. To see what happen at r = 0, we consider the integration of r 2 (1/ r) over a sphere of radius R centered at the origin: ( ) 2 d V 1 r 3 r ( ) =. 1 d 3 r (1.107) V r ( ) = ˆn. 1 ds (1.108) r where S is the surface enclosing the volume V of the sphere. The integrand in the last integral can be expressed as ( ) ˆn. 1 = d ( ) 1 = 1 r (1.109) r d r r 2 S 23

27 which is constant on the surface S, and hence we obtain 16 This implies that V 2 ( 1 r V 2 ( 1 r Hence, we obtain the following identity ) ) d 3 r = (4πR 2 ) ( 1R ) = 4π (1.113) 2 d 3 r 4π, if the origin is in V; = 0, if the origin is not inside V. (1.114) 2 ( 1 r) = 4π δ (3) ( r) (1.115) With the use of this relation, the Laplacian of the potential field reads 2 Φ( r) = k e ρ( r r)( 4πδ( r)) d 3 r (1.116) V = 4πk e ρ( r) which is identical to the expression in (1.104) that we derived using Gauss s law. Example 1.5: Potential of Infinite conducting wire Let λ be the linear charge density of an infinitely long wire with radius R, where the charge is on distributed on its surface (as we will see in the next chapter this is a conductor). We would like to find the potential at distance r from the wire 16 The above integral was performed over a surface of a sphere. What about an arbitrary surface S enclosing a volume V?. If we consider that V contains the sphere of volume V, and we denote by V the volume in between S and S, then using Gauss s theorem (this is not gauss s law), we can write ( ). 1 d 3 r ( ) ( ) = ˆn. V r 1 ds ˆn. S r 1 ds (1.110) S r However, this integral is zero since 2 (1/ r) vanishes everywhere in V. Thus, ( ) ( ) ˆn. 1 ds = ˆn. S r 1 ds, S (1.111) S r which is equivalent to V 2 ( ) 1 r d 3 r ( ) = 2 d V 1 r 3 r = 4π, V (1.112) 24

28 by solving Laplace s equation. Because of the cylindrical symmetry of this wire, the potential depends solely on r, and so, the equation 2 φ(r) = 0, reads which has the solution 1 r d dr ( r dφ ) = 0 (1.117) dr φ(r) = A ln r + B (1.118) where A and B are constant to be determined from boundary conditions. It is important to note that one can not set φ( ) 0 as a boundary condition. This is because the wire is assumed to be infinite. Instead, we assume that there is a point at r = r 0 at which the potential is zero. So there two regions in which the potential need to be determined: φ(r) = { φ in = A in ln r + B in φ out = A out ln r + B out for r R for r R (1.119) Since the potential inside the wire is constant (you can easily convince yourself using Gauss s law), we deduce that A in = 0 φ in = B in = constant (1.120) Now we use the discontinuity of the normal component of the electric field at the surface of wire, which reads lim ɛ 0 [ dφout dr ] + dφin r=r+ɛ dr = 4πk e σ (1.121) r=r ɛ where σ = λ is the surface charge density. Using (1.120) the expression of φout 2πR in (1.119) we obtain A out = 2λ (1.122) We show in the previous section that the potential is continuous, i.e. at r = R we have φ in (R) = φ out (R) B in = 2λ ln r + B out (1.123) 25

29 So now we have only one unkown constant, B out common to both φ in (r) and φ out, and which does not contribute to the electric field. Thus, φ(r) = { φ in = 2λ ln r + B in φ out = B out for r R for r R (1.124) and E( r) = φ( r) = { 2λ r for r R 0 for r R (1.125) 1.9 Green s Theorem Let us consider the divergence theorem applied to a vector field A: [ ] [ ] d 3 r. A( r) = d 2 r A( r).ˆn V S (1.126) where S is the surface bounding the volume V, and ˆn is a unit normal vector to the element of surface ds. Now, we consider the special case where Then we have and A( r) = φ( r) Ψ( r) (1.127). A( r) = φ( r). Ψ( r) + φ( r). 2 Ψ( r) (1.128) ˆn. A( r) [ = φ( r) ˆn. Ψ( r) ] = φ( r) Ψ( r) n (1.129) where Ψ/ n is the outward normal derivative of Ψ at the surface. Substituting (1.128) and (1.129) into the divergence theorem we get [ φ( r). Ψ( r) + φ( r). Ψ( r)] 2 d 3 r = φ( r) Ψ( r) n d2 r (1.130) V The above equation is known as Green s identity. We may also starts from a vector field A( r) = Ψ( r) Φ( r), and we get [ Ψ( r). φ( r) ] + Ψ( r). 2 φ( r) d 3 r = d 2 rψ( r) φ( r) n d2 r (1.131) V By subtracting (1.131) from (1.130), yields S S 26

30 V [ φ( r). 2 Ψ( r) Ψ( r). 2 φ( r) ] d 3 r = S which is known as Green s theorem 17. [ φ( r) Ψ( r) n Ψ( r) φ( r) n ] d 2 r (1.132) 1.10 Appearance of Boundary Condition in Poisson Equation In this section we will use the results of the previous section, but we make a particular choice for the scalar function Ψ Ψ = 1 r r 1 R (1.133) where r in the function Ψ is to be regarded as the point at which we evaluate the potential. Substituting (1.133) in to the Green s theorem and changing the integration variables from r to r, we obtain V [ ( ) 1 φ( r ) 2 1 ] R R 2 φ( r ) d 3 r = or, equivalently, 4πφ( r) + 4πk e V S [ φ( r ) (1/R)) 1 n R ρ( r [ ) r r d3 r = Φ( r ) (1/R)) 1 S n R ] φ( r ) n ] φ( r ) n d 2 (1.134) r d 3 r (1.135) where we have assumed that r is inside the volume V and used the identity 2 ( 1 R) = 4πδ( r). Rearranging (1.135), we obtain φ( r) = k e V ρ( r ) r r d 3 r + 1 4π S [ 1 r r φ( r ) n ] φ( r ) 1 n r r d 2 r (1.136) The first term on the right-hand side is the familiar volume integral over the charge density, however notice that it is not all over the space. The charge outside the volume V contributes to φ( r), and it is taken into account by the integral over the surface surrounding V. Thus, if besides ρ( r ) in some region of space, also the distribution potential φ( r ) is known on the surface of this region, then one can determine the potential at any point within the volume V. There are two limiting cases that can be applied to Eq(1.136): The boundary of V at infinity (i.e. S ) At large distances the integrant in the surface integral goes faster to zero than 17 In some textbooks tit is called 2 nd Green s theorem. 27

31 1/r 2. Hence the second term on the right-hand side of (1.136) vanishes, and we obtain the known expression for the scalar potential: φ( r) = k e V=R ρ( r ) r r d3 r (1.137) No charges in V (i.e. ρ( r V) = 0) In this case the first term on the right-hand side of (1.136) becomes zero, and the scalar potential is determined by the surface integral of the potential and its normal derivative. From Eq (1.136) it seems that in order to compute φ( r), one needs to specify both φ and φ/ n on the boundary of S. However, as we shall see, it is possible to find φ( r) from the charge density in V and from either φ( r ) OR Φ( r ) n on V. Hence, one may require one of the following boundary conditions 18 : 1. The Dirichlet boundary condition, where φ S is specified ( φ/ n S is given by the solution); 2. The Neumann boundary condition φ/ n S is specified (φ VS is given by the solution). Various combinations are also possible, such as Dirichlet conditions on part of surface S and Neumann conditions on on the remainder, known as mixed boundary conditions The Uniqueness of the Solution in Dirichlet and Neumann B.Cs Here we will show that it is sufficient to know either Φ S or Φ n S in order to obtain a unique solution to Poisson equation in a volume V bounded by S. Let Φ 1 and Φ 2 be two solutions to the Poisson equation in V satisfying the same boundary conditions on S. We define µ = Φ 2 Φ 1 (1.138) This function satisfies Laplace equation µ( r) = 0, r V and is subject either to or µ S = 0 (Dirichlet) (1.139) µ n S = 0 (Neumann) (1.140) 18 Specifying both Φ V and φ/ n V is called Cauchy boundary conditions. We will show that for the Poisson equation, no solution exist for Cauchy boundary conditions. 28

32 If we put φ = Ψ = µ in the first Green s identity, then we get ( µ 2 µ + µ. µ ) d 3 r = µ µ ds (1.141) n V Now, since µ satisfies Laplace equation, the first term in the bracket on the LHS side vanishes. Also, from the B.C from either the Neumann or the Dirichlet condition, the RHS vanishes, Hence, Eq (1.141) reads µ 2 d 3 x = 0 (1.142) Since µ 2 0, it follows that V S µ = 0, r V (1.143) Therefore, µ( r) is a constant in V and so the two solutions Φ 1 and Φ 2 are the equivalent up to a constant. For the Dirichlet conditions, this constant is zero since the two functions Φ 1 and Φ 2 are the same on the boundary. However, for the Neumann conditions, the value of the constant is not fixed, and the solution is unique up to a constant. But this is actually the gauge ambiguity inherent in the definition of Φ. Hence, in this case too, the solution is uniquely specified. The above proof also shows that the uniqueness of the solution for mixed boundary conditions. It also implies that no solution exist for the Cauchy B.Cs. However if Cauchy B.Cs are given on part of S, then that can be sufficient to give a unique solution for the scalar potential Solution of the Electrostatic Boundary-Value Problem with Green Functions It is evident that the expression of Φ( r) given in (1.136), which involves surface integrals of both the potential and its normal derivative, is not very effective way to solve an electrostatic boundary-value problem. If we had made a better choice of Ψ( r ) at the outset, we could have come up with a better result. In the earlier attempt we used Ψ = 1/ r r in the Green s theorem, but all we really needed was is that 2 Ψ = 4πδ( r r ). So, we could also have used a more general function with such that G( r, r ) = 1 r r + F ( r, r ) (1.144) 2 F ( r, r ) = 0 (1.145) 29

33 2 G( r, r ) = 4πδ (3) ( r r ) (1.146) The function G( r, r ) that satisfies the above equation is called the Green function. Thus, the Green function solves the Poisson equation for a unit charge (in Gaussian unit-system) at point r within a volume V, and so requires boundary conditions specified on S. These B.Cs should be simple and generic and independent of the detailed form of the B.Cs on Φ. Note that even though the particular choice ψ( r, r ) = 1/ r r is a solution of Poisson s equation (1.146), it does not satisfy Dirichlet or Neumann boundary conditions, except when the surface lies at infinity. For G( r, r ), the boundary conditions can be imposed via the function F. Now by setting φ = Φ (i.e. the scalar potential field), and Ψ = G( r, r ) in the expression of the Green s theorem, we get [ ] Φ( r) = k e V ρ( r )G( r, r )d 3 r + 1 4π S G( r, r ) Φ n Φ( r ) G( r, r ) n ds (1.147) Furthermore, we can use the freedom of choosing the function F in G to eliminate either the Φ S or Φ n S dependent terms from the surface integral. So there are two cases: Dirichlet case: For the Dirichlet B.Cs on Φ we demand that This implies that G D ( r, r ) = 0, r S (1.148) Φ( r) = k e V ρ( r )G D ( r, r )d 3 r 1 4π S Φ( r ) G D( r, r ) n d 2 r (1.149) Then we have, in principle, the information we need to complete the integration and so find Φ( r) in which case Eq (1.149) is an integral solution for the scalar potential as opposed to an integral equation for the potential. The pressing question now is: does G D exist? In other words, is it possible to find the Dirichlet Green s function G D which satisfies Poisson equation and the condition (1.148)? The answer is that this function does exist; moreover, it is unique. We know this because, as we will see later when we discuss the method of images, G D is just the scalar potential at r given a unit point charge 30

34 at r inside of a cavity with conducting walls coincident with S and held at zero potential. Note that the Dirichlet Green s function is strongly dependent function on S but it is not dependent of any other properties of the system. It means that we can solve any Dirichlet problem for a given geometry if we can solve the point charge with grounded conducting surface problem for the same geometry 19. An important property of the Dirichlet Green s function is that it is invariant under the interchange of r and r, i.e. G D ( r, r ) = G D ( r, r). To show this, let φ( R) = G D ( r, rr) and Ψ( R) = G D ( r, R). Then inserting these functions into Green s theorem (with Y as an integration variable) gives [ G D ( r, R) 2 G D ( r, R) G D ( r, Y ) 2 G D ( r, R) ] d 3 R = (1.150) V V [ G D ( r, R) G D( r, R) n G D ( r, R) G D( r, R) n By making use of the properties of the Dirichlet Green s function, we obtain G D ( r, r ) is symmetric under the exchange of r and r. Neumann case: Naively, one may set G N / n r S= 0, similar to what we did for the Dirichlet case. But, Gauss law gives V 2 G N ( r, r ) d 3 r = S G N ( r, r ).ˆn ds = Hence, such boundary condition on G does not exist. The next simplest possibility for a boundary condition on G N is S ] d 3 R G N n ds = 4π (1.151) G N ( r, r ) n r S = 4π (1.152) A S where A S is the area of the surface S which bounds the volume V. Given such a function, we use it in Green s Theorem and we obtain Φ( r) =< Φ > S + V G N( r, r )ρ( r )d 3 r + 1 4π S Φ n G N ( r, r )d 3 r (1.153) where < Φ > S is the average of the potential over the surface S: < Φ > S = 1 Φ( r )d 2 r (1.154) A S S One can understand the necessity of the presence of this term from the fact that Neumann boundary condition problem can only be solved up to an arbitrary constant as we have shown before. 19 In the sense that we can reduce the solution to a quadrature, i.e. to an integral. 31

35 2.1 The Electric Field Screening 2 The Conductors An insulator, such as glass or paper is a material in which electrons are attached to some particular atoms and can not move freely. On the other hand, inside a conductor, electrons are free to move around in response to electric field. However, in electrostatics there should be no such motion 20, so that everywhere inside the conductor the electric field should vanish: E = 0 (2.1) Another way to see this is as follows. By applying a constant external electric field E ext (see figure below) to an isolated conductor the positive charges will be driven toward the direction of E ext, and the negative ones to the opposite direction 21, thereby inducing an electric field, E ind, which points in the opposite direction of E ext. Since charges are mobile, they will continue to move until E ind completely cancels E ext inside the conductor. Then the electric field inside the conductor vanishes and an electrostatic equilibrium will be reached 22. However, outside the conductor, the electric field is non-vanishing. The vanishing of E everywhere inside the conductor have the following implications: any net charge must reside on the surface of the conductor: This is because.e = 0 inside the conductor, and hence the charge density is zero. The electric field just outside the conductor is normal: We have seen (see Eq (1.5) ) that the tangential component of electric field, E t, remains continuous across the surface. However, since E (inside) = 0, we conclude that Then Eq (1.72) yields 23 E t = 0 (2.2) 20 The special case of DC current will be discussed in another chapter in these notes. 21 In practice it is only the negative charges (electrons) that do the moving, but after they depart the side they were in before applying an external electric field, is now left with a net positive charge (the nuclei). 22 Actually, the whole process is practically instantaneous. 23 In SI and Gaussian unit system, the strength of E n just outside a conductor is 4πσ, in Gaussian units; E n = σ ɛ 0, in MKS units. 32

36 E (just outside) = 4πk e σ ˆn (2.3) where ˆn is the unit vector normal to the surface of the conductor at point r. Note that the above result holds for a conductor of arbitrary shape. All points on a conductor are at the same potential. If A and B are two points on the surface of a conductor, then we have V B V A = B A E.dl = 0 (2.4) where we have used the fact that E.dl = E t dl = 0 which is a consequence of Eq (2.2). Thus, the surface of a conductor in electrostatic equilibrium is an equipotential surface. Now, let us suppose that there is some cavity in the conductor, and within it there is some charge q. In this case the field inside the cavity will not be zero. Moreover, no external fields penetrate the conductor, they are cancelled at the outer surface by the induced charges. Similarly, the fields due to charges within the cavity is killed off, for all the exterior points, by the induced charge on the inner surface (see the figure below). However, the compensating charge left over on the outer surface of the conductor effectively communicates the presence of q to the outside world. Incidentally, the total charge induced on the cavity surface is equal in magnitude and opposite in sign to the charge inside. If we surround the cavity with a Gaussian surface, then E. d s = 0 ( since the electric field vanishes inside the conductor), and so the net enclosed charge must be zero. Hence, the induced charge is q ind = q. 2.2 Surface Charge and Force on a Conductor Consider a small patch of a charge on a conducting surface, as show below. The total electric field anywhere outside the surface can be written as E = E patch + E (2.5) where E patch is the electric field due to charge on the patch, and E is the electric field due to all other charges. Since by Newton s third law, the patch can not exert a force on itself, the force on the patch must come solely from E. Assuming the patch to be a flat surface, from Gauss s law, the electric field created by the patch is +2πk e σ, z > 0; E patch = 2πk e σ, z < 0. 33

37 By superposition principle, the electric field above the conducting surface is given by similarly, below the conducting surface,we have E above = 2πk e σẑ + E (2.6) E below = 2πk e σẑ + E (2.7) Note that the electric field E is continuous across the boundary. This is due to the fact that if the patch were removed, the field in the remaining hole exhibits no discontinuity. Combining (2.6) and (2.7), we find 1 2 (E above + E below ) E avg (2.8) However, in the case of a conductor E below = 0 and E above = 4πk e σẑ, and hence we obtain Then the force acting on the patch is E avg = 2πk e σẑ (2.9) F patch = qe avg = 2πk e σ 2 Aẑ (2.10) where A is the area of the patch. Note that irrespective of the sign of the charge density σ, the force tends to pull the patch into the field. Using the result above, we may define the electrostatic pressure on the patch as 24 P = F A = E2 8πk e (2.11) where here E denotes the magnitude of the electric field just above the patch. This pressure is transmitted to the surface of the conductor via the electric field. 24 In SI and Gaussian unit system, the electrostatic pressure on a conductor is P = 2πσ 2 = E2 8π, in Gaussian units; σ 2 2ɛ 0 = 1 2 ɛ 0E 2, in MKS units. 34

38 2.3 Capacitance Consider a single conductor in free space. According to (2.1), all its volume should have a constant electrostatic potential Φ cond. Another quantity that is constant is the total charge Q = ρd 3 r = σds (2.12) V where the latter integral is over the whole surface of the conductor. In general, what is the relation between Q and Φ? At Q = 0, there is no electric field in the system, and so it is natural to choose the arbitrary constant in the electrostatic potential to have Φ = 0. Then, if the conductor is charged with a finite Q, according to the Coulomb law, the electric field in any point is proportional to Q. Hence, the electrostatic potential everywhere, including its value Φ cond on the conductor, is also proportional to the charge Q: S Φ cond = pq (2.13) The proportionality constant p, which depends on the conductor size and shape, not on Q, is called the reciprocal capacitance 25. Usually, Eq (??) is presented as Q = CΦ cond ; C 1 p (2.14) where C is called self-capacitance (or just capacitance). The electrostatic energy of a single conductor is U = 1 ρφ cond d 3 r = Φ condq (2.15) V where Φ cond was taken out of the integral sign since it is constant in all the volume of the conductor. Moreover, using the relation (2.14). we can rewrite the above equation in either of two more forms U = Q2 2C = C 2 Φ2 cond (2.16) 25 Some times it is called the coefficient of potential 35

39 Example 2.1: Capacitance of conducting sphere Let us consider a conducting sphere of charge Q and radius R. We know that the electric field inside the conducting sphere vanishes, while outside is E(r R) = k e Q/r 2 Q, and so the potential distribution in space is Φ(r) = k e. Thus, on the r surface of the sphere we have φ sphere = k e Q R (2.17) Using Eq (2.14), we find that the self-capacitance of a conducting sphere is C sphere = 1 2k e D (2.18) where D = 2R is the diameter of the sphere. In SI system of units we get C (SI) sphere = 2πɛ 0D (2.19) with the Farad (abreviated as F) as the unit of capacitance. To get the feeling of how big is a Farad, the self.capacitance of Earth 26 is approximately F. So 1 F is a huge considered a huge capacitance. Although the expression (2.19) was derived for a spherical conductor, it can be used parametrically as an estimate of a conductor of arbitrary shape as C (SI) 2πɛ 0 d (2.20) where d represents the scale of the linear size of the conductor 27. Now for a system of two conductors, one constant C is insufficient to describe such system. indeed, here we have two, generally different conductor potentials, Φ 1 and Φ 2, which depend on both conductor charges Q 1 and Q 2. Using the same arguments as for the one conductor case, we may conclude that the dependence is always linear: Φ 1 = p 11 Q 1 + p 12 Q 2, (2.21) Φ 2 = p 21 Q 1 + p 22 Q 2, Thus, the system of two conductors is, in general, described by the so called reciprocal capacitance matrix 28 : ( ) p11 p ˆP = 12 (2.22) p 21 p It can be shown that the matrix ˆP is always symmetric. 36

40 Substituting (10.52) into the definition of the electrostatic energy U = 1/2 2 i=1 Q iφ i, gives U = 1 2 p 11Q (p 12 + p 21 ) p 11 Q 1 Q p 22Q 2 2 (2.23) The term in the middle in the RHS of the above equation describes the electrostatic coupling of the conductors. Systems with p 12 = p 21 << p 11, p 22 are called weakly coupled systems. For n conductors, (10.52) can be generalized to: Φ i = n p ij Q j (2.24) j=1 Its inversion yields the charges Q i as a linear combinations of the potentials Φ j : Q i = n C ij Φ j (2.25) j=1 The coefficients C ij are called coefficients of capacitance. Thus, we see that for n 2 with arbitrary charges, the system properties can not be reduced to just one coefficient. Let us consider three particular cases when such reduction is possible: Pair of conductors with Q 1 = Q 2 = Q: So the system as a whole is electrically neutral. In this case the most important function of Q is the difference of conductor potentials: V = Φ 1 Φ 2, frequently called voltage, which can be written as where the coefficient C m is given by C m = 1 p 11 + p 22 p 12 p 21 = V = Q C m (2.26) 1 p 11 + p 22 2p 12 (2.27) usually called the mutual capacitance between the conductors. The electrostatic energy of this system,which according to (2.23), is given by or, equivalenly, U = 1 2 (p 11 + p 22 2p 12 ) Q 2 (2.28) U = Q2 2C m = C m 2 V 2 (2.29) As an example, we will calculate the mutual capacitance for a system of two parallel metal surfaces of area A held at a distance d apart.if we put a charge 37

41 +Q on the upper plate and charge Q on the lower one (see the figure), then they attract each other and sit entirely on the sides of the narrow gap. We assume that the separation distance d is much smaller than the linear size of the plate. Let us choose the Gauss surface to be a pillbox whose area is a small part of the gap but much larger than d 2, with one of the plane of the pillbox inside a conductor and another one inside the gap. Then applying Gauss s law, we find that the electric field within the gap is E = σ ɛ 0 (2.30) where σ is the surface charge density on the palate. Integrating this field across the gap, gives Hence, we obtain V = σd ɛ 0 = Q Aɛ 0 d (2.31) C m = ɛ 0A d (2.32) Few comments about the expression (2.32) are in order: (a) It is valid even if the gap is not quite planar, as long as the curling is on a scale larger than the separation distance d. (b) It is only valid if A >> d 2, because its derivation ignores the electric field deviations from uniformity 29 at distances of order d near the gap edges. (c) The condition A >> d 2 assures that C m is much larger than the selfcapacitance of each of the conductors, which is of great practical importance. For example, a 3 nm layer of high-quality Aluminium oxide 30, Al 2 O 3,with area 0.1 m 2 provides C m 1 m F, larger than the self-capacitance of the whole planet Earth. In the previous example of the two plates, the electrostatic coupling between the two conductors is evidently strong. As an example of weakly coupled system, consider two sphere of same radius R, separated by a distance d much larger than R. In this case the diagonal elements of the reciprocal matrix ˆP may be approximately found from Eq (??) by neglecting the coupling terms altogether: p 11 = p 22 k e R (2.33) 29 Such deviations from uniformity leads to what is called fringe fields which results in additional stray capacitance C m ɛ 0 A. 30 Which may provide a nearly perfect electric insulation between two thin metallic films. 38

42 To see why neglecting p 12 and p 22 is justifiable in the limit d >> R, suppose had just one sphere, say sphere 1, then electric potential at distance d from its center will be given by Φ 1 = k e Q 1 /d. If we move into this point a small sphere (R << d) without its own charge, we may expect that the value its potential would not be too far from this result, so that Φ 2 KQ 1 /d. Comparing this result with (10.52) (and taking Q 2 = 0), we get This implies that the mutual capacitance is which in SI system of units reads p 12 = p 21 k e d << p 11, p 22 (2.34) C m 1 p 11 + p 22 (2.35) C m 2πɛ 0 R (2.36) Thus, in this case C m does not depend on the distance between the spheres, i.e. does not involve their electrostatic coupling. Pair of conductors with Q 1 0 and Q 2 = 0: In this case Eq (10.52) yields Φ 1 = p 11 Q. Now, if we define C j = 1 p jj (2.37) as the partial capacitance of conductor number j, we see that it has nothing to do with the mutual capacitance C m defined in Eq (2.27). For example, in the case of the two conducting spheres discussed above we have C 1 = C 2 4πɛ 0 R 2C m. Pair of conductors with Q 1 = Q and Φ 2 = 0: From Eq (10.52) we obtain Q 2 = p 21 p 2 2 Q 1 (2.38) Substituting this relation into the first equation in (10.52), yields ( ) Φ 1 = p 11 p2 12 Q 1 (2.39) p 22 Thus, if we treat the expression in the parentheses as the reciprocal, then we can define ( ) 1 C eff 1 = p 11 p2 12 (2.40) p 22 as the effective capacitance of the first conductor. different from C m and C 1 = 1/p 11. Note that it is generally 39

43 In summary, the actual capacitance of a conductor in a two-conductor system very much depend on what exactly is being done with the second conductor when the first one is charged. This is also true for multiconductor systems. In that case even the mutual capacitance between any two conductors may depend on the electrostatic conditions of the other components of the system. 40

44 3 Boundary-Value Problem in Electrostatic In this section, our goal is to develop some general techniques to solve the Poisson or Laplace equations subject to appropriate boundary conditions. 3.1 The Method of Images Th idea of this method is that the solution for the potential in a finite volume V with specific charge density and potentials on its surface V can be found by replacing the B.Cs on V by a different charge elsewhere to be chosen such that the new system of charges satisfy the same B.Cs. Then, according to the uniqueness theorem of the solution to the Poisson equation with given B.Cs, the solution of this new system of charges, if it exists, is the same solution to the original problem. Note that the initially undetermined charges which will replace the B.Cs, called image charges must be external to V otherwise the will not satisfy the Laplace equation (or Poisson equation) in V. Thus, if one has found the image charges, then the potential is simply given by Φ( r) = k e V ρ original ( r ) r r d 3 r + k e V V ρ image ( r ) r r d 3 r (3.1) where ρ original ( r) is the original charge density and ρ image ( r) is the charge density of the image charges. This method for finding the potential of a system is useful for charge distribution in the presence of grounded conductor of simple geometry. It can also be used to find the Green s function for these same geometries. Below, we present few examples. 1. Point charge above a conducting plane: The simple example of the method of images is that of a point of a point charge q at a distance d from an infinite conducting plane which is held at zero potential. Thus, the problem of finding the potential at some point above the conducting plane, amounts to solving Laplace equation with the Dirichlet boundary condition Φ = 0 on the conducting plane. For convenience, let the plane be at z = 0 and the charge q above it on the positive z-axis. We know that near a grounded conducting body the charge q attracts counter charges such that the potential and the electric field vanishes inside. That is, the potential due to the charge induced on the plane must precisely cancel the potential of the point charge for z < 0. 41

45 Figure 3. Point charge q above a conducting plane held at zero potential. Thus, we must determine the image charge located in the z < 0 such that the total potential satisfy the boundary condition Φ(z = 0) = 0. Denoting the potential due to the induced charges on the plane by Φ ind, we must have q Φ ind + k e = 0; z = 0 (3.2) r dẑ The simplest way for this to be satisfied is that the the potential due to the induced charges on the plane can be represented by a point charge q at the same position as q. However, as we mentioned earlier, the image charge must be external to the region V ( i.e. z 0). Moreover, the total potential trivially vanishes everywhere in space, which is clearly inconsistent in the region z > 0. The other way is that Φ ind can be represented by a point charge q placed at the mirror image position " dẑ". This is due to the symmetry of the problem about the plane z = 0. Thus, the potential at a point above the conducting plane can be represented by a superposition of Φ ind with the point charge potential: or, equivalently [ ] q Φ( r) = k e r dẑ q r + dẑ (3.3) 42

46 [ q Φ( r) = k e x2 +y 2 +(z d) 2 ] q x2 +y 2 +(z+d) 2 (3.4) which is the potential of a dipole. In addition, it satisfies Poisson s equation for a point charge q at (0, 0, d), in the region z > 0. Of course the potential in (3.4) can not be applied to the z < 0, since the grounded plate shields this region where z < 0 from the point charge q so that Φ = 0 in this region. In real system there is some surface charge density σ(x, y) on the conducting plane, which gives rise to Φ ind. It can be calculated using the expression of the normal component of electric field at the surface of the plane (see (2.3)): Substituting (3.4) into (3.5), yields σ = 1 4πk e E n = 1 4πk e Φ z z=0 (3.5) σ = q d (3.6) 2π (l 2 + d 2 ) 3/2 where l 2 = (x 2 + y 2 ). The total charge induced on the plate is Q ind = S σ ds = 2π 0 σ(l)l dl = qd 0 Making a change of variable l ξ = l 2 /d 2, gives l dl (3.7) (l 2 + d 2 3/2 ) Q ind = q 2 0 dξ (1+ξ) 3/2 = q (3.8) The force on the point charge q due to the induced charge on the plane is given by Coulomb s law F q = k e qq (2d) 2 ẑ = k e q 2 4d 2 ẑ (3.9) 43

47 The electrostatic potential at the location of the charge q, after ignoring the infinite potential created by the charge itself so that the remaining potential is that due to the image charge is Φ image (dẑ) = k e q 2d (3.10) It is tempting to conclude that the potential energy of the point charge q-tosurface interaction is W = qφ image = k e q 2 /2d. However, that would be incorrect! The reason is that Φ image is not independent of the potential charge q, but it is induced by this charge. Moreover, the electric field is finite in region z < 0 for the q-image charge system whereas it is zero in the q-conducting plane system. Thus, one expect that the correct potential energy is half times the potential energy of the charge-image charge system. This can be seen as follows. For the q q system we have W q q = 1 8πk e all space E 2 q q d3 r = 1 8πk e [ Eq q 2 d3 r + z>0 z<0 E 2 q q d3 r ] (3.11) However, in the charge-image charge system, the fields on either side of the plane are mirror images of one another, so Eq q 2 (x, y, z) = E2 q q (x, y, z), and we get 1 W q q = 2. 8πk e z>0 E 2 q q d3 r (3.12) For the charge- conducting plane, i.e. the real physical system, we have E q q, z > 0 E(x, y, z) = 0, z < 0 and so, (3.13) W q plane = 1 Eq q 2 8πk d3 r = 1 e z>0 2 W q q (3.14) Another way by which we can obtain the above result in (3.14) is to calculate the work required to gradually move the charge q from infinity towards the plane along the z-axis until it reaches the position (0, 0, d), i.e. or, equivalently, W q plane = d F q (z) ẑdz = k e d q 2 2 dz (3.15) (2z) 44

48 W q plane = k e q 2 4d (3.16) It is true that the induced charge is moving over the conductor, but this costs nothing in terms of energy since the whole conductor is at zero potential. But in the other way, when q is moved from infinity we do work only on q. By contrast, if we simultaneously move the q q system (with no conductor(, we do work on both of them, and the total work done is twice as great. 2. Point charge outside a grounded conducting sphere: Consider a point charge q at a position Y = Y ˆn outside a grounded (i.e Φ = 0) conducting sphere of radius a. Thus, the potential due to the induced charges on the sphere must precisely cancel that of the point charge q for r = a. This can be achieved by representing the potential due to the induced charges by an image charge q placed at Y = Y ˆn 31 (see figure.3) such that the total potential at a point P outside the sphere with vector position r = r ˆn is given by 32 Figure 4. Point charge outside a grounded conducting sphere. [ ] [ q Φ( r) = k e r Y + q r Y = k e q rˆn Y ˆn + q rˆn Y ˆn ] ; r a (3.17) 31 From the symmetry of the problem, q must be on the same axis as the charge q 32 As we noted earlier, the trivial possibility of having an image charge q = q at Y ˆn is not acceptable since then q will be within the volume V and cancels the real charge. Also, since we want to solve the boundary problem at r a we can place the image charge only inside the sphere. 45

49 and which must satisfy the boundary condition Φ(a) = 0. This requires that q a 2 + Y 2 2aY cos α = q a 2 + Y 2 2aY cos α (3.18) which should be satisfied for any value of α, the angle between ˆn and ˆn. Squaring both sides of (3.18) yields q 2 Y = q 2 Y, (3.19) q 2 (a 2 + Y 2 ) = q 2 (a 2 + Y 2 ), Dividing the second equation by the first one yields or, equivalently a 2 Y 2 a 2 Y 2 + Y Y 2 Y Y 2 = 0 (3.20) Omitting the solution (Y, q ) = (Y, q), we find Y = a2, Y (Y Y ) ( a 2 + Y Y ) = 0 (3.21) (3.22) q = a q, Y Hence, from the uniqueness theorem, we conclude that the potential of this system is Φ( r) = k e [ q r ˆn Y ˆn a Y ] q, r a (3.23) r ˆn a2 Y ˆn For r inside the sphere (in which the method of images can not be used), the potential can determined by solving Laplace equation 2 Φ = 0 with the boundary condition Φ(a) = 0, independent of the presence of the charge q outside. Hence, the solution is the same as for q = 0, and so Φ( r) = 0 inside the sphere. The surface charge distribution that gives rise to the above potential is given by Using the expression of Φ in (3.23) yields σ = 1 4πk e E. ˆn = 1 4πk e ˆn. Φ (3.24) 46

50 σ = q ( q ) (1 a 2 /Y 2 ) 4πa 2 Y (3.25) [1+ a2 Y 2 2 a Y cos α ] 3/2 Then, the total charge induced on the surface of the sphere is Q surface = σd 2 r = a Y q = q (3.26) sphere The point charge q experience an attractive force F q = k e qq Y Y 2ŷ = k eq 2 a 2 (a/y ) 3 (1 a2 which at large distances, i.e. a/y 0, goes as (a/y ) 3. Y 2 ) 2 ŷ (3.27) 3. Point charge outside a charged insulated conducting sphere: Suppose now, that a point charge q at a position Y = Y ˆn outside an insulated conducting sphere which carries a total charge Q. The potential at r is independent of how the charge Q is transferred to the sphere. Thus, we can split the charge Q = q + (Q q ) and use the superposition principle to calculate the potential. We start with the grounded conductor of the previous example which carries charge q distributed such that Φ( a) = 0. Now, we insulate the sphere and add a charge (Q q ) to it at its centre. Since the surface was at zero potential, this additional charge distributes itself uniformly over the surface of the sphere. Outside the sphere, this is equivalent to a point charge (Q q ) at the centre. Therefore, the image charges are q at Y and (Q q ) at the centre (i.e. the origin). So the potential at a point r is given by [ ] q Φ( r) = k e r Y + q r Y + (Q q ) ; r a (3.28) r Using q = a/y q and Y = a 2 /Y, we obtain Φ( r) = k e [ q r Y ] aq Y r a2 + (Q+aq/Y ) Y 2 Y r ; r a (3.29) 47

51 The force felt by the charge q is [ ] qq F q = k e Y Y ŷ + q (Q q ) ŷ 2 Y 2 (3.30) or, equivalently [ ] q F q = k e Y Q a3 q(2y 2 a 2 ) ŷ 2 Y (Y 2 a 2 ) 2 (3.31) We note that : (a) For Y 0, F q qq/y 2 (as expected). (b) For Q 0 and q > 0, the force on q is always attractive. (c) For Q > 0 and q > 0, the force is repulsive at large distances. However, at small distances y a, the second term in (3.31) dominates and the force becomes attractive. When very close to the surface of the sphere, it experiences a very large attractive force even when both carry charges of the same sign. This is the reason why elementary charges do not leave the surface of charged conductor. As soon as a point charge is moved away from the surface of the conductor, an attractive force is developed due to the induced charges on the sphere. 3.2 The Green s Function Method for the Sphere In this section we would like to calculate potential distribution outside an empty spherical cavity of radius a (i.e. ρ( r) = 0, r V), given the potential field on its surface and some charge distribution outside. That is: 2 Φ( r) = 4πk e ρ( r), r > a (3.32) Φ( r ) S = Φ(a, θ, φ ) Φ(a, Ω ), This defines a Dirichlet problem with solution given by (see (1.149)) Φ( r) = k e ρ( r )G D ( r, r )d 3 r 1 Φ( r ) G D ds (3.33) 4π n where V 2 G D ( r, r ) = 4πδ( r r ), G D ( r, r ) r =a = 0, S (3.34) 48

52 Clearly G D ( r, r ) is the solution for the potential at r due to a unit charge at r in the presence of a grounded conducting sphere. This corresponds to the second example that we solved earlier using the method of images. Thus, G D ( r, r ) is given by G D ( r, r ) = 1 r r a r r a2 r 2 r which in spherical coordinates reads [ ] G D ( r, r 1 ) = (r 2 + r 2 2rr cos α) 1 1/2 (r 2 r 2 + a 4 2a 2 rr cos α) 1/2 (3.35) (3.36) We see that G D ( r, r ) is symmetric under the exchange of r and r, and it vanishes at the the surface of the sphere. The derivative of G D along the direction ˆn = ˆr 33 is G D n S = G D (r 2 a 2 ) r r =a = (3.37) a [r 2 + a 2 2ar cos α] 3/2 For ρ( r) = 0 inside the sphere, the solution to Laplace equation is Φ( r) = 1 a(r 2 a 2 ) 4π Φ(a, Ω )dω (3.38) (r 2 +a 2 2ar cos α) 3/2 Here cos α = ˆn.ˆn depends on the angles θ, θ, φ and φ and it is given by cos α = (sin θ cos φ, sin θ sin φ, cos θ).(sin θ cos φ, sin θ sin φ, cos θ ) (3.39) = sin θ sin θ (cos φ cos φ + sin φ sin φ ) + cos θ cos θ or cos α = cos θ cos θ + sin θ sin θ cos (φ φ ) (3.40) Example 3.1: Two hemisphere at constant potentials Suppose that the potential at the surface of a sphere of radius a is given by V 0, 0 θ π/2 Φ(a, θ, φ ) = V 0, π/2 θ π (3.41) 33 Since the volume V is external region of the sphere, the normal vector ˆn points the centre of the sphere, and hence ˆn = ˆr. 49

53 where V 0 is a constant. Since in this example the potential on the surface is independent of the angle φ, we may set it equal to zero. Defining ɛ = a/r, we have Φ(ɛ, θ) = V [ 2π π/2 0 4π ɛ(1 ɛ2 ) dφ 0 0 = V 0 4π ɛ(1 ɛ 2 [ ) 2π 1 dφ (1 + ɛ 2 ) 3/2 0 0 sin θ dθ π (1 + ɛ 2 2ɛ cos α) 3/2 π/2 d cos θ (1 2ɛ 1+ɛ 2 cos α) 3/2 0 1 ] sin θ dθ (1 + ɛ 2 2ɛ cos α) 3/2 ] d cos θ (1 2ɛ cos α) 1+ɛ 3/2 2 By making the change of variable θ π θ and φ φ +π in the second integral, we obtain Φ(ɛ, θ) = V 0 ɛ(1 ɛ 2 ) 2π 4π (1 + ɛ 2 ) 3/2 0 dφ 1 0 [ d cos θ (1 2ɛ ] 1 + ɛ cos 2 α) 3/2 (ɛ ɛ) (3.42) The integral is difficult to perform analytically in the general case. However, for θ = 0, i.e. the point P is along the z-axis, the integration is straightforward and the result is [ ] [ ] Φ(z) := Φ(ɛ, 0) = V ɛ2 1+ɛ = V z2 a 2 z z 2 +a 2 (3.43) An alternative approach, valid for a/r << 1, is to expand the integrand in (3.42) in power series of ɛ and then complete the integration term by term, and obtain 34 Φ(ɛ, θ) = 3V [ 0 2 ɛ2 cos θ 7 ( 24 ɛ2 5 cos 3 θ 3 cos θ ) ] +... (3.44) Note that the expression of Φ(ɛ, 0) in (3.43) when expanded up to order ɛ 4 agrees with the above equation for θ = 0. 50

54 4 Solving the Laplace Equation 4.1 The Orthogonal Functions Let ξ be a variable over an interval [a, b], and consider a set of complex or real function U n (ξ) for, say, n = 1, 2,.. over this interval. These functions are said to form an orthogonal set over the interval [a, b] if: b a dξu n(ξ)u m (ξ) = 0; n m (4.1) These functions can always be normalized to unity so that b Furthermore, the set is said to be complete if a dξu n(ξ)u m (ξ) = δ nm (4.2) Un(ξ )U m (ξ) = δ(ξ ξ) (4.3) n=1 If such a set of functions is given, then any function f(ξ) over the interval [a, b] can be expanded in the basis of this orthogonal and complete set of functions 35 : f(ξ) = a n U n (ξ) (4.4) The ortho-normality property of the functions U n (ξ) allows us to write n=1 a n = b a dξu n(ξ)f(ξ) (4.5) and the completeness condition in (4.3) insures the self-consistency of the expansion: f(ξ) = ( b n=1 a ) dξ Un(ξ )f(ξ ) U n (ξ) = b a dξ f(ξ )δ(ξ ξ) (4.6) The above can be generalized in a straight forward way to more than one variable, for example: 35 This is much in the same way as any vector can be expanded in orthogonal basis. 51

55 f(ξ, η) = a nµ U n (ξ)v µ (η) (4.7) n=1 µ=1 where a nµ = b a dξ d c dηu n(ξ)v µ f(ξ, η) (4.8) with the set of orthogonal functions V µ (η) are defined on the interval [c, d]. A common example of the above expansion is in terms of Fourier series. in this case the orthogonal functions are U n (x) = 1 a e i2πn x a (4.9) with n = 0, ±1, ±2,.. over the interval a/2 x a/2. These functions are solutions to the partial differential equation 2 U(x) x 2 + k 2 U(x) = 0 (4.10) for k 2 = (2πn/a) 2. The orthogonality and completeness relations read 1 a/2 )x/a a a/2 dxei2π(m m = δ mm, 1 a m= )/a dxei2πm(x x = δ(x x ), (4.11) which serves as a very useful representation of the Kronecker and the Dirac-delta distributions. The expansion of a function f(x) is given by where f(x) = 1 q + A n = 1 a a/2 a/2 n= A n e i2πnx/a (4.12) dxf(x)e i2πnx/a (4.13) 4.2 Separation of Variables: Laplace Equation in Rectangular Coordinates Laplace equation in rectangular coordinates reads 2 Φ x Φ y Φ z 2 = 0 (4.14) 52

56 We wish to look for the separation of variable solution 36 : Φ(x, y, z) = X (x)y(y)z(z) (4.15) With this factorization, Laplace equation takes the form (after dividing the entire equation by Φ), 1 2 X X x Y 2 Y y Z 2 Z z = 0 (4.16) 2 Each term on the left hand side is an function of an independent variable and in principle could be varied without affecting the other two terms. So, for the above equation to be true, each term should be separately constant, i.e. 1 2 X = α 2 X x X = e ±iαx, 1 2 Y = β 2 Y y 2 Y = e ±iβy, (4.17) 1 2 Z = γ 2 Z = e ±γz, Z z 2 where we have chosen α and β to real, and γ 2 = (α 2 + β 2 ). Since the factors X, Y and Z solve Laplace equation for any value of α and β, the general solution for Φ can be constructed as a product of these factors summed over all these values: Φ(x, y, z) = + ] + α= β= eiαx iβy e [Ã(+) α,β e α2 +β 2z + Ã( ) α,β e α 2 +β 2 z (4.18) where Ã(±) α,β are arbitrary constants to be fixed by the boundary conditions of the problem. The summation could be either discrete or continuous, depending on the nature of their indices. One can see that the solution has the form of an expansion in orthogonal functions (Fourier series as we shall see later). Now we would like to consider Φ to be the potential inside a rectangular box bounded by the three planes perpendicular to the coordinate axes at x = a, y = b and z = c. We impose the boundary conditions that Φ vanishes on all sides of the box except for the face at z = c, where it is given by Φ(x, y, z = c) = V (x, y). Thus we have The condition φ(x = 0, y, z) = 0, x, y implies that + α= Ã (±) α,β e± α 2 +β 2z = 0 (4.19) 36 Nothing guarantees that Φ will always factor as (4.15). In fact, there are certainly many solutions to the Laplace equation that are not of this form. However, there are many examples for which the B.Cs are specified such that the solution turn to be of the form (4.15). 53

57 or, equivalently α 0 which gives the solutions [Ã(±) α,β + Ã(±) α,β ] e ± α 2 +β 2z = 0 (4.20) The condition φ(x, y = 0, z) = 0, à (±) α,β = Ã(±) α,β ; y, z implies that Ã(±) 0,β = 0 (4.21) à (±) α, β = Ã(±) α,β ; Ã(±) α,0 = 0 (4.22) As a result of the above two conditions we can write the term in the summation over α and β as + + α= β= e iαx e iβy à (±) α,β = The condition φ(x, y, z = 0) = 0, α>0,β>0 x, y yields (2i) 2 à (±) α,β sin (αx) sin (βy) (4.23) Thus, the potential reads Φ(x, y, z) = α>0,β>0 where I have omitted the superscript (+) on A α,β à ( ) α,β = Ã(+ α,β (4.24) A α,β sin αx sin βy sinh ( α 2 + β 2 z) (4.25) The condition Φ(x = a, y, z) = 0, and Φ(x, y = b, z) = 0 yield α n = nπ, a n 2 γ nm = π β m = mπ, a + m2 (4.26) 2 b 2 b The condition Φ(x, y, z = c) = V (x, y) allows to determine A nm as A nm = 4 ab sinh (γ nm ) c a 0 dx b 0 dyv (x, y) sin(α n x) sin β m y (4.27) where we used the orthonormality property of the trigonometric functions L 0 sin nπx L mπx sin L dx = 1 L sin nπx 2 L L mπx sin L dx = Lδ nm (4.28) 54

58 Thus, the final solution has the form of Fourier expansion Φ(x, y, z) = n,m Z A nm sin nπ a mπ x sin b y sinh π n 2 a 2 + m2 b 2 z (4.29) with the coefficients A nm can be determined by the integral in Eq (4.27) which completely fixes the solution of the Laplace equation. 4.3 Separation of Variables: Laplace Equation in Spherical Coordinates The Laplace equation in spherical coordinate system is 37 1 r 2 r (rφ) r 2 sin θ ( sin θ Φ ) + θ θ 1 2 Φ r 2 sin 2 θ φ = 0 (4.30) 2 To solve the above equation, we use the method of separation of variables and assume that the solution has the form Φ(r, θ, φ) = U(r) P(θ)Q(φ) (4.31) r Substituting in (4.30), multiplying both sides of the equation by r 2 sin 2 θ, and dividing by UPQ yields r 2 sin 2 θ 1 U d 2 U dr + sin θ ( d sin θ dp ) + 1 d 2 Q 2 P dθ dθ Q dφ = 0 (4.32) 2 Since the last term in the above equation depends only on the variable φ and so it must be equal to a constant we get 37 A simple way to obtain the above equation in spherical coordinate is to use the expression for the Laplacian in curved space in terms of 2 Φ = 1 ( ) i det gg ij j Φ det g where g ij is the metric of the curved space. In spherical coordinates an infinitesimal element of distance is given by ds 2 = g ij dξdξ j = dr 2 + r 2 dθ 2 + r 2 sin 2 θdφ 2 Thus, the non zero components of the metric can be read of from the above expression as g rr = 1; g θθ = r 2 ; g φφ = r 2 sin 2 θ Substituting in the expression of of 2 Φ leads to the Laplace equation in spherical coordinates. 55

59 r 2 d 2 U U dr sin θ P ( d sin θ dp dθ dθ The φ equation has pretty simple solution given by 1 d 2 Q Q dφ = 2 m2 (4.33) ) m2 sin 2 θ P = 0 (4.34) Q(φ) = e imφ (4.35) If the full azimuthal range is allowed (i.e., 0 φ 2π), then the single valued-ness of the solution, Q(φ) = Q(φ + 2π), restricts m to be an integer. Similarly, the first term in (4.34) depends only on the radial coordinate r we can set it to a constant, which we write as l(l + 1), i.e. d 2 U l (l + 1) U = 0 (4.36) dr2 r 2 which implies that the second term in (4.34) satisfies the equation ( 1 1 d sin θ dp ) ) + (l(l + 1) m2 sin θ P dθ dθ sin 2 P = 0 (4.37) θ For the solution to the radial equation, one can try power law solution of the form U(r) = α C αr α. Substituting into (4.36) implies that α (α 1) = l (l + 1) (4.38) which gives the solutions α = (l + 1) or α = l. Thus, we obtain U(r) = Ãlr l+1 + B l r l (4.39) where Ãl and B l are arbitrary constants. However, if the solution is required to remain finite at r = 0 (or r ) for say l > 0 (or l < 0), then we must have B l = 0 (or à l = 0). The θ-equation is known as the generalised Legendre equation and its solutions are given by P(θ) = P m l (cos θ) (4.40) known as the associated Legendre polynomials. These polynomials are well defined over the range 1 cos θ 1 provided l = 0, 1, 2,.., and for each l, the range 56

60 of m is restricted to m = l, l + 1,.., l. The Legendre equation and the associated Legendre polynomials is discussed with some details in the Appendix. Combining these factors, i.e. Eq4.35, Eq(4.39), and Eq(4.40), the full solution for the potential Φ can be written as Φ(r, θ, φ) = l l=0 m= l (Ãlm r l + B ) lm r (l+1) Pl m (cos θ)e imφ (4.41) It is customary to combine the two angular factors into new functions Y lm (θ, φ) = (Normalization) P m l (cos θ)e imφ (4.42) called the spherical harmonics, and then we write Φ(r, θ, φ) = l=0 l m= l ( Alm r l + B lm r (l+1)) Y lm (θ, φ) (4.43) The Y lm (θ, φ) form a complete orthonormal set on the surface of a unit sphere (see Appendix). The constants A lm and B lm are determined by the B.Cs on Φ specified, say, on surfaces at fixed r. Since there two sets of constants, one needs boundary conditions specified on two surfaces one of which could be at the origin or at infinity. As an example, consider the solution (4.43) of the Laplace equation in a region bounded by two sphere of radii r = a and r = b, and let the potential on the sphere be specified as Φ(r = a, θ, φ) = V a (θ, φ). Then, using the orthonormality of Y lm (Ω), the B.Cs on Φ yield A lm a l + B lm a l 1 = A lm b l + B lm b l 1 = 2π 0 2π 0 π dφ dφ 0 π 0 Y lm(ω )V a (Ω ) sin θ dθ (4.44) Y lm(ω )V b (Ω ) sin θ dθ Note that if either surface moved to infinity with the condition that the potential on it remains finite, then A lm = 0. Similarly, if either surface is moved to the origin with the condition that the potential does not blow up, then one gets B lm = 0. If the system with rotational symmetry about the z-axis. The solution to the potential problem then should not depend on the angle φ. Hence its expansion only contains m = 0 mode, and so 57

61 Φ(r, θ) = l=0 ( Al r l + B l r l 1) P l (cos θ) (4.45) In a given problem, once the coefficients A l and B l are determined by the B.Cs, then the above expression is the unique expansion of the potential. Thus, by knowing only the r-dependence of the solution, the θ-dependence can be inserted by multiplying each power of r l and r l 1 by P l (cos θ). It is important to note that if m 0, which is the case in the absence of azimuthal symmetry, then the same power of r could correspond to different values of m and the uniqueness will be lost. 4.4 Expansion of point-like potential in spherical harmonics Suppose there is a charge q at a r = aẑ, in which case we know that the potential at a point P at the position r is given by q Φ(r, θ) = k e r2 + a 2 2ar cos θ (4.46) and for θ = 0, i.e. the point P is along z-axis, is simply Φ(r, 0) = k e q r a At distance r < a, we can expand the above expression in power series as q 1 Φ(r, 0) = k e a 1 r/a = q [ 1 + r ( r ) 2 ( ] r 3 a a a a) (4.47) (4.48) from which it follows that A l = k e q/a ( l + 1), and hence Φ(r, θ) = k e q a l=0 For r > a, expanding in powers of a/r yields Φ(r, θ) = k e q r r l a l+1 P l(cos θ), r < a (4.49) ( a ) l Pl (cos θ), r > a (4.50) r From equations (4.49) and (4.50), we deduce that in general we have l=0 1 r r = l=0 r< l r> l+1 P l (cos γ) (4.51) where r < (r > ) is the smaller (larger) of r and r, while γ is the angle between r and r. If (r, θ, φ) and (r, θ, φ ) are the spherical coordinates of r and r, then than angle γ can be written as (see Eq (3.39)) cos γ = cos θ cos θ + sin θ sin θ cos (φ φ ) (4.52) 58

62 The function P l (cos γ) can be written in terms of the variables θ, φ, θ and φ as (see Appendix) and so we obtain P l (cos γ) = 4π 2l + 1 m=l m= l Y lm(ω )Y lm (Ω) (4.53) 1 r r = l m= l 4π 2l+1 r< l r> l+1 Y lm (Ω )Y lm (Ω) (4.54) Example 4.1: A sphere with a specified potential We consider an isolated sphere of radius a centered at the origin, and with the potential at its surface given by Φ(a, θ, φ) = V 0 cos 3 θ (4.55) where θ is the polar angle. This problem has azimuthal symmetry, and so in general the potential has the form given in (4.45). Since our volume contains all r > a, the finiteness of the potential at infinity requires that A l = 0 for all l. The constants B l are then determined by matching the terms in the series with the B.C on the surface of the sphere. Recalling that P 1 (x) = x and P 3 (x) = 1 2 (5x3 3x), we can write the B.C (4.55) as [ 3 Φ(a, θ, φ) = V 0 5 P 1(cos θ) + 2 ] 5 P 3(cos θ) (4.56) Thus, [ 3 ( a ) 2 Φ(r, θ, φ) = V 0 P1 (cos θ) + 2 ( a ) 4 P3 (cos θ)] 5 r 5 r (4.57) 59

63 Example 4.2: Hemispheres of opposite potentials Suppose that the potential on a surface of a sphere of radius a is given by V 0, 0 θ π/2 Φ(a, θ) = (4.58) V 0, π/2 θ π We would like to solve Laplace equation inside the sphere. Since the origin is included, the constants B l vanish for all l, and the expansion of the potential takes the form. Φ(r, θ, φ) = V 0 l=0 A l ( r a) l Pl (θ) (4.59) where for convenience we introduced the factor V on the righthand side, along with the use of the powers of a in the sum. To find the coefficient A l, we multiply the above equation by P l (cos θ) and integrate over cos θ. Making use of the orthonormality of the Legendre polynomials, we find or, equivalently, π 0 Φ(a, θ)p l (cos θ) sin θdθ = V 0 ( 2 2l + 1 A l = 2l [ 1 dxp l (x) ] dxp l (x) ) A l (4.60) (4.61) Using the inversion property of the Legendre polynomials. P l ( x) = ( 1) l P l (x), we find 0, for l even A l = Φ(a, θ) = (2l + 1) (4.62) 1 dxp 0 l(x), for l odd To complete the integration for the case of odd l we use the recurrence relation so that, A l = 1 0 dp l+1 dx = (2l + 1) P l + dp l 1 dx (4.63) [ dpl+1 dx dx dp ] l 1 = P l 1 (0) P l+1 (0) (4.64) dx From the Rodrigues formula, one can show that for l even we have l/2 (l 1)!! P l (0) = ( 1) l!! (4.65) 60

64 where the "double factorial" sign means l!! = l(l 2)(l 4)...(2 or 1). Hence (l 1)/2 (l 2)!! A l = ( 1) (2l + 1) (4.66) (l + 1)!! Now setting l = (2m + 1), with m = 0, 1, 2,.., we can write the solution of Laplace equation as Φ(r, θ, φ) = V 0 l=0 C m ( r a) 2m+1 P2m+1 (θ) (4.67) where m (2m + 1)!! C m = ( 1) (2m + 3) (4.68) (2m + 2)!! Example 4.3: Conducting sphere in a uniform external electric field Here we will consider a conducting sphere of radius a placed in a uniform electric field directed along the z-axis, i.e. E ext = E 0 ẑ Since the spheres assumed to be a conductor, the electric field must be normal at its surface; hence we expect that it will distort the external field lines in its vicinity. However, far away from the sphere the external field is unaffected by the sphere, and so or, equivalently, lim E( r) = E 0 ẑ r lim Φ( r) = E 0 z + const. = E 0 r cos θ + const. r which indicates that all A l vanish except the terms corresponding to l = 0 and l = 1. Thus, we can write Φ( r) = A 0 E 0 r cos θ + l=0 B l r (l+1) P l (cos θ) (4.69) 61

65 The term B 0 /r indicates that the conductor carries a net charge, which is not the case in this example. Thus B 0 = 0. Furthermore, the potential on the surface of a conductor must be constant, which in the case of sphere means that Φ(r = a, θ) must be independent of the angle θ. This implies that and B l = 0, l > 2 (4.70) Thus the potential reads E 0 a + B 1 a 2 = 0 B 1 = E 0 a 3 (4.71) ) Φ( r) = Φ S E 0 r cos θ (1 a3 r 3 (4.72) where Φ S = A 0 is the potential on the surface of the sphere. The electric field can be obtained with the the use of E = Φ, and it is given by ) ) E = E 0 cos θ (1 + 2 a3 ˆr E 0 sin θ (1 a3 ˆθ (4.73) r 3 The surface charge distribution on the sphere can be found by noting that r 3 4πk e σ ˆr = lim r a E( r) σ(θ) = 3 4πk e E 0 cos θ (4.74) Note that this surface charge density is consistent with our assumption that there is not net charge on the sphere. This can be see easily by integrating σ over the surface of the sphere, i.e. Q = σds = 3 2π E 0 R 2 4πk e 0 π dφ cos θ sin θdθ = 0 (4.75) Expansion of Green s Functions in Spherical Coordinates. Let us consider the Green function for the exterior problem with a spherical boundary at radius r = a on which the potential is specified. Previously we have found it, using the method of images, to be 62

66 G( r, r ) = 1 r r + q r Y (4.76) where q = a/r and y = a 2 /r. By expanding both terms with the use of (4.54), and noting that for the second term r > Y and Y r, we can write G( r, r ) = l=0 l=0 [ 4π 2l+1 r l < r l+1 > 1 a ( ) ] l+1 a 2 rr Y lm (Ω )Y lm (Ω) (4.77) We see that the behaviour of the radial part of G( r, r ) changes at r = r as ( ) 1 r l a2l+1, r < r r l+1 r l+1 ( ) 1 r l a2l+1, r > r r l+1 r l+1 (4.78) This behaviour has a simple explanation given that G( r, r ) can be regarded as the potential at r due to a unit charge at r. In this case, as r increases beyond r, the potential due to the unit charge as well as the one due to the induced charge on the sphere keep on falling. But for a < r < r, as the potential due to the unit charge decreases, the one due to the sphere increases, or vice-versa. The reason for the change in behavior at r = r is because the above expansions are obtained on the assumption that G( r, r ) satisfies Laplace equation, which is not true at r = r. Now, let us generalise the previous case, and find an expansion for the Dirichlet Green s function for the region V bounded by r = a and r = b, with b > a. In general we can write G( r, r ) = 1 r r + F ( r, r ) (4.79) where 2 F = 0 in the region V. With the use of equations (??) and (4.43) it is possible to write G( r, r ) = + 4π r< l 2l + 1 r> l+1 r (A l lm l=0 m=l l=0 m= l m=l m= l b l+1 + B lm Y lm (Ω )Y lm (Ω) + (4.80) a l r l+1 ) Y lm (Ω) where A lm and B lm can be functions of r. The first term on the righthand side is 1/ r r and the second is a general solution to Laplace equation. The coefficients are determined by requiring the B.C G( r, r ) = 0 for r on one of the two bounding spherical surfaces: 63

67 1. At r = a (r < = r = a, r > = r ) we have 0 = [ ] 4π a l < Y 2l + 1 lm(ω a l ) + A r l+1 lm l,m > b + B 1 l+1 lm Y lm (Ω) (4.81) a Using the orthogonality of the spherical harmonics it follows 4π 2l + 1 a l < r > l+1 2. Similarly, At r = a (r < = r = a, r > = r ) we find that 4π 2l + 1 r < l b> l+1 a l Ylm(Ω ) + A lm b + B 1 l+1 lm a = 0 (4.82) Ylm(Ω 1 ) + A lm b + B a l lm = 0 (4.83) bl+1 Combining (4.82) and (4.83) we obtain We see that the behavior of the radial part of G( r, r ) changes at r = r and as A lm = 4π B lm = 4π 1 2l+1 [1 ( a b ) 2l+1] Ylm (Ω ) 1 2l+1 [1 ( a b ) 2l+1] Ylm (Ω ) [ ] a 2l+1 r l b l r l+1 b l [ ] a l+1 r l al+1, b 2l+1 r l+1, (4.84) Substituting (4.84) into (4.80) we get G( r, r ) = 4π/2l + 1 [ l,m 1 ( ) a 2l+1 ]Ylm(Ω )Y lm (Ω) (4.85) b [ ( a ) ] 2l+1 r l ( { 1 < a ) 2l+1 r l + b r> l+1 b r r l r l ( a ) 2l+1 l+1 b + r l 2l+1 b r a2l+1 l+1 r l+1 r } l+1 The quantity in the bracket {...} can be written in a compact form. To see this suppose that r > = r and r < = r, then ) ( {...} = (r l a2l+1 1 r l+1 r l+1 (r l< a2l+1 r l+1 < r l b 2l+1 ) ( 1 rl > r> l+1 b 2l+1 ) ) (4.86) If, now r > = r and r < = r, then we have 64

68 ) ( {...} = (r l a2l+1 1 r l+1 r l+1 (r l< a2l+1 r l+1 < rl b 2l+1 ) ( 1 rl > r> l+1 b 2l+1 ) ) (4.87) Thus, we can write G( r, r ) as G( r, r ) = l,m ) ( ) 4π/2l + 1 [ 1 ( ) a 2l+1 ]Ylm(Ω )Y lm (Ω) (r l< a2l+1 1 rl > (4.88) r< l+1 r> l+1 b 2l+1 b To obtain the Green function for the interior of a sphere of radius b we take the a 0, and one finds G( r, r ) = l,m = ( ) 4π r /2l + 1 Y lm(ω l )Y lm (Ω) < rl <r> l r> l+1 b 2l+1 1 r r l,m 4π /2l + 1 Y lm(ω )Y lm (Ω) b r r l (b 2 /r ) l+1 (4.89) or, equivalently, G( r, r ) = 1 r r b/r r r R (4.90) where r R = (b2 /r, θ, φ ) is the spherical coordinates of the charge q = b/r. The Green function of the interior of a sphere of radius a can be obtained by taking b 0 in (4.88), which leads to the same result in equation (4.77). 4.6 Fields and Potentials Around Conical Hole In this section we will be interested in finding the electric potential near the edges of of tips with azimuthal symmetry, such as the sharp corner of a cone. Let β be conical angle and 0 θ β, and we assume that the conducting surface at θ = β is held at zero potential. Around such point, r is small enough that we can write the solution of the Laplace equation as Φ(r, θ) r ν P ν (cos θ) (4.91) with P ν is the solution to the Legendre equation (4.37), which can be re-cast as 65

69 [ d ξ(1 ξ) dp ] ν + ν(ν + 1)P ν = 0 (4.92) dξ dξ where we made the change of variable ξ = sin 2 θ/2. We seek solutions for P ν as an expansion in powers of y, i.e. P ν = ξ α Then, substituting in (4.92), yields k=0 a k ξ k (4.93) [ ak (α + k) 2 ξ α+k 1 + a k (α + k)(α + k + 1) + ν(ν + 1)ξ α+k] = 0 (4.94) k=1 which implies that and a 0 α 2 ξ α 1 = 0 α = 0 (4.95) a k+1 = k(k + 1) ν(ν + 1) (k + 1) 2 a k (4.96) Hence, [ P ν (ξ) = a ( ν) ] ( ν)( ν + 1)(ν + 1)(ν + 2) 1!1!ξ + ξ ν + 1 2!2! (4.97) If ν = l = 0, 1, 2,..., one can check that the above series is just the Legendre polynomial, as expected. However, ν does not have to be integer, but it needs to be zero or positive for the series to terminate, and P ν represents a generalisation of the Legendre polynomial, called Legendre function of the first kind and order ν. Furthermore, to have a finite potential at the origin (see below), ν is required to be positive. Since the potential is zero at the surface of the conductor, we must have P ν (cos θ) = 0 (4.98) which provides the allowed values of ν. In fact, the above equation has an infinite number of solutions, which we denote by ν k, with k = 1, 2,... Without loss of generality 66

70 we arrange the sequence ν k with order of increasing magnitude, i.e. ν 1 < ν 2 < ν 3 <... Thus, the complete solution at finite value of r can be written as Φ(r, θ) = A k r ν k P νk (cos θ) (4.99) k=1 Since we are interested on the behaviour of the potential near the conical hole, i.e. r very small, the dominant contribution to Φ is given by the k = 1 term. Hence, Φ(r, θ) = A 1 r ν 1 P ν1 (cos θ); r 0 (4.100) Around the conical hole, the components of the electric field and the surface charge density are given by E r = Φ r A 1ν 1 r ν1 1 P ν1 (cos θ) E θ = 1 Φ r θ A 1ν 1 r ν1 1 sin θ dp ν 1 (cos θ) d cos θ σ(r) = 1 E θ θ=β 1 A 1 ν 1 r ν1 1 sin β dp ν 1 (cos θ) 4πk e 4πk e d cos θ θ=β 67

71 5 Multipole Expansion A s we saw previously, the scalar potential of a charge distribution ρ( r) contained in a volume V, is defined as Φ( r) = k e d 3 r ρ( r ) (5.1) r r V Unless the charge distribution has some symmetry, this integral is very complicated and usually it is done numerically. However, If the observation point r lies outside the region of the charge distribution and at a distance r large compared to V 1/3, then we can expand the denominator in the integrand in terms of powers of 1/r. 5.1 Dipole Moment of pair of charges +q and q Before studying the general case of a continuous charge distribution, we will first consider the potential of a pair of opposite charge at a distance d apart. For convenience, we choose the origin of coordinates at the midpoint of the two charges. Thus, the potential due two this pair of charges +q and q at a point r is [ ] 1 Φ( r) = k e q r 1 d 1 r + 1 (5.2) d 2 2 = k eq r d2 r. d 4r 2 r d2 + r. d 4r 2 r 2 (5.3) where + d/2 and d/2 are the position vectors of the charges +q and q, respectively. If r >> d, then we can make a Taylor expansion of the square roots in the above expression in terms of powers of 1/r and, as a reasonable approximation, we keep just the lowest order 38 : Φ( r) k eq r [( ( which can be written in the form d 2 4r r. d 2 r 2 ) +.. ) ( The Taylor expansion of a function f(x) around a point x 0 is: f(x) = n=0 f (n) (x 0 ) n! (x x 0 ) n ( d 2 4r + r. d 2 r 2 ) +.. )] (5.4) where f (n) (x 0 ) is the n th derivative of f(x) evaluated at x 0. For a function of the form (1 + x) m we get where m is a real number. (1 + ɛ) m = 1 + mɛ + m(m 1) ɛ 2 + 2! m(m 1)(m 2) ɛ ! 68

72 Φ( r) = k e p. r r (5.5) where p = q d (5.6) Z +q d /2 X q Y Figure 5. The electric dipole moment of equal but opposite charges. The vector p is called the dipole moment of the two charges +q and q 39. If d 0, but p = qd is held constant, then all higher order terms vanish, since they contain powers of d and not qd, and hence, the first term in the above expansion is the 39 Dipole moments are often quoted in Debyes (D), where 1 D = C.m. For instance, two charges q 1 = e and q 2 = e, respectively, separated by a distance of 1 A 0, have a dipole moment of p = C.m = 4.8 D. 69

73 exact result. In this case we call this system of charges an ideal dipole of moment p 40. The electric field due to these charges can be obtained using the relation E =.Φ, which gives or, equivalently, E = k e (p. r) r 3 + 3k e (p. r) r r (5.7) E = k e 3 (p.ˆn) ˆn p r 3 (5.8) where ˆn = r/r is a unit vector directed from the midpoint of the dipole to the point where the point r where the electric field is calculated. Note that, the electric field due to a dipole decreases with r as 1/r 3, unlike the 1/r 2 behaviour for a point like charge. This is not surprising since the net charge of a dipole is zero, and hence the absence of the 1/r 2 term. The figure below shows the field lines of a dipole. In deriving the above expression we assumed that the point r is not at the origin of the coordinates, since in this case our approximation fails. However, for the case a point-like dipole, i.e. an ideal dipole, to find the electric field at the point r = 0, we use the fact that 2 (1/r) =. ( r/r 3 ) =, and we obtain E = k e 3(p. ˆn)ˆn p r 3 4π k e 3 p δ( r) (5.9) where at r = 0 only the δ( r)-term contributes since in deriving the first term, it was assumed that the point r is not at the origin. 5.2 Linear Quadrupole A linear quadrupole consists of two equal charges of strength q a distance 2d apart and a charge 2q placed at the midpoint of the line joining the two. So, a linear quadrupole can be regarded as an inverted pair of electric dipoles of equal dipole moment magnitude p = qd with centers displaced by the dipole separation d along their common axis. The potential at point r is ( q Φ( r) = k e r d + q r + d 2q r For r >> d, we can use Taylor expansion in terms of powers of d/r and write 40 It should be noted that for finite d, the system has other terms beside the dipole moment. ) (5.10) 70

74 Figure 6. The dipole electric field lines. Φ( r) = k eq r = k eq r ( [( 1 + d2 r 2 r. d 2 r r. d r ) 1/2 ( ) + d d 2 (5.11) d 2 r + 3 ( r. d) 2 3 d 2 ( r. ) ] d) + 3 d 4 ( ) 2 2 r 4 2 r 4 8 r... + d d 2 4 By keeping only the lowest order terms, the potential take the following form where Φ( r) = k e 3 i,j=1 Q ij x i x j (2) +... (5.12) r 5 Q ij (2) = q ( 3d i d j d 2 δ ij) (5.13) which is called the quadrupole moment. This is a symmetric traceless matrix, and so only five components of Q ij (2) are really independent. In the limit where d 0, but with qd i d j is kept fix, the quadruple term in (5.12) represents the exact result for the potential. In this case, such charge arrangement is called ideal quadrupole of moment Q ij. 71

75 5.3 Multipole Expansion of a Charge Distribution in Rectangular Coordinates Let us first consider a point charge q located at position r, not necessarily at the origin of the coordinate system. The potential at some point r very far from the location of the charge q, i.e. r >> r, can be expanded around the origin of coordinates as q Φ q ( r) = k e r r = k ] 1/2 eq [1 + r 2 r r 2 2 r. r (5.14) r [ 2 = k eq 1 1 ( ) r 2 r 2 r 2 2 r. r + 3 ( ) r 2 2 r 2 r. +...] r 2 8 r2 r 2 which can be arranged in the form where q Φ q ( r) = k e r + k p. r e r 3 + k 1 e 2 3 i,j=1 Q ij x i x j (2) +... (5.15) r 5 p = q r ; Q ij (2) = q ( 3x i x j r 2 δ ij) (5.16) Now let suppose there are a set of N charges, {q a, a = 1,..N}, localized at positions { r a, a = 1,..N}. The potential due to these charges at a point r, with r >> r a, can be obtained using the expression (5.15) and it reads Φ( r) = k e N a=1 which can be re-written as q a r + k e N a=1 q a r a. r 1 + k r 3 e 2 N a=1 q a 3 ( r a. r) 2 r 2 a r 2 r (5.17) Q Φ( r) = k e r + k p. r e r 3 + k e r. ˆQ(2) r +... (5.18) 2 r 5 where Q = a q a is the total charge, also called the monopole moment, p = a q a r a is the total dipole moment, and ˆQ (2) is the quadrupole matrix of this charge distribution given by ˆQ (2) = N a=1 [ ] q a 3 r a r a r a 2 Î (5.19) 72

76 where denotes the tensor product and Î is the 3 3 unit matrix. In component form, the quadrupole matrix elements read Q ij (2) = N a=1 q a [ 3x i a x j a r 2 a δ ij] (5.20) For a continuous charge distribution, ρ, we replace the summation over the over the N charges by integral over the volume, V, in which the charges are localized, i.e. Φ( r) = k e r V ρ( r )d 3 r r + k e r. ρ( r ) r d 3 r x i x j + k 3 e ρ( r ) ( ) 3x V r ix 5 j r 2 δ ij d 3 r (5.21) V Thus, the potential has the same form as the one in (5.16) with the monopole, dipole and components of the quadrupole matrix given by Q = V ρ( r )d 3 r ; (5.22) p = V ρ( r ) r d 3 r (5.23) Q ij (2) = V ρ( r ) [ 3 x i x j r 2 δ ij] d 3 r (5.24) Thus, the potential at a point r far away from the charge distribution (i.e. r >> r ) can be regarded as an infinite series of ideal multipoles placed at the origin. Note that, a spherically symmetric charge distribution has only monopole moment. So, higher order terms are measures of the deviation from spherical symmetry of the system. In the case where the total electric charge is non-zero, i.e.q 0, the electrostatic potential is well approximated by the monopole term Φ monopole = k e Q/r, characterised with a single quantity, the total charge. When Q = 0, however, the potential is dominated by Φ dipole = k e p. r/r 3, which is characterised by 3 quantities, the components of the dipole moment p. When both the total charge and dipole moment of the distribution are zero, the potential is approximated by Φ quad = k e i,j Qij (2) x ix j /2r 5, characterized by 5 independent components of the quadrupole moment tensor. One can always diagonalise the quadrupole tensor by rotating the reference frame. Hence, to specify a quadrupole moment one needs to give its orientation and two of its diagonal components (usually Q 11 (2) and Q33 (2)). In particular, for an axial symmetric charge distribution two of the diagonal components of the tensor Q ij (2) are equal, and so it can be specified by a single component. 73

77 In general, the dipole and the higher order terms depend on the choice of origin of the coordinate system. This means that if the Q 0, one can always choose an origin of coordinates for which the dipole vanishes. However, if the monopole term vanishes then the dipole term is independent of the choice of coordinates. To see this, one can substitute ( r + a) for r in the dipole expression and find that the dipole term remains unchanged if Q = 0. In general, one can show that only the leading non-vanishing multipole is independent of the origin. 5.4 Multipole Expansion of a Charge Distribution in Spherical Coordinates We know that the scalar potential, Φ, due to a charge distribution, ρ, satisfies Laplace s equation at points outside it. Thus, we can expand Φ in spherical harmonics using the expression in Eq (4.43) Φ(r, θ, φ) = l=0 l m= l 4π 2l+1 q lm 1 r l+1 Y lm (θ, φ) (5.25) where we have used the notation q lm instead of B lm and all the A l terms vanish so that the potential for r does not diverge. The task is to determine the constants q lm in terms of the properties of the charge charge distribution ρ. Using the definition of the potential in Eq (5.1), and the expression of 1/ r r for r >> r in terms of spherical harmonics (see Eq (4.54)), we can write Φ(r, θ, φ) = l l=0 m= l from which it follows that 4π 1 2l + 1 r Y lm(θ, φ) l+1 Y lm(θ, φ ) ρ( r ) r l d 3 r, r >> r q lm = V ρ(r, Ω ) r l Y lm (Ω ) d 3 r (5.26) where in V is the volume in which the charge distribution is localised, and d 3 r = r 2 dr Ω is the volume element written in the spherical coordinates, with dω = sin θ dθ dφ is the infinitesimal element of solid angle. The constant q lm are called the multipole moments of the charge distribution ρ( r ). Note that the moments with m 0 are related to the moments with m < 0 via 41 q l, m = ( 1) m q lm (5.27) 41 We have used the following property of the spherical harmonics: Y l, m (Ω) =... 74

78 By comparing the powers of r in (5.21) to the ones in (5.25) we see that q lm is the 2 l -pole moment of the charge distribution in the spherical coordinates. The first three terms in the multipole expansion are Monopole moment (l = 0) Here there is only one component: q 00 = ρ( r ) r 0 Y00(Ω ) d 3 r = 1 ρ( r ) d 3 r 4π Hence, q 00 = 1 4π Q (5.28) Dipole moment (l = 1) Here there are three components q 11, q 1, 1 = q 1,1, and q 10. So, we need to calculate only the components q 11 and q 10. The moment q 11 reads q 11 = ρ( r ) r Y11(Ω ) d 3 r (5.29) V 3 = ρ( r ) r (sin θ cos φ i sin θ sin φ ) d 3 r 8π V 3 = ρ( r )(x iy ) d 3 r 8π Thus, V q 11 = 3 8π (p x ip y ) (5.30) For the moment q 10 we have q 10 = ρ( r ) r Y10(Ω ) d 3 r (5.31) V 3 = ρ( r ) r cos θ d 3 r 4π V 3 = z ρ( r ) d 3 r 4π V 75

79 and we obtain q π p z (5.32) From the above expressions of q 1,m, it follows that q 1, 1 q 1,0 q 1,1 = 3 4π Û p x p y p z (5.33) where Û = i (5.34) 1 i 0 Thus, up to a unitary transformation Û, the vector q := {q 1, 1, q 1,0, q 1,1 } is equivalent to the dipole moment vector p. Quadrupole moment (l = 2) Here there are five components: q 22, q 2, 2 = q 22, q 21, q 2 1 = q 21, and q 20. Thus, we need to compute only q 22, q 21, and q 20. They are given by q 22 =.. (5.35) Now, we derive the expressions of the components of the electric field in spherical coordinates by applying the gradient operator [ ] = e r r + e 1 θ r θ + e 1 φ r sin θ φ (5.36) to the potential in Eq (5.25), which yields E( r) = l l=0 m= l [ (l + 1) e r e θ θ e im φ sin θ ] Ylm (θ, φ) r l+2 (5.37) where e r, e θ, and e φ are the unit vectors in spherical coordinates. The θ-derivative of the spherical harmonics can be expressed as θ Y lm(ω) = m cot (θ) Y lm (Ω) + l (l + 1) m (m + 1)Y l,m+1 (Ω)e iφ (5.38) 76

80 Hence, we obtain l E r = E θ = E φ = i l=0 m= l l=0 m= l l + 1 r l+2 Y lm(θ, φ) (5.39) l l l=0 m= l m cot (θ) Y lm (Ω) + l (l + 1) m (m + 1)Y l,m+1 (Ω)e iφ m 1 Y lm (θ, φ) r l+2 sin θ Example 5.1: Quadrupole moment of 4 charges on the corners of a square I am still working on it... coming soon. Example 5.2: Quadrupole moment of a uniformly charged ellipsoid Consider a uniformly charged ellipsoid of revolution, with semi-major axis a along the z axis and semi-minor axis b having total charge Q. let us first find the zz component of the quadrupole moment tensor, i.e. Q 33. The equation of the ellipsoid is given by where ξ = x 2 + y 2 denotes the cylindrical radius. Hence and where ξ 2 b 2 + z2 a 2 = 0 (5.40) ( Q 33 = 3z 2 r 2) z=a ξ=b 1 (z/a) 2 ρd 3 r = ρ ( 2z 2 ξ 2) 2πξdξdz (5.41) V z= a ξ=0 ( z=a ) )] 2 = 2πρ [z 2 b 2 1 (1 z2 a z2 dz a 2 z= a = 8 15 πρab2 ( a 2 b 2) = 2 5 Q ( a 2 b 2) Q = z=a ξ=b 1 (z/a) 2 z= a ξ=0 ρ2πξdξdz = 4π 3 ρab2 (5.42) is the total charge of a uniformly charged ellipsoid. For the other components we note that because of the azimuthal symmetry of the charge distribution we have 77

81 Q 11 = Q 22, whereas the off-diagonal elements vanish. Using the traceless property of the quadrupole tensor and the above expression of Q 33, we get Q 11 = Q 22 = 1 2 Q 33 (5.43) = 1 5 Q ( a 2 b 2) The above expressions of the components of the quadrupole moments were calculated with respect to the center of the ellipsoid. Since the net charge Q is not zero, then their values are not unique, and depend on the origin with respect to which they will be calculated. 5.5 Energy of a Charge Distribution in an External Filed As we have seen in chapter one, the electrostatic energy U of a localized charge distribution within a region of volume V subject to an external potential Φ ext is given by U = ρ( r)φ ext ( r)d 3 r (5.44) V Note that there is no factor of 1/2 since no double counting exists here (see Eq (1.98)). If we expand the external potential in Taylor series about a point, r 0, inside the volume V, then, we have Φ ext ( r) = Φ ext ( r 0 ) + ( r r 0 ). Φ ext r0 + 1 x i x j 2 Φ 2 x i x j r = Φ ext ( r 0 ) ( r r 0 ).E ext ( r 0 ) 1 2 = Φ ext ( r 0 ) ( r r 0 ).E ext ( r 0 ) 1 6 i,j j,j i,j x i x j Ej ext x i r ( 3x i x j δ ij r 2) E j ext x i ( r 0 ) +... where we used the fact that.e ext vanishes inside the localised charge distribution since the external field is assumed to be generated by sources that are outside the volume V. Thus, the electrostatic energy of this charge distribution is U = Q Φ ext ( r 0 ) p. E ext ( r 0 ) 1 6 i,j Q ij Ej x i ( r 0 ) +... (5.45) where Q is the total charge inside the volume V, and p and Q ij are the dipole and the component of the quadrupole moments of the charged distribution calculated with respect to the point r 0. Hence, we can interpret the terms in the above series as 78

82 the interaction energy of the various multipoles with the external field. For instance, if the total charge is zero, we have U p.e ext ( r 0 ), and so for the system to minimise its potential energy, the dipole has to align itself along the direction of the external electric field. As an example, consider the interaction energy between two independent electric dipoles p 1 and p 2, located at points r 1 and r 2, respectively. The energy of dipole p 1 in the external field E 2 of p 2 is U Dipole 12 = p 1 E 2 (5.46) Using the expression of the electric field of a dipole given in (5.8), we obtain U Dipole 12 = k e p 1 p 2 r 2 (p 1. r)(p 2. r) r 5. (5.47) where r = ( r 1 r 2 ) is the vector connecting the two dipoles. Note that the above expression is symmetric with respect to exchange of p 1 and p Force and Torque on a Charge Distribution in External Electric Field Consider a charge distribution ρ( r), with total charge Q, in some region of space where there is an external field E ext. Then, the net force, F, acting on this charge is the sum of the forces on the individual parts of it, i.e. F = V ρ( r) E ext ( r)d 3 r (5.48) Expanding E ext ( r) around the point r 0, we get [ ( F = ρ( r)e ext ( r 0 ) + ρ( r ) ( r r 0 ). ) ] E ext ( r 0 ) +... or, equivalently, V (5.49) F = QE ext ( r 0 ) + ( p. ) E ext ( r 0 ) i,j Q ij 2 x i x j ( r 0)... (5.50) Hence, we see that in a uniform external electric field, and in the absence of a monopole term, i.e. Q = 0, the net force on the charge distribution vanishes. So, in a uniform external electric field, only the monopole term feels the a net force where as the higher 79

83 order multipoles do not 1. The net torque on the charge distribution is defined by τ = V ( r r 0 ) ρ( r) E ext d 3 r (5.53) Using the expansion of the external electric field around r 0 and keeping the leading term, we get τ p E ext ( r 0 ) (5.54) where p is the dipole moment with respect to the point r 0. Note that even if the total charge vanishes, and the external electric field is constant, there is a non zero torque acting on the dipole. Moreover, when Q = 0, τ does not depend on the origin of coordinates. This torque tends to turn the charge distribution so that the dipole moment align with the external electric field. 1 The above expression of the net force can be also obtained from the definition of the force using the potential energy: F = U (5.51) If we consider the motion of the localized charge distribution as a whole, then we can take r 0 as a coordinate system instead of r, and so the gradient is taken with respect to the coordinate r 0, and we write F = Q 0 Φ ext ( r 0 ) + 0 (p.e ext ( r 0 )) E j Q ij x i ( r 0) +.. (5.52) i,j = Q 0 E ext ( r 0 ) + (p. ) 0 E ext ( r 0 ) i,j Q ij 0 ( E j x i ( r 0) ) +.. In obtaining the second equation above, we used the fact that the multipole moments are constant with respect to 0. 80

84 6.1 Introduction 6 Electrostatic in Dielectric Media In previous chapters we have been studying the potential and the electric fields due to charge distributions in vacuum or on the surface of conductors. Here, we will discuss electrostatics in matter, in particular the electric field equation inside insulators, a very poor conductors, generally known as dielectrics. Examples of dielectrics include wood, plastic, and glass (amorphous SiO 2 ). Since microscopically a dielectric material consists of atoms, we expect that its macroscopic electrostatic properties arise from the collective contributions of its microscopic constituents. The fact that the valence electrons in dielectric material are bound to ionic cores of the atoms, the average charge density vanishes. Therefore, one is tempted to conclude that the electrostatics in a dielectric is determined by the extra free charges that one add to the substance. However, that is not true, due to the phenomena of the polarization which can give rise to regions of net charge, as will explained later. Let us consider the simplest atom, the hydrogen atom, in its ground state. In the absence of an external electric field, i.e. E ext = 0, we can model the hydrogen atom as a spherically electron charge distribution orbiting the proton which is sitting at the centre. The strength of the electric field felt by the electron due to the proton electric charge is E (e) p = k e e r 2 (6.1) Considering the typical proton-electron distance to be about m, we find that E (e) p V olts/m (6.2) This is a huge electric field compared to the one usually used in laboratory experiments, which is typically in the range ( ) V olts/m. Thus, one can treat the effect of the external electric field on the atom as a perturbation to its internal electric field. For example, we place the hydrogen atoms between the two parallel plates of a capacitor with an electric field, say, E ext 10 3 V/m ˆx, where ˆx is a unit vector along the distance between the two plates, then the nucleus and the electron feel an equal, but opposite, force with strength F p = F e = e E ext (6.3) The net effect of external electric field is that the nucleus is now displaced by a small distance, d/2, from its initial position, along the direction of E ext, and the electron 81

85 Element Z α/4πɛ 0 [m 3 ] H (Alkali Metal) He (Noble Gas) Li (Alkali Metal) Be (Alkali metal) C (Alkali Metal) Ne (Nobel Gas) Na (Alkali metal) Ar (Noble Gas) cloud is moved by the same amount in the opposite direction. As a result of this displacement, an electric dipole moment p = e d ˆx = e d is induced in the atom. Thus, the electric field inside the dielectric material becomes smaller by this shielding but not zero as in conductors since charges can not move far enough to completely cancel the external electric field 42. Since E ext << E p (e), one expect, to a good approximation, that dipole moment is linear in applied electrical field, i.e. p = α E ext (6.4) The constant α is known as the atomic electric polarizability. In SI unit system, one often expresses the polarizability of atoms in terms of α/4πɛ 0 so that it has units of (meter) 3. In the table below we list the polarizability of some atoms. For molecules, often it happens that they polarize more in some directions than others, even with no applied electric field. This is for example the case of carbon dioxide molecule, CO 2, where the atoms arrange them selves in a straight line. Because of this the dipoles in the linear CO2 molecule cancel each other out, and so the CO2 molecule is a non-polar molecule 43. On the other hand, for water molecule, H 2 O, the electrons tend to cluster around the Oxygen atom, leaving a negative charge at the vertex, and a net positive charge at the opposite end 44. However, since H 2 O molecule is bent, (where the two H-O bonds make approximately 105 0, a net dipole moment is induced 45. Hence, water molecule is a polar molecule. Other examples of polar molecules include HCl and NH It is easy to show that displacement is much smaller that the atomic size, and so E p (e) does not decrease sufficiently to get cancelled by E ext. 43 However, not every molecule has vanishing dipole. For instance, COS is a linear molecule but the C-O bond is much dipolar than C-S and hence is net effect is a dipolar molecule. 44 We say that oxygen is more electronegative than hydrogen. 45 The dipole moment of water molecule is C.m, unusually large. It is this fact that accounts for its effectiveness as a solvent. 1 Materials exhibiting large spontaneous polarization are known as ferroelectrics (clearly there must be polar molecules involved). Ferroelectric materials are known as electrets. The best known 82

86 6.2 Macroscopic description of dielectric In this section we wish to derive the equations that describe the behavior of the electric field E inside a dielectric. At the molecular scale, we have E = 0 and.e = 4πk e ρ. Due to the possible polarization of the molecules, the charge density ρ varies wildly, and hence there will be large fluctuations in the magnitude and the direction of the electric field, from one point to another within the dielectric material. However, on a macroscopic scale we see no evidence for wild fluctuations in the fields, and so one can consider averaging ρ and E over volumes that are much larger than the atomic scale but are still small compared to the size of the substance. Thus, we define spatially averaged quantity Q( r) by Q( r) = f( y)q( r + y)d 3 y (6.5) V where V is a small volume centered around the point r and which encloses a huge number of molecules 46, and f( r) is a positive and single valued function, which vanishes for distances larger than ( V) 1/3, and it is normalized to unity 47. If we denote by ρ (n) the charge distribution within the n th molecule, centered at the point r n, then the potential outside the n th molecule is given in the cgs unit system by Φ (n) ( r) = k e V Mn ρ (n) ( ξ) r r n ξ d3 ξ (6.6) where the integration volume V Mn is the space within the n th molecule. Considering the multipole expansion of (6.7), yields Φ (n) ( r) = k e [ q (n) r r n + p(n). (n) ( ) ] r r n (6.7) where (n) is the gradient with respect to r n, q (n) and p (n) are the total charge and the electric dipole moment of the n th molecule, respectively, given by q (n) = ρ (n) ( ξ)d 3 ξ; V Mn p (n) = ξ ρ (n) ( ξ) d 3 ξ V Mn (6.8) example of a ferroelectric crystal is BaTiO3. 46 The size of the region can be about few µm. 47 An example of a profile that has these properties is 1 V, r < L = ( V )1/3 ; f( r) = 0, r L.. 83

87 Note that even if a molecule is neutral, i.e. q (n) = 0, its dipole moment will be not be zero if the charge density ρ (n) ( r) is not distributed in a spherical symmetrical way. The electric field created by the i th molecule can be calculated from the gradient of the potential, i.e. E (n) = Φ (n). So the total microscopic electric field is obtained by adding up the field of each molecule in the substance: E (micro) (x) = k e n;molecules [ q (n) r r n + p(n). (n) ( 1 )] r r n (6.9) where we assumed that quadrupole and higher terms in the series of (6.7) can be neglected. The above discrete sum can be written as an integral E (micro) ( r) = k e V D [ ( ) ] ρ( r ) r r + p( r ). 1 d 3 r r r (6.10) where the integration is taken over the whole volume of the dielectric material, V D, and we introduced the charge density, ρ, and dipole moment density, p, of the molecules in the substance, defined by ρ( r ) = n;molecules q (n) δ (3) ( r r n ); p( r ) = n;molecules p (n) δ (3) ( r r n ) (6.11) Now, by averaging out over the volume V, we obtain the macroscopic electric field E (Macro) ( r) = k e V D [ ρ( r ) r r + P( r ). ( ) ] 1 d 3 r (6.12) r r where ρ( r ) is the average charge density within the volume V at a point r, and P( r ) is the average electric dipole moment density, called the Polarization of the substance. Explicitly they read ρ( r ) = P( r ) = V V f( y) ρ M ( r + y) d 3 y N ( r ) q M ( r ), (6.13) f( y) p M ( r + y)d 3 y N ( r ) p M ( r ) Here N ( r ) represents the number density of the molecules at point r, and q( r ) and p( r ) denote the average charge and dipole moment per molecule within V, respectively. For the ease of notation, in what follow we will denote the macroscopic electric field simply by E, and the average of the charge density and the polarization without 84

88 the bar sign on top of them. Now, taking the divergence of E above, gives.e( r) = = 4π V D V D [ ( ) ( )] 1 ρ( r ) 2 + P( r ). r r 1 2 d 3 r (6.14) r r [ ρ( r )δ (3) ( r r ) + P( r ). ] δ (3) ( r r ) d 3 r or, equivalently. E( r) = 4πk e [ρ( r).p( r) ] (6.15) The above equation can be re-written as. D = ρ (6.16) where the vector D is called the electric displacement, defined as D = 1 4πk e E + P (6.17) For a the medium that is composed of different types of molecules, with number densities N i, the expressions in (6.16) and (6.17) still hold, with the charge density and the polarization replaced by ρ = i N i q Mi ; P = i N i p Mi (6.18) Usually, the molecules that constitute the dielectric material are electrically neutral, i.e q i = 0, and so ρ = 0. However, if there are charges embedded in the dielectric, with charge distribution ρ free, then in this case ρ = ρ free. To summarize, the electrostatic field in a medium is described by the following equations.d( r) = ρ free ( r); E = 0 (6.19) which are known as Maxwell s equations in electrostatics. If we include the quadrupole term, then Eq (6.19) still hold with the displacement given by 85

89 D( r) = 4πk e E( r) + P( r) Q ( r) (6.20) where Q = (6.21) To see if it was a reasonable approximation to ignore the quadrupole terms, we let a 0 be the length scale characterizing the size of a molecule in the dielectric, l the typical spacing between molecules, and L the length scale of the space that we are averaging over, i.e. ( V) 1/3. Then, the magnitude of the dipole moment is p a 0, and so the polarization is roughly P a 0 l 3 (6.22) Now let us calculate the expression of the macroscopic electrostatic potential of a dielectric. If we assume that the molecules in the substance are neutral, the first term in Eq(6.12) vanishes, and we deduce that 48 Using integration by parts yields ( ) Φ( r) = k e P( r ). 1 d 3 r (6.24) V D r r Φ( r) = k e V D [ ].P( r ) d 3 r + r r S P( r ).ˆn ds (6.25) r r where ˆn is the outward unit normal vector to the surface S that is enclosing the volume V D. We can recast (6.25) in the form where Φ( r) = k e V D ρ b ( r ) r r d3 r + S σ b ( r ) ds (6.26) r r 48 In general, if the molecules were charged and some external charge distribution were added to the dielectric substance, the potential reads Φ( r) = k e V D [ N ( r )q M ( r ) + ρ ext ( r ) r r + P( r ). ( 1 r r )] d 3 r (6.23) 86

90 ρ b ( r) =. P( r); σ b ( r) = P( r). ˆn (6.27) This means that ρ b and σ b act like a volume and surface charge densities, respectively, even though we have considered a neutral dielectric medium with no free standing charges. Thus, the electrostatic potential, and so the electric field, produced by a polarized dielectric is the same as that produced by a bound charge density ρ b =.P and a bound surface charge density σ b = P.ˆn. Example 6. 1 Consider a cylindrical dielectric of length L and radius R, with a constant polarization P along its axis of symmetry, call it z-axis. So In this case the volume charge density vanishes (see Eq (6.27)). The electric field at a point along the z-axis within the dielectric is given by E(z) = k e S (P.ˆn) (zẑ r ) ds zẑ r Choosing the base of the cylinder to be in the xy plane, we can re-write the above integral as 2π R 2π R P (zẑ ρˆρ) P [(z L) ẑ ρˆρ] E(z) = k e ρdρdφ + K 0 0 (z 2 + ρ 2 3/2 [ ) 0 0 (z L) 2 + ρ 2] ρdρdφ 3/2 = (2πK) P ẑ z z2 + R z L 2 (z 2 L) 2 + R 2 Here ρ = (x 2 + y 2 ), and (ˆρ = xˆx + yŷ) /ρ. We see that inside the cylinder the first two terms inside the bracket are smaller than one, and hence the electric field will be pointed in the negative direction of the z-axis. The polarization of molecules at a point within a small volume V, is related to the electric field, E, in their neighborhood, and it is expected to vanish in the absence of external field. Experimentally it has been found that in the regime of weak electric field, the polarization is linear in the electric field 1 P i = 3 j=1 χ ije j ; (6.28) 1 One should keep in mind that there are a number of crystals where contribution of the non-linear terms in the electric field can not be neglected. 87

91 Material Dielectric Constant Vacuum 1 Dray air Water Vapor (100 0 ) Teflon 2.1 Benzene 2.28 Polystyrene 2.56 Paper 3.7 Pyrex glass 5.6 Diamond 5.7 Porcelain 6 Water 80 where the indices i, and j label the coordinate axis x, y, and z, P i denote a component of the polarization, and χ ij is the electric susceptibility tensor. For homogeneous materials, χ ij is constant. If in addition, the dielectric is isotropic, then and so we have χ ij = χ e δ ij (6.29) P = χ e E (6.30) Substituting the above expression in Eq (6.17), we can write the displacement vector as D = ɛe; ɛ = 1 4πk e (1 + χ e ) (6.31) where the quantity ɛ is called the dielectric constant of the material. Thus, inside a homogeneous and isotropic dielectric, the differential form of Gauss s law in the cgs unit system reads.e = ρ free ɛ (6.32) So the electric field due to some charge distribution, ρ free, inside the dielectric can be obtained by rescaling the the electric field due to ρ in the vacuum but by a factor of 1/ɛ. Thus, by knowing the dielectric constant we can solve for the macroscopic fields in medium. In the table below we list the dielectric constant of some of the substances. 88

92 All the expressions derived above are given in the Gaussian system of units. In the MKS system, Eq (6.19) reads where the electric displacement D (SI) is defined as.d (SI) (x) = ρ free (x) (6.33) D (SI) = ɛ 0 E + P; (6.34) For an isotropic dielectric, the constant of proportionality between the polarization and the electric field is defined by the relation P = χ (SI) e ɛ 0 E (6.35) where χ (SI) e = 4πχ e However, instead of using the dimensionless constant χ e, it is more comment to use the parameter ɛ r := 1 + χ (SI) e ; (6.36) which is called the relative permittivity or the relative dielectric constant 49. With this definition, we can write the polarization in the form Thus, substituting the above expression in (6.38), yields P = (ɛ r 1) ɛ 0 E (6.37) D (SI) = ɛ (SI) E (6.38) where ɛ (SI) = ɛ 0 ɛ r is the permittivity of the dielectric in the MKS 50. Thus, combining (6.38) and (6.33), we find that in the SI units, the differential form of Gauss s law in the dielectric with free charge distribution ρ free reads.e = ρ free ɛ (SI) (6.39) As it was argued before, in the weak field approximation, the dipole of each molecule is 49 Note that the relative permittivity is equal to the dielectric constant in the Gaussian system of units, i.e. ɛ r = ɛ (cgs). 50 Note that 89

93 proportional to the applied electric field E, which in the SI units reads p M = 4πɛ 0 α M E, where is molecular polarizability. So we have P = 4πɛ 0 α M ne ɛ r = 1 + 4πα M n (6.40) To get an idea about the magnitude of ɛ r, we will use a crude model to represent a dielectric material. We assume that it consists of a set of identical conducting spheres of radius a, separated from each other by a distance large enough so that n << 1/a 3. This means that their dipole moments can be calculated independent of each other. In example 4.3 we have already computed the potential of a conducting sphere in a uniform electric field E 0 along the z-axis, and found that the potential outside the sphere is (in SI units) Φ(r, θ) = E 0 r + E 0 a 3 cos θ (6.41) r2 The first term describes the applied external field, whereas the second term describes the field of the surface charge induced on the sphere by E 0. Thus, the potential a produced by the the sphere itself is Φ(r, θ) = E 3 0 cos θ, which may be written as r 2 Φ induced = p.r r 3 (6.42) with p = 4πɛ 0 E 0 a 3 ẑ. Thus, α M = a 3, and ɛ r = 1 + 4πa 3 n. Now, we can use this expression to get a rough estimate of the relative dielectric constant of air at normal atmospheric pressure P Pa, and at a temperature T 300 K. At these conditions, the density of the air molecules can be calculated from the equation of state of an ideal gas, i.e. n air = P 2.5 k B T 1025 m 3. Taking the size of a molecule in the air to be of order m, we obtain ɛ r 1, which is consistent with the value given in the table Boundary Conditions for E and D at medium interfaces Let us now consider the boundary conditions of the electric field at the interface between two materials, of dielectric constants ɛ 1 and ɛ 2. Since E = 0, then applying Stokes s theorem implies that E.dl = 0 (6.43) C 2 A more accurate estimate consider R = m, which is Vander-Waals radius of the main component of the air, the nitrogen molecule N 2. This value yields ɛ r

94 where C is a closed curve. Taking C to be a small rectangular loop through the boundary surface between the two materials, with infinitesimal width, we get E 2 E 1 = 0 (6.44) where the superscript means parallel to the surface. Thus, the tangential components of the E is continuous at the boundary. The above equation can be rewritten in the equivalent form ˆn (E 2 E 1 ) = 0; (6.45) where ˆn is the unit normal vector at the interface. Now, let us choose a closed surface S in the form of a small cylinder which goes through the interface between the two medium. Gauss s law in (6.16) can be written as S Dds = 4π k e q free (6.46) where q free is the net free charge inside S. Taking the height of the cylinder to be infinitesimally small, the above equation reads ˆn. (D 2 D 1 ) = 4πk e σ free ; (6.47) with σ free being the free surface charge density at the interface. Hence, in the presence of free charges on the boundary surface the normal component of D is discontinuous. Note that at the interface between a conductor and a dielectric the boundary conditions are E Interface = 0; D Interface = 4πk e σ free ; (6.48) where E and D represent the electric and the displacement fields in dielectrics, where as they vanish inside the conductor. Example 6. 2 Consider a capacitor made of two parallel plates with charges +Q and Q on each and with dielectric slab in between them. 91

95 6.4 Boundary-value problems in dielectrics 1. Point charge between two dielectrics: 2. Slab of dielectric in a uniform electric field: 3. Dielectric sphere in a uniform electric field: Consider a dielectric sphere of radius a and dielectric constant ɛ s, placed inside a medium of dielectric constant ɛ m and subject to uniform a electric field E 0 directed along the z-axis. Since there are no free charges, the scalar potential satisfies Laplace s equation Thus for r < a, we have 2 Φ = 0 (6.49) Φ < (r, θ) = A l r l P l (cos θ) (6.50) l=0 and, for r > a, Φ > (r, θ) = ( Bl r l + C l r (l+1)) P l (cos θ) (6.51) l=0 where we have omitted terms like r (l+1) from the expansion of Φ < (r, θ) since they diverge as r 0. The continuity of the tangential and the normal components of the electric and displacement fields, given in (6.45) and (6.47) (with σ f = 0), yield Φ < θ r=a = Φ > θ r=a (6.52) ɛ s Φ < r r=a = ɛ m Φ > r r=a (6.53) Moreover, at distances far away from the sphere the electric field reduces to E 0, which means that lim Φ(r, θ) E 0r cos θ = E 0 r(cos θ) (6.54) r 92

96 This condition implies that B 1 = E 0, and B l 1 = 0. Equation (6.52) can be integrated over θ to give or, equivalently, Φ < (r = a) = Φ > (r = a) (6.55) E 0 a cos θ + l=0 C l a (l+1) P l (cos θ) = l=0 A l a l P l cos θ (6.56) For the equation above to be valid for all values of θ, we must have and E 0 a + C 1 a 2 = A 1a (6.57) A l = C l ; l 1 (6.58) a2l+1 Note that this matching of the coefficients of P l (cos θ) was possible due to the orthogonality of these functions. The boundary condition (6.53) yields ( A 1 ɛ s = E 0 a 2 C ) 1 ɛ a 3 m (6.59) C l la l ɛ s = (l + 1) a ɛ m; (2l+1) l 1 (6.60) Equations (6.58) and (6.62) leads to A l = C l = 0 for l 1. The combination of (6.57) and (6.59) gives ( ) 3ɛm A 1 = ɛ s + 2ɛ m C 1 = a 3 ( ɛs ɛ m ɛ s + 2ɛ m E 0 (6.61) ) E 0 (6.62) So, the potential of a dielectric sphere of dielectric constant ɛ s immersed in a medium with dielectric constant ɛ m is given by ( ) 3ɛm Φ < (r, θ) = E 0 r cos θ, (6.63) ɛ s + 2ɛ m ( ) Φ > (r, θ) = E 0 r cos θ + E 0 a 3 ɛs ɛ m cos θ ɛ s + 2ɛ m r. (6.64) 2 93

97 Note that (6.64) shows that the electric field in the interior of the sphere is uniform and of magnitude ( ) 3ɛm E(r < a) = E 0 (6.65) ɛ s + 2ɛ m The electric field outside the sphere is a combination of the constant external field E 0 and that of a pure dipole of moment E 0 a 3 (ɛ s ɛ m ) /(ɛ s + 2ɛ m ). The polarization is given in terms of the scalar potential as follow 51 Inside the sphere, this gives P = ɛ 1 4π E = ɛ 1 4π Φ (6.66) P(r < a) = 1 3ɛ m (ɛ s 1) E 0 (6.67) 4π ɛ s + 2ɛ m and so the polarization surface charge density on the sphere is σ = P(r < a).ˆr = 1 3ɛ m (ɛ s 1) E 0 cos θ (6.68) 4π ɛ s + 2ɛ m which is positive on the upper hemisphere and negative on the on the lower one. 6.5 Clausius-Mossotti Relation Consider a region of a dielectric subject to an external electric field E ext. In the previous subsection, the expression of the electric dipole moment of a molecule and the one of the polarization of the dielectric were given by p M = [4πɛ 0 ]α M E; P = [4πɛ 0 ]χ e E (6.69) However, there is a subtlety in the above relations. The electric fields in the definition of p M and the one in P are not the same. The one in the expression of dipole moment of the molecule, is a sum of E ext and the electric field due to the induced dipole itself. On the other hand, E in the formula of the polarization, includes both E ext and the electric field due to the dipoles that were averaged in P. As we will show, in the approximation where the molecular density is very low, the electric fields in both expressions coincide, and in that case the relation between the macroscopic quantity 51 The expression in SI units is given in (6.37) 94

98 ɛ r and α M (the microscopic quantity) given in Eq (6.40) is valid. Therefore, in general we should define two different electric fields as follows: p M = [4πɛ 0 ]α M E loc ; P = [4πɛ 0 ]χ e E (6.70) where E loc is the electric field at the site of the molecule due to the external field and the field of the surrounding dipoles, and E is the average field in the medium which is a combination of E ext and the filed sue to the polarization of the charges in the dielectric. If the density of molecules in the dielectric material is n, then P = np M, and so p M = [4πɛ 0 ]nα M E loc (6.71) Thus, if we can relate the fields E loc and E, then we can find a relation between dielectric constant and the molecular polarizability. 6.6 Electrostatic energy in dielectric media 6.7 Application of the method of images to dielectric media 95

99 Exercise 1 7 Exercises on Electrostatics Suppose the electric field in some region of space is found to be E = kr 3ˆr, where k is some constant. (i) Find the charge density ρ. (ii) Find the total charge contained in a sphere of radius R, centred at the origin. Exercise 2 (Qualifying exam, University of Wisconsin) Suppose that, instead of the Coulomb s force law, one found experimentally that the force between any two charges q 1 and q 2 was F 12 = k e q 1 q 2 ( 1 αr12 ) r 12 ˆr 12 where α is a constant r 12 is the distance between the two charges. (i) What is the expression of the electric field created at some distance r by a point charge q. (ii) Choose a path around this point charge and calculate the line integral E.d l. Compare with Coulomb result. (iii) Find E.d A over a spherical surface of radius r 1 with the point charge at this centre. Compare with the Coulomb result. (iv) Repeat (c) at radius (r 1 + ) and find.e at a distance r 1 from the point charge. Compare with the Coulomb result. Note that is a small quantity. Exercise 2 (Qualifying exam, SUNY, Buffalo) A static electric charge is distributed in a spherical shell of inner radius R 1 and outer radius R 2. The electric charge density is given by ρ = a1br, where r is the distance from the centre, and zero everywhere else. (i) Find an expression for the electric field everywhere in terms of r. (ii) Find expressions for the electric potentials and energy density for r < R 1. Take the potential to be zero at r. Exercise 4 Four identical particles with charge q are placed at the corners of a square of side a. A particle of charge q and mass m is placed at the center of the square, and constrained to move only in the plane of the square. Show that, if displaced slightly from the centre and released, it will have simple harmonic motion and find its frequency. 96

100 Exercise 5 A charge q sites at the back corner of a cube. What is the flux of the electric field E through the side parallel to the xy plane? Exercise 6 (Qualifying exam, University of Berkeley) A sphere of radius R 1 has charge density ρ uniform within its volume except for a small spherical hollow region of radius R 2 located a distance d from the centre. (i) Find the electric field E at the centre of the hollow sphere. (ii) Find the potential φ at the same point. Exercise 7 Find the electric field at a height z above the centre of a square sheet (of side a) carrying a uniform surface charge density σ. Check your result when a and z >> a. Exercise 6 (a) The Uranium nucleus has Z = 92 protons and N = 146 neutrons uniformly distributed over a radius R m. Calculate, in the SI units, the total electrostatic energy of the Uranium nucleus. (b) A symmetric U fission creates two P d daughter nuclei (not that fission actually prefers to be symmetric). Using the fact that nuclear radii obey R A 1/3 where A = (Z + N), find the electrostatic energy of daughter nuclei compared to U. Exercise 8 The time-averaged potential of a neutral hydrogen atom is given by: Φ = q e αr ( 1 + αr ) 4πɛ 0 r 2 where q is the magnitude of the electronic charge, and α 1 = a 0 /2, with a 0 being the Bohr radius. Find the distribution of charge (both continuous and discrete) that will give rise to this potential. Exercise 9 Suppose that the electric field of a point charge q i were E = q ˆr, δ << 1 r2+δ rather than having the Coulomb form E = qˆr/r 2. 97

101 (a) Calculate. E and E for r 0. Also, find the potential for such a point charge. (b) Two concentric spherical conducting shells of radii a and b < a are joined by a thin conducting wire. If a charge Q a resides on the outer shell, find the charge on the inner shell. Exercise 10 Two uniform infinite sheets of electric charge densities +σ and σ intersect at right angles. Find the magnitude and direction of the electric field everywhere. Exercise 11 Demonstrate the Earnshaw s Theorem which states that "A charge can not be held in equilibrium solely by an electrostatic field". Exercise 12 Show that the mean value of the potential over a spherical surface is equal to the potential at the center, provided that no charge is contained within the sphere. Exercise 13 (problem 2.5 form Jackson) (a) Calculate the work done to remove the charge q from a distance r > a to infinity against the force, Eq(3.27) in the notes, of a grounded conducting sphere. Relate your result to the electrostatic potential given in Eq(3.23) in the notes and the discussion on the electrostatic potential energy in section 1.7 of theses notes. (b) Repeat the calculate of the work done to remove the charge q from a distance r > a to infinity against the force, Eq(3.31) in the notes, of a an isolated charged sphere. Relate your result to the electrostatic potential given in Eq(3.29) in the notes and the discussion on the electrostatic potential energy in section 1.7 of theses notes. Exercise 13 An electric dipole of strength p is oriented perpendicular to and at a distance d from an infinite conducting plane. Calculate the force exerted on the plane by the dipole. Exercise 14 A capacitor is made of two concentric cylinders of radii R 1 and R 2 (R 1 < R 2 ) and length L >> R 2. The region between R 1 and R 3 = R 1 R 2 is filled with a 98

102 cylinder of length L and dielectric constant κ e ɛ/ɛ 0 ; the remaining volume is an air gap of dielectric constant unity. (a) Determine the capacitance of the system. (b) calculate the dipole moment p in the dielectric (R 1 < r < R 3 ) and in the air-gap (R 3 < r < R 2 ) expressed in terms of the potential difference V between R 1 and R 2. (c) How many work needs to be done to remove the dielectric cylinder while maintaining this constant potential. Exercise 15 Find the Green s function of the 1-dimensional operator d2 dx 2. Exercise 16 Find the potential V (r) such that its value on a sphere of radius R is V (R) = V 0 cos 4 θ where V 0 is a constant. Exercise 17 A thick spherical shell,with inner radius a and outer radius b, is made of dielectric material with a polarization p( r) = α ˆr (7.1) r where α is a constant and r is the distance from the center. We assume that there are no free charges in the dielectric material. find the electric field in the three regions: r < a, a < r < b, and r > b. Exercise 18 Consider a line charge of length 2l and charge density A z δ(x) δ(y), z [ l, l]; ρ( r) = 0, elsewhere. where A is a real constant with unit of charge per unit area. Find its dipole moment. 99

103 Exercise 19 Consider a hemi-spherical shell with the polar angle θ subtends 0 cos θ 1, of radius R and has a total charge q which is uniformly distributed over its surface. (a) Calculate the dipole and the quadrupole moments. (b) Suppose a point charge q is placed at the position (0, 0, R). Find its dipole, quadrupole, and octupole moments. Exercise 20 A hollow spherical shell of radius R is made of insulating material, and has a surface charge distribution σ given by σ(θ, φ) = σ 0 (cos θ + 1 sin θ cos φ) 2 (a) Find the potential for r < r and R > r. (b) Determine the dipole moment in the cartesian coordinates. (c) Find the electric field for r < R in cartesian coordinates. Exercise 21 The potential at the surface of a sphere of radius R is given by V (R) = k cos θ where k is a constant. (a) Find the potential inside and outside the sphere. (b) Determine the charge density σ(θ, φ). Exercise 22 A very long, right circular, cylindrical shell of dielectric constant ɛ and inner and outer radii a and b, respectively, is placed in a previously uniform electric field E 0 with its axis perpendicular to the field. The medium inside and outside the cylinder has a dielectric constant ɛ 0. Determine the potential and the electric field in the three regions. Exercise 23 A slab material with dielectric constant ɛ is partially inserted in a square parallel plate capacitor. The plates have length L apart a distance h, and each has a charge Q. 100

104 (a) Assuming a small gap thickness, find the force on the dielectric if the capacitor is isolated. (b) What changes if instead the capacitor is held at constant voltage V 0. Exercise 24 Apply the Neumann boundary condition to a concentric sphere with inner and outer radii a and b, respectively, in which the normal electric field is E r (r = b) = E 0 cos θ; E r (r = a) = 0 (a) Calculate the electrostatic potential inside the volume V between the concentric surfaces r = a and r = b.. (b) Fin the components of the electric field in the spherical coordinates, i.e. E r and E θ. (c) Calculate the components of the electric field in the cylindrical coordinates, i.e. E z and E ρ. 101

105 Exercise 1 8 Answers to the Exercises on Electrostatics (a) The charge density is ρ(r) = 5 k 4πk e r 2 (b) The total charge contained in a sphere of radius R, centered at the origin is Exercise 2 The frequency of the oscillation is Q total = k k e R 5. (8.1) 2q 2 ω = 2πɛ 0 a 3 m Exercise 3 The flux of the electric field through the side parallel to the xy plane is E. d s = q. S xy 24ɛ 0 Exercise 4 Exercise 5 (a) the total electrostatic energy of the Uranium nucleus in SI is given by U E = 3 20πɛ 0 (Ze) 2 R J. (b) the electrostatic energy of daughter nuclei compared to U U (daughter) E = 0.63U E 460 MeV. Hence, electrostatic energy released in fission is about 270 M ev, compared the observed release of 200 MeV. Exercise 6 The distribution of charge is ) ρ(r) = q (δ (3) ( r) α3 8π e αr = q ( ) 2δ( r) α 3 e αr. 8π r 2 This can be interpret it as follows. The first term corresponds to the proton charge, and the second term corresponds to the negatively charged electron cloud of the 1 s orbital around the proton. 102

106 Exercise 7 (a) The divergence and the curl of the electric field at r 0 are given by The potential is. E = q δ ; E = 0. r3+δ Φ(r) = 1 q (δ + 1) r. δ+1 (b) The the charge on the inner shell is given in terms of Q a as Exercise 8 Q b Q aδ [2b ln (2a) (a + b) ln (a + b) + (ab) ln (a b)]. 2(a b) The magnitude of the electric field everywhere is Exercise 11 E = 1 2 σ ɛ 0 The force exerted on the plane by the dipole is F = 3p2 8d 4 ˆx. Exercise 12 (a) The the capacitance of the system is ( ) ( ) 4πɛ 0 κe C =. ln R 2 R 1 κ e + 1 (b) The dipole moment in the dielectric (R 1 < r < R 3 ) and in the air-gap (R 3 < r < R 2 ) are given ( ) (κe ) 2ɛ p (R 1 < r < R 3 ) = ln R 2 R 1 κ e + 1 ρ êρ, p(r 3 < r < R 2 ) = 0. (c) The mechanical work that needs to done is ( W = ɛ 0 ln R 2 R 1 ) (κe ) 1 V 2. κ e

107 Exercise 13 (problem 2.5 form Jackson) (a) The work done to remove the charge q from r > a to infinity of a grounded conducting sphere is given by W = k e q 2 q 2 2(r 2 a 2 ) (8.2) (b) The work done to remove the charge q from r > a to infinity of an isolated charged sphere is given by [ q 2 q 2 W = k e 2(r 2 a 2 ) q2 a 2 2r qq ] (8.3) 2 r Exercise 13 The Green s function of the 1-dimensional operator d2 dx 2 G(x, x ) = 1 2 x x. is Exercise 14 The potential V (r) is V (r < R) = 8V ( 0 r ) 4 35 P 4V ( 0 r ) 2 4(cos θ) + R 7 P V 0 2(cos θ) + R 5, V (r > R) = 8V ( ) 5 0 R 35 P 4(cos θ) + 4V ( ) 3 0 R r 7 P 2(cos θ) + V 0 r 5 Exercise 15 The electric fields in the three regions are Exercise 16 E(r < a) = 0, E(a < r < b) = α ɛ 0 r ˆr, E(r > b) = 0. The electric dipole moment of this line charge is ( ) R. r p x = p y = 0, p z = A 2l3 3. Exercise

108 (a) The dipole and the quadrupole moments are given by p = 1 2 qrẑ, Q ij (2) = 0. (b) dipole, quadrupole, and octupole moments for a point charge q at position (0, 0, R) are Exercise 18 p = 1 2 qrẑ, Q ij (2) = qr2 2 (3δ i3δ 3j δ ij ), Q ijk (3) = qr2 2 ( 3δ i3δ jk + 5δ i3 δ j3 δ k3 ). (a) The potential for r < r and R > r is given by V (r < R) = 1 σ(θ, φ)r, 3 V (r < R) = 1 σ(θ, φ)r3 3 r. 2 (b) The dipole moment in the cartesian coordinates is given by p = 1 3 σ 0R 3 (ẑ ˆx). (c) The electric field for r < R in cartesian coordinates is E = 1 3 σ 0(ẑ ˆx). Exercise 19 (a) The scalar potential inside and outside the sphere is given by V (r < R) = 3k ( r 5 P 1(cos θ) R) V (r > R) = 3k 5 P 1(cos θ) (b) The charge density σ(θ, φ) is given by + 8k ( r ) 3 5 P 3(cos θ), R ( ) 2 R + 8k r 5 P 3(cos θ) σ(θ, φ) = ɛ 0 9k 5R P 1(cos θ) + ɛ 0 56k 5R P 3(cos θ). ( ) 4 R. r 105

109 Exercise 20 Applying the matching conditions on the fields D n and E t (equations (??) and (??) at ρ = a and ρ = b, one finds that the scalar potential in the three regions has the form Dρ cos φ, ρ < a); Φ = (Bρ + Cρ 1 ) cos φ, a < ρ < b; (Aρ 1 E 0 ρ) cos φ. ρ > b. The electric field can obtained by taking the gradient of Φ, i.e E ρ = Φ D cos φ, ρ < a); ρ = ( B + Cρ 2 ) cos φ, a < ρ < b; (Aρ 2 E 0 ) cos φ. ρ > b. and E φ = 1 Φ ρ φ = Dρ sin φ, ρ < a); (B + Cρ 2 ) sin φ, a < ρ < b; (Aρ 1 E 0 ρ) sin φ. ρ > b. The constants A, B, C, and D are determined from the matching at ρ = a and ρ = b, and they are given by A = E 0 b 2 (1 ɛ [ 0/ɛ) 1 (a/b) 2 ] [ (1 + ɛ 0 /ɛ) ( ) ɛ0 /ɛ 1+ɛ 0 /ɛ (a/b) 2] ɛ 0 /ɛ B = 2E 0 (1 + ɛ 0 /ɛ) [ 1 ( 1 ɛ0 /ɛ 1+ɛ 0 /ɛ 1 ) 2 (a/b) 2] (1 ɛ 0 /ɛ)ɛ 0 /ɛ 1 C = 2E 0 [ (1 + ɛ 0 /ɛ) 2 ( ) ɛ0 /ɛ (a/b) 2] 1+ɛ 0 /ɛ (ɛ 0 /ɛ 1 D = 4E 0 [ (1 + ɛ 0 /ɛ) 2 ( ) ɛ0 /ɛ (a/b) 2] 1+ɛ 0 /ɛ Exercise 21 (a) The force on the dielectric in the case where the capacitor is isolated is F = 1 Q 2 ɛ 0 L (ɛ 1). 2 C 2 h (b) One has to include the work done b the battery to keep the voltage constant. In this case the force on the dielectric is F = dw mec = 1 dx 2 V 0 2 dc dx = 1 2 V 0 2 ɛ 0 L (ɛ 1). h 106

110 Exercise 22 (a) (b) E r = E 0 1 p 3 ) r cos θ Φ( r) = E 0 (1 + a3. 1 p 3 2r 3 (1 a3 r 3 ) ) r sin θ ; E θ = E 0 (1 + a3. 1 p 3 2r 3 (c) ) cos θ E z = E 0 (1 + a3 (ρ 2 2z ) ; E 1 p 3 2(ρ 2 + z 2 ) 5/2 ρ = 3E ( 0 1 p 3 a 3 zρ 2(ρ 2 + z 2 ) 5/2 ). (8.4) 107

111 9.1 Introduction Some History of Magnetism... 9 Magnetostatics There is a fundamental difference between electricity and magnetism... No magnetic monopole... In the 16th century William Gilbert discovered that the Earth is a huge magnet with the lines of force field are directed not towards the geographic poles but rather towards the magnetic poles (The magnetic north pole is located near the geographic south pole.). They are also directed (except at the equator) towards or away from the center of the earth 52. In this chapter we will study magnetostatics in vacuum. This is does not mean that charges are static but rather the magnetic field and electric current are constant in time. In the next section we will discuss the continuity equation, then in section three we define magnetic forces and give an example of motion of a charged particle in a crossed electric and magnetic field. 9.2 Current and Continuity Equation In a conductor, there are electrons which are free to move when an external electric field is applied. The amount of charge flowing through a point in the wire per unit time is called current. It has unit of Coulomb-per-second, which is called ampers (A). For a line charge distribution of density ρ moving in the wire with a (drift) velocity vector v, the amount of charge δq crossing a small surface δ s = δsˆn, where ˆn is the unit vector perpendicular to to the surface ds, during a time δt, is δq = ρ ( vδt).δ s, which can be written as where δq = J. δ s (9.1) J = ρ v (9.2) is called the electric current density, which is a vector field in the direction of the flow and has unit of Coulomb-per-square meter-per-second. Thus a current crossing a surface S is 52 The intensity of the field at the surface is on the order of one Gauss. Sediments of magnetic materials (iron, cobalt, nickel) can drastically change the local pattern of this field which has been carefully mapped, most recently with the use of satellites. 108

112 I = S J. d s (9.3) It is worth noting that for a linear charge distribution λ passing along a thin filament, the current is a vector quantity and it is given by I = λ v (9.4) Now, suppose that S is a closed surface bounding a volume V. Then the total charge charge in V is Q = ρ( r) d 3 r (9.5) S V Since the total charge is conserved (this is an experimental fact), the rate at which the charge "passing out of" S must equal to the "minus" of the rate of change of Q, i.e. J. d s = dq dt = ρ t d3 r (9.6) Using the divergence theorem to Eq (9.6), we get ( ) ρ t +.J d 3 r = 0 (9.7) V For the above equation to be true for arbitrary l volume V, we must have V ρ t +. J = 0 (9.8) The above equation states that an increase in charge density can only be achieved by having more current arrive at the point than leaves it. Thus, it represents the local charge conservation, and it is known as the continuity equation. In magnetostatics, we require that ρ/ t = 0, which corresponds to a study state. Thus, in magnetostatics the conservation of current leads to which is the definition of steady current.. J = 0 (9.9) 109

113 9.3 Magnetic Force In 1820, the Danish Physicist Hans Christian Oersted ( ), showed that magnetic effects could be produced by an electric current flowing in a circuit. He observed that when placing a compass over a wire that had electrical current, its needle moved in one direction, and when the it is placed beneath the wire, the needle moved in the opposite direction. The needle was neither attracted to the wire nor repelled from it, instead, it stands at right angles. About the same time, Andre Marie Ampere realized that if a current in a wire exerted a magnetic force on a compass needle, two such wires also should interact magnetically. He found that when considering two parallel straight wires in which current in both is flowing in the same direction, the wires are found to attract each other. However, if the currents are flowing in opposite direction, the wires are found to repel each other. The force responsible for this effect is called the magnetic force. These experimental facts can be understood if we postulate that moving charges generate a magnetic field B which acts with on a charge q moving with velocity v with a force, given in SI units by 53 F mag = q ( v B ) (9.10) which can be taken as the definition of the magnetic field in a similar way as we defined the electric field from the observed Coulomb s force in electrostatic. The SI unit of magnetic field is the tesla (T), defined as 1 T = 1 Newton second Coulomb meter (9.11) In the cgs system of units, the magnetic field unit is the gauss (G), where 54 1 T = 10 4 G (9.12) Note that the magnetic force force is very different then the Coulomb s force in that it depends on the velocity of the particle and it is perpendicular to the magnetic field. 53 In the Gaussian system of units, there is a factor of 1/c in front of the velocity, i.e. ( ) v F (Gaussian) mag = q c B 54 Compare with the Earth s magnetic field of about 10 4 T, a bar of magnet has a typical magnetic field of about 10 2 T, in MRI is about 1 T, where as the superconducting magnets used at the Large Hadron Collider (LHC), in Geneva, have a magnitude of about 8 T, and neutron stars can have magnetic field of order 10 8 T. 110

114 Hence, the magnetic field does no work on a moving charge 55. Thus, the speed of the particle will not change, and the effect of the magnetic force on a moving particle is to bend its trajectory 56. As an example, consider a particle of mass m and charge q moving with initial some velocity v 0 which enters a region where there is a static uniform magnetic field B = B 0 ẑ. Then, the equations of motion is m d v ( dt = q v B ) Multiplying both sides of the equation by v, yields (9.13) m v. d v dt = d ( ) 1 dt 2 m v2 = 0 (9.14) Consequently, the kinetic energy, and thus the speed, remain constant. Writing (9.20) in terms of components read dv x = ω c v y (9.15) dt dv y dt = ω cv x dv z dt = 0 where we have introduced the so called cyclotron frequency 57 ω c = qb m (9.16) The above system of equations can be solved to give v x (t) = v 0 cos ω c t (9.17) v y (t) = v 0 sin ω c t v z = v 0 55 dw magn = F mag.d ( l = q v B ). vdt = The speed remain constant if the magnetic field is uniform, i.e. does not depend on time, otherwise, as we will see later on these notes, a time dependent magnetic field, generate a an electric field, which accelerates the particle. 57 Note that the cyclotron frequency is given in terms of the strength of the magnetic field, the charge of the particle, its mass, and independent of its speed. 111

115 where we have chosen the origin of time of our coordinate system so as to eliminate any constant offsets in the above equations. Here v0 and v 0 are the perpendicular and the parallel components of the initial velocity, respectively. The time-integration of the the velocity components yields x(t) = v 0 ω c sin ω c t + x 0 (9.18) y(t) = v 0 ω c cos ω c t + y 0 z(t) = v 0t + z 0 where x 0, y 0 and z 0 represent the initial position of the particle. Thus, we see that the projection of the trajectory in the x-y plane is given by ( ) v (x x 0 ) 2 + (y y 0 ) 2 2 = 0 (9.19) which corresponds to a circle of radius R L = v 0 /ω c = mv 0 / q B, which is known as Larmor radius (or cyclotron radius) 58. Since the the velocity along the direction of the magnetic field is constant, the trajectory is a helix. If in addition to a magnetic field, B, there is an electric field, then the equation of motion reads m d v ( ) dt = q E + v B (9.20) ( ) where the force F = q E + v B is called Lorentz force. Assume that the particle is initially at rest, and that the electric and magnetic fields be perpendicular to each other, say E = E 0ˆx and B = B 0 ẑ, with E 0 and B 0 are time-independent. As the electric field exerts a force on the particle, it acquires a velocity in the direction of E, and the magnetic force now start to act on it.. As there is no component of the force along the z-direction, the velocity of the particle remains zero in this direction. Hence, the motion occurs in the x-y plane. Writing Eq (9.20) in terms of the x and y components of the velocities, and combining them, gives ω c v x (t) = A sin ω c t + B cos ω c t (9.21) v y (t) = E B + A cos ω ct B sin ω c t where A and B are constants of integration. Using the initial conditions v x (0) = v y (0) = 0, we get B = 0, and A = E/B. Hence, we obtain 58 In some textbooks it is called gyroradius. 112

116 v x (t) = E B cos ω ct (9.22) v y (t) = E B (cos ω ct 1) If we assume that initial the particle was at the origin (i.e. x 0 = y 0 = z 0 = 0), after integration the equations above, we find that the trajectory of the particle reads ( ) 2 ( ) ( x E Bω c + y Et 2 B = ) 2 E Bω (9.23) c which is the equation of a circle of radius R = E/Bω c, whose centre is moving in the negative direction of the y-axis with a constant speed v center = E/B. This trajectory is the same as that of a point on the circumference of a wheel of radius R, moving along the y-axis with a speed v center. This trajectory is called cycloid. 113

117 Example 9.1: The charge-mass ratio of the electron In 1897, Joseph John Thomson ( ), in his cathode tube (which is now part of the TV tube sets) experiment, demonstrated that the rays emitted from the cathode are charged particles. These were the first "elementary particles" to be discovered, which later were named electrons by other scientists. Thomson used a Crookes tube, which is a sealed glass container in which two electrodes are separated by a vacuum 59. When a voltage is applied across the electrodes, cathode rays are generated, creating a glowing patch where they strike the glass at the opposite end of the tube. He discovered that the rays could be deflected by an electric field. He also subjected the rays to a magnet and found that they bend under the effect of magnetic field. He concluded that these rays, rather than being a form of light, were composed of charged particles he called "corpuscles" which were detached from the cathod. To measure the charge-to-mass ratio of the electrons, he subject them to electric and magnetic fields at the same time. Let consider these cathode rays being particles, electrons, of charge e and mass m e, being emitted from a cathode C and then accelerated toward a slit A due to a potential difference between V between A and C. The change in potential energy is equal to the external work done in accelerating the electrons; and so by energy conservation, the kinetic energy gained is and, therefore, acquire velocity 1 2 m eυ 2 = e V (9.24) υ = 2 V e m e (9.25) When these electrons exit the slit A, they pass through a uniform electric field E between two parallel plates directed in the upward z-direction. The effect of this electric field is deflecting the electrons downward. If in addition to the electric field, a magnetic field B directed out of the the page (i.e in the negative direction of the y-axis) is also applied, then the electrons will experience an additional magnetic force (??) directed in the negative direction of the z-axis. Thomson adjusted the magnetic field such that the electric force and the magnetic force cancel each other, i.e. e E = e υb, which implies that υ = E B (9.26) So, only electrons with velocity given by the equation above will be able to move on a straight line. This experimental setup is some times it is called velocity 114

118 selector. After that, he turned off the electric field, so the particle only feels the magnetic force which is always perpendicular to the velocity of the particles. This gives the central force for the cyclotron motion of the electron and we have or, equivalently, m e υ 2 R = e υb (9.27) e m e = υ BR (9.28) where R is the the radius of curvature of the beam, as deflected by he magnetic field alone. Using the expression of the velocity from (9.26), we obtain e m e = E B 2 R (9.29) The ratio e /m can also be given in terms of the potential difference between the cathode and the anode by substituting the result of (9.25) in equation (9.28), and get e = 2 V m e BR (9.30) We could also combine (9.25) and (9.26), and obtain e m = E2 2 V B 2 (9.31) Therefore, by measuring one the following sets: (E, B, R), ( V, B, R), or(e, B, V ), one can determine the charge-mass ratio. Since Thomson experiment, there has been many measurements of the ratio e /m and It is found to be approximately e m = C/kg (9.32) Thomson was awarded the Nobel Prize for physics in

119 9.4 Biot-Savart law We have seen that currents (charges in motion) are a source of magnetic force, and hence magnetic field. In 1820, Biot and Savart experimentally determined the relation between the current and the magnetic field it creates 60. They found that for a current I flowing through an element segment d l with position vector r, the magnetic field db created at a point P with position vector r has the following properties : It is perpendicular to both d l and the distance vector from dl to P, i.e. ( r r ), db proportional to dl and to I and the angle between d l and ( r r ), db Inversely proportional to the distance r r. which give the following expression for db at point r: where k m = db( r) = k m Id l ( r r ) r r 3 (9.34) µ 0 4π, SI system; 1 c Gaussian system. (9.35) where µ 0 = 4π 10 7 N/A 2 is called the permeability of free space, and c = m/s is the speed of light in vacuum. To convert formulae from the S.I. system to the Gaussian system one can use the following transformation: ɛ 0 1 4π (9.36) µ 0 π c 2 B B c So, to obtain the total magnetic field, one integrates over the whole length of the wire, C,: 60 In the same year, Ampere was studying the force between two current circuits. After many experiments he found that the force that one circuit exert on the other, say C 1 on C 2 is F 2 1 = µ d 0 4π I 1I 2 C 1 C2 [ l2 d ] l 1 ( r 2 r 1 ) r r 3 (9.33) where the indices 1 and 2 label the circuit one and two, respectively, and F 2 1 is the force that. It is straight forward to show that F 1 2 = F 2 1, which is consistent with Newton s third law. He then concluded that all magnetic phenomena are due to currents. In fact, using this expression for the magnetic force between two circuits, one can derive the Biot-Savart law given in (9.44). 116

120 B( r) = k m I C d l ( r r ) r r 3 (9.37) This formula is called Biot-Savart law, which allows the calculation of magnetic field due to a steady current. Note that the magnetic field falls off like 1/r 2, as in the case of electric field of a point like charge, however, there are no experimental evidence for the existence of isolated magnetic monopoles. For a single point-like charge q moving with a velocity υ, there will be a current I associated with its motion. Suppose that this current passes in an infinitesimal element of a cylinder of cross sectional area δa and length δl can be written as I q = qn q Aυ == q δt (9.38) where we used the fact that the number density of a single charge is n q = 1/δAδl and wrote that the speed υ is just δ/δt. Substituting the above expression of the current in (9.34), we find that the magnetic field at a point P is B = µ 0 q υ R (9.39) 4π R 3 where R is the vector which point from the charge to the point P. However, because a point charge does not constitute a steady current, the above equation only holds in the non-relativistic limit, where v << c, so that the effect of "retardation" can be ignored 61. Now, if the conductor is not infinitesimally thin, and has a cross sectional area through which a current passes, then with the use of (9.3), we can write the law of Biot-Savart as 62 B( r) = k m V J( r ) ( r r ) r r 3 d 3 r (9.40) 61 This effect will be discussed in later chapters in these notes. 62 For surface current densities, J s (units A/m), the Biot-Savart law reads B( r) = k m S J s ( r ) ( r r ) r r 3 ds 117

121 Example 9.2: Magnetic field due to a line- current Consider a straight wire of length L, carrying current I. We would like to calculate the magnetic field at a point P a distance h from the wire. The vector dl ( r r ) / r r points out of the page, say the z-axis, and has magnitude dl cos θ. So we have d l ( r r ) r r = d l cos θ ẑ (9.41) where θ is the angle between the distance h and the vector ( r r ). We also have l = h tan θ, and so dl = Substituting these results in (9.44), we get B(h) = k m I = k mi h θ2 θ 1 h dθ (9.42) cos 2 θ ( cos 2 θ h 2 (sin θ 2 sin θ 1 ) ẑ ) ( ) h cos θdθ ẑ (9.43) cos 2 θ To obtain the magnetic field created by infinite line, we take θ 1 θ 2 = π/2, and,in the SI unit system, gives = π/2 and B(h) = µ 0I 2πh ẑ (9.44) Note that for the case of infinite line, the system possesses a cylindrical symmetry, and the magnetic field lines are circular, with the direction determined by the righthand rule by directing the right thumb along the direction of the current in the wire. Example 9.3: Magnetic field due to a circular loop current Consider a circular loop of radius R carrying current I lies in the x-y plane with its center at the origin. We wish to calculate the magnetic field at a point on the z-axis. In cylindrical coordinates, we can write

122 J = Iδ(r R)δ(z ) ˆφ; ( r r ) = z ẑ R ˆr (9.45) So, the expression of the magnetic field in (9.40) reads B(z) = k m Iδ(r r )δ(z )(z ˆr + R ẑ) (R 2 + z 2 ) 3/2 r drdφ dz (9.46) = 2k m πr 2 I (R 2 + z 2 ) 3/2 ẑ where we have used 2π 0 ˆrdφ = 0. We see that the magnetic field is along the z-axis, and this is expected due to the cylindrical symmetry of the system. Furthermore, the 1/z 3 dependence suggests that this a magnetic dipole field. 9.5 Magnetic Force between two currents The magnetic force on an infinitesimal element of length dl of a wire (a very thin filament) carrying a current I is given by df m = ( v B) λdq (9.47) = (I B) dl where λ is the linear density of the charge distribution in the wire. Using the fact that Idl = Id l, we can write the above expression as ( df m = I d ) l B (9.48) Hence, the force on a whole segment L of a wire reads in the SI unit system 64 F m = L I (dl B) (9.49) For a closed loop, C, in which a uniform current is passing through. If the magnetic field is constant (9.50) can be written as F m = I C ( d ) l B (9.50) 64 In the cgs system of units we have F m = 1 c L ( I d ) l B 119

123 Since the set of differential length elements d l form a closed polygone, then their sum is zero, i.e. d l = 0. Hence we conclude that the net magnetic force exerted on a closed loop carrying a constant current in a uniform magnetic field is zero. However, such a loop will experience a torque. For a an infinitesimal element of the loop, the torque relative to the origin reads and so the net torque on the loop is d τ = r df (9.51) For constant magnetic field,... τ = C r (d r B) (9.52) Now let us consider two closed current loops, C 1 and C 2, carrying currents I 1 and I 2, respectively. Then using the Biot-Savart law, the magnetic force that, say, C 1 exerts on C 2 reads F 2 1 = µ d 0 I 1 I 2 4π C 1 C2 [ l2 d ] l 1 ( r 2 r 1 ) (9.53) r r 3 which for steady currents reduces to the expression (9.33) given by Ampere. Using the vector identity for the cross product of three vectors the above equation can be written as A ( B C) = ( A. C) B ( A. B) C (9.54) [ µ d ] l 2. ( r 2 r 1 ) d [ l 1 d l 2.d ] l ( r 2 x 1 ) 0 F 2 1 = I 1 I 2 (9.55) 4π C 1 C 2 x x 3 For a more general case in which the filament has a cross sectional area S, the element of current Idl is replaced by J( r )dlds = J( r )d 3 r, and we obtain Hence, the force density at a point r is F m = J( r ) B( r )d 3 r (9.56) F m ( r) = J( r) B( r) (9.57) 120

124 The total work for the volume element of current J( x )d 3 x with respect to the origin is and so the total torque is d τ = x (J( x ) B( x )) (9.58) τ = V x (J( x ) B( x )) d 3 r (9.59) 9.6 The divergence and the curl of B Computing the magnetic field using Biot-Savart law can be very cumbersome and in some cases not possible analytically. However, according to Helmholtz theorem, if we know the divergence and the curl of a vector field, then it be can determined uniquely given appropriate boundary conditions. For that we will use the vector identity (1/r) = r/r 3 to re-write the magnetic field due to a current distribution J in a volume V given in (9.40) as B = k m V ( ) 1 J( r )d 3 r (9.60) r r where = ( / x, / y, / z ). Using the fact that f( r r ) = f( r r ), we can rewrite the above equation as B = k m However, since the divergence of any curl is zero, it follows that V J( r ) r r d3 r (9.61).B = 0 (9.62) which is the analogue of Gauss s law for the electric field. It also says that the magnetic flux through any closed surface is zero, which implies that there exists no isolated magnetic monopoles. To evaluate the curl of B, we use the vector identity (9.54), and write 121

125 B = µ 0. 4π V = µ 0 4π. = µ 0 4π. ( J( r ). 1 V V r r ( J( r ). 1 r r [ ] d 3 r J( r ) r r ) d 3 r µ 0 4π ) d 3 r µ 0 + µ 0 4π. ( ) 1 J( r ) 2 d 3 r (9.63) V r r ( ) 1 J( r ) 2 d 3 r 4π V r r d 3 r.j( r ) r r µ ( ) 0 1 J( r ) 2 d 3 r 4π r r where is the operator gradient with respect to the coordinates (x, y, z ). The first term in the above equation can be converted to a surface integral, which vanishes for localized current distribution. The second term also vanishes since.j for steady current distribution. The last term can be easily calculated by using the identity 2 (1/r) = 4πδ( r), and we obtain 65 V V B = µ 0 J ( r) (9.64) Using Stokes theorem, we can convert (9.65) to a line integral and get C B. d l = µ 0 S J( r). d s = I enc (9.65) where C is the closed curve enclosing the surface S, and I enc is the current crossing S in the direction of a right-hand normal to the direction in which the contour integral around C is done. This equation is called Ampere s law. In situations where the current distribution posses a high symmetry, Ampere s law provides the easiest and quickest way of calculating the magnetic field, in the same way Gauss s law in electrostatic does for highly symmetrical charge distribution. So in general, given a set of current distributions J( r) and with boundary conditions for the densities and the fields at infinity, one can determine the magnetic field by solving equations (9.62) and (9.65). Example 9.4: Applying Ampere s law to an infinite line current This is an example where one can use Ampere s law to determine the magnetic field. We consider an infinite-length line current I lying along the z-axis. Thus, the magnetic field lines are circles, C r, of radius r in the xy plane. According to Ampere s law, we have 65 In cgs system of units, the curl of B is given by B = 4π c J( r). 122

126 B. d l = C r Bdl = µ 0 I C r (9.66) From the symmetry of the system, the magnitude of the magnetic field is constant on a given path C r, so that So, B dl = B(2πr) = µ 0 I C r (9.67) B = µ 0I 2πr which is in agreement with the result obtained in example 7.2. (9.68) Example 9.5: Applying Ampere s law to a surface current Let us determine the magnetic field produced by a uniform surface current on the xy plane flowing in the ˆx direction, i.e J = J 0ˆx, with J 0 a positive constant. We can apply Ampere s law to this system, and chose the closed path to be rectangular, with its centre at the origin, hight 2h, and with 2w, as shown in the figure below. Then, we have C B. d l = 1 B. d l + 2 B. d l + 3 B. d l + 4 B. d l = I enc (9.69) where I enc is the current enclosed within the rectangular loop. By symmetry, the magnitude of the magnetic field is constant, B 0, on the horizontal segments 2 and 4. We can also show, based on the symmetry of the system, that the magnetic field above the surface current is everywhere in the ŷ direction, whereas below the surface is along the +ŷ direction. So we can write B 0 ŷ, z > 0; B = (9.70) B 0 ŷ, z < 0. This implies that the integrals along the paths 1 and 3 vanish. So, the left.hand side of equation (9.69) reads w w B. d l = B 0 ŷ). ( dyŷ + B 0 ŷ. (dyŷ) = 4B 0 w (9.71) C w w 123

127 Since J is uniform, the enclosed current is I enc = J 0 2w. Putting these results in (9.69), yields B 0 = J 0 2 Therefore, the magnetic field of this system is B = J 0 2 ŷ, z > 0; J 0 2 ŷ, z < 0. (9.72) (9.73) 9.7 Vector potential It is tempting to write the magnetic field as B = Φ m, where Φ m is some scalar field, in analogy with the electrostatic field E which derives from potential. However, the curl of B is generally not zero, and so it can not be expressed as the gradient of some scalar function 66. On the other hand, since the magnetic field is divergence free, it can alway be written as the curl of a vector: B = A (9.74) and we call the vector field A the magnetic vector potential, in analogy with the scalar potential Φ in electrostatics. Comparing the above equation with (9.60) one might deduce that A( r) = µ 0 4π V J( r ) r r d3 r (9.75) However, the relation (9.74) does not define the ( vector magnetic potential uniquely for a given magnetic field, since both A and A + ψ ), with ψ an arbitrary scalar function, will lead to the same B field. The transformation: A A = A + ψ (9.76) is called gauge transformation of the magnetic vector potential. By substituting B = A into (9.65) and using the identity (9.54), we find 2 A (. A) = µ 0 J (9.77) 66 If the current density vanishes, then the magnetic field can be written as the Φ m, and the scalar function Φ m is called the magnetic potential. 124

128 The above equation would of been simpler if the term (. A) vanished. This can be achieved by using the gauge freedom of the vector potential, i.e. if A is not divergence free, we can shift it by ψ, for an appropriate choice of ψ, such that.a vanishes. In general, specifying.a is called choosing the gauge of the vector potential. For instance, the divergence of a generic potential is [.A = V ] J( r ) r r d3 r + ψ( r) (9.78) or, equivalently,.a = V.J( r ) r r d3 r + 2 ψ( r) (9.79) where we used the fact that the operator acting on 1/ r r can be replaced by " ", performed the integration by parts and assumed that the current densities J( r ) are localized in space, i.e. it vanishes on the surface of the volume of the integral at infinity. Since we are considering steady currents (.J = 0), the above equation reads.a = 2 ψ( r) (9.80) So, if we want to make the divergence of the vector potential be some function f( r), then one needs to choose ψ( r) such that it is a solution to the Poison equation 2 ψ( r) = f( r), which has the solution ψ( r) = 1 4π V f( r ) r r d3 r (9.81) Note that one can make gauge transformation without changing the gauge, by just changing the gauge parameter ψ( r) with a function that satisfies Laplace equation. The condition.a = 0 is called the Coulomb gauge. According to (9.78), (9.80), one particular choice of the vector potential in the Coulomb gauge is the one given in Eq (9.75). This result can be also obtained by solving (9.77) in the Coulomb gauge, i.e. which is equivalent to the three Poison s equations: 2 A = µ 0 J (9.82) 2 A i = µ 0 J i (9.83) 125

129 where the index i labels the three cartesian coordinates. However, since.j = 0, not all the three components J i of the vector current density are independent. With the boundary conditions chosen such that at infinity the fields and current densities vanish, then the solution to (9.83) is A i ( r) = µ 0 4π V which yields the Biot-Savart law for the magnetic field. J i ( r ) r r d3 r (9.84) 9.8 Magnetic field of a circular current loop Consider a circular current loop of radius R carrying a current I. This is an example of a system which does not have enough symmetry to be able to determine the magnetic field directly from Eq (??) of Ampere s law. So, we will first evaluate the vector potential from the integral in equation (9.84), and then use the relation (9.74) to determine the magnetic field. For convenience, we place the loop on the xy plane with its center at the origin of the coordinate system. Then the current flows along the ˆφ direction is J φ ( r ) = I R δ(r R)δ(cos θ ) (9.85) The presence of the delta-functions in the above expression is to restrict the current to the loop (r = R) on the plane (θ = π/2). Since this system has an azimuthal symmetry ( invariant under the rotation with respect to the z-axis ), the vector potential will also have that symmetry, i.e. A = A φ ˆφ, and Aφ is independent of the angle φ. Hence, we can chose a convenient observational point for φ to evaluate the integral (9.84). The simplest choice will be φ00, which corresponds to having r lying in xz plane. This implies that the vector potential is in the direction of the y-axis, and so we write where (in cgs units) A( r) = A φ (r, θ, φ = 0) ˆφ = A φ (r, θ)ŷ (9.86) A φ (r, θ) = I Jy ( r ) c r r d3 r (9.87) = I δ(r R)δ(θ π ) cos φ 2 d 3 r (9.88) c r r 126

130 where we substituted J y = cos φ J φ in the last expression. Now, we use of the expansion of 1/ r r in spherical coordinates given in Eq (4.54) to write A φ (r, θ) = I r 2 dr dω δ(r R)δ(cos θ ) cos φ 4π r< l Y cr (2l + 1) r l+1 lm (Ω)Ylm(Ω ) (9.89) l,m > [ = I cr Re r 2 dr dω δ(r R)δ(θ π ] 4π r< l 2 )eiφ Y (2l + 1) r l+1 lm (Ω)Ylm(θ, 0)e imφ l,m > where r < and r > refer to the smaller and larger of r and r = R. The integration over r and θ is trivial due to the delta functions. Using the orthonormality relation we obtain 2π 0 dφ e i(m 1)φ dφ = 2πδ m1 (9.90) A φ (r, θ) = 4π c IR 2π (2l + 1) l=1 r< l r> l+1 Y l1 (θ, 0)Yl1( π, 0) (9.91) 2 The spherical harmonics Y l1 can be expressed in terms of the Legendre polynomials as 2l + 1 Y l1 = 4πl(l + 1) P1 l (cos θ) (9.92) The polynomial Pl 1 (0) is zero for l even, and for odd l it is given by P2n+1(0) 1 n+1 (2n + 1)!! = ( 1) (2n)!! (9.93) Using these results in (9.91), the vector potential reads A φ (r, θ) = 2π c IR n+1 (2n + 1)!! ( 1) (2n)!! n=1 r< l r> l+1 P 1 l (cos θ) (9.94) Far from the current loop (i.e. r >> R), we can keep only the leading term (n = 0) in the expansion, to get A φ (r, θ) = πir2 c sin θ r 2 (9.95) Since A = A φ ˆφ, the magnetic filed has no component along the ˆφ direction, and so we write 127

131 B( r, θ) = 1 r sin θ θ (sin θa φ) ˆr 1 r r (ra φ) ˆθ (9.96) Substituting the expression of A φ in Eq (9.95) into the equation above, yields B( r, θ) = 2πR2 I cos θ c ˆr r + πr2 I sin θ 3 c ˆθ r 3 (9.97) 9.9 Magnetic moment of a localized current distribution From the previous section, we found that the expression of the vector magnetic potential due to a localized current densities J( r) is given by A( r) = µ 0 4π V J( r ) r r d3 r (9.98) Suppose that the origin of coordinates is within the current distribution. At a point r far away from the current distribution, i.e. r >> r, we can expand the denominator in (9.75) in terms of r /r and retain up to the dipole term. We obtain A( r) = 1 r V J( r )d 3 r + 1 r 3. V ( r. r ) J( r )d 3 r (9.99) The first term on the right hand side the equation above vanishes as a consequence of the fact that the current distribution is localized and divergence free. This can be shown as follows. V. (fjg) d 3 r = (fjg) d s = 0 (9.100) V where we assumed that the current densities are localized. Now, using J = 0, the left hand side of the above equation reads By taking f = 1, g = x i, we get V [( ( )] f).jg + fj. g d 3 r = 0 (9.101) V J( r )d 3 r = 0 (9.102) 128

132 Hence, the magnetic monopole associated with current density vanishes. (9.99) is given now by Thus, Eq A i ( r) = 1 r 3 = 1 2r 3. V ( ) xj x j Ji ( r )d 3 r (9.103) ( ) x j J i + x ij j ( r )d 3 r + 1 V 2r. ( ) x 3 j x j J i x ij j ( r )d 3 r V where summation over repeated indices is understood. Now, in Eq (9.101) let we take f = x i and g = x i. Then, we obtain ( Ji x j + x ij j ) d 3 r = 0 (9.104) Substituting (9.104) in equation (9.103) yields 67 which can be written as 68 V A i ( r) = µ 0 1 ( ) x 4π 2r 3 j x i J j x jj i ( r )d 3 r (9.105) V = µ 0 1 [( r J( r )) r] 4π 2r 3 i d 3 r V where m = 1 2 A( r) = µ 0 m r 4π r (9.106) 3 V ( r J( r )) d 3 r (9.107) is called the magnetic dipole moment of the current distribution vector potential over the volume V. We can also define M = 1 2 ( r J( r )) (9.108) as the magnetic dipole moment density. Note that the magnetic dipole moment m is independent of the location of the origin. To see this we replace r by ( r + d), where d is a constant vector. Then in this new coordinates, the magnetic moment reads 67 We have use the vector identity C i D j C j D i = ɛ i jk( C D) k, to obtain the second equation in (9.105). 68 The expression of A in SI unit system can be obtained by multiplying the right hand side of (9.106) by an over all factor µ 0 /4π. 129

133 m = 1 2 = 1 2 = m where we used Eq (9.102) in the last equality. V V ( ( r + d) ) J d 3 r (9.109) ( r J) d 3 r + 1 d 2 Jd 3 r For a very thin wire carrying a current I, the current density element Jd 3 r will be replaced by line current element Id l, and the magnetic dipole moment reads V m = 1 2 r Id l (9.110) For example, a current loop with radius R, the magnetic dipole moment is m = IAˆn (9.111) where A = πr 2 is the area of the circular loop. In fact the result of Eq ((9.111)) this generalizes for a planar current loop of arbitrary shape since 1 2 r d l is the area of the loop times an outward normal vector, ˆn, with its orientation determined by the right hand rule with respect to the direction of the current. Now that we have the expression of the vector potential, we can evaluate the magnetic field. We have 69 ( ) B = m r (9.112) r ( ) 3 = m r ( m. r r ) 3 r 3 which far away from the origin gives 69 We used the identity: B = 3( m.ˆn)ˆn m r 3, ˆn = r r 3 (9.113) For a = m, a. a = 0 and ( a b) = a(. b) b(. a) + ( b. ) a ( a. ) b ( b. ) a =

134 Note that this expression for the magnetic field has exactly the same form as the one of the electric dipole moment field (see equation (5.8) in these notes). Hence the name magnetic dipole moment for the quantity m is appropriate 70. However, the expression of the magnetic field in (9.113) is not complete for a point like magnetic dipole. This case there is an additional δ( r) term that takes into account the field at the origin. To show that let us write (9.112) in component form as B i = m i j ( xj r 3 ) m j j ( xi r 3 ) (9.114) Since j ( xj r 3 ) = 4πδ( r), we can deduce write the derivative term of the second term as j ( xi r 3 ) = 4π 3 δ ijδ( r) + j ( xi r 3 ) r 0 (9.115) = δ ij r 3x ix j 3 r 5 Substituting this expression in (9.114), it follows that + 4π 3 δ ijδ( r) B = 3( m.ˆn)ˆn m r 3 + 8π 3 mδ( r) (9.116) Again, it has the same form as the one of the electric dipole moment. This expression of the magnetic field has been used to verify whether particles like the electron, which posses an intrinsic magnetic dipole moment, contain two opposite monopoles leading to a dipole or rather it is a circulating current. By probing the magnetic field strength at r = 0, experiments gave evidence in favour of current loops origin of the electron s magnetic moment. Example 9.6: Magnetic dipole of moment of rotating charged sphere In this example, we would like to calculate the magnetic moment m for a spherical, uniform volume charge distribution ρ and radius R, rotating about its axis with angular frequency ω. In this case, the current density is given by J( r ) = ρ υ = ρ( ω r ) (9.117) 70 Equation (9.113) suggests that permanent magnets (which exhibit dipole magnetic field) are made out of "subatomic" current loops. However, this kind of thinking at the atomic scale will require a quantum mechanical description. Although, one might easily imagine that an electron in a "circular" orbit around the nucleus causes a current loop, it is also found that an isolated electron has a magnetic moment of size e /2mc. However, experimentally it has not been possible to assign a circular current loop to an electron charge. 131

135 So, r J( r ) = ρω [ r 2 ẑ r cos θ r ] (9.118) = ρω [ r 2 ẑ r 2 cos θ ẑ r 2 cos θ sin θ e ρ ] where in the first equality we used the vector identity (9.54), and in last expression we decomposed r into a component along ẑ and another along the unit cylindrical basis vector e ρ. Substituting the above expression in (9.107), we get m = 1 [ R 2 ρω ẑ r 4 dr 0 = 4πρωR5 ẑ 15 dω R 0 r 4 dr cos 2 θ dω ] (9.119) where we used the fact that e ρ dφ = 0. The above expression can be also written as where q is the total charge of the sphere. m = qωr2 ẑ (9.120) Force on a current distribution in an external magnetic field Consider a local current distribution, J, subject to an external magnetic field B (ext). We take the origin of the coordinates to be within the current distribution. Let us assume that B (ext) varies slowly over the region J is non zero. Then we can expand the external field about some point within the current distribution, B (ext) ( r) = B (ext) (0) + ( r. ) B (ext) r=0 +.. (9.121) Then according to Eq (??), the components of force that the external magnetic field exerts on the current distribution is F i = ɛ ijk B (ext) k (0) J j ( r)d 3 r + ɛ ijk B (ext) k (0) J j ( r. )d 3 r +.. (9.122) V V As we showed before (see (9.102)), the first integral vanishes for a steady localized current distribution. The integrant of the second integral can be manipulated as follows: 132

136 ɛ ijk J j ( r. ) = ɛ ijk J j x l l B (ext) r=0 (9.123) = 1 2 ɛ ijk (J j x l J l x j ) l B (ext) r=0 = 1 2 ɛ ijkɛ ljn ( r J) n l B (ext) r=0 = ɛ ijk ɛ ljnmn l B (ext) r=0 = ɛ ijk ( M ) j B (ext) k r=0 where l denotes the partial derivative with respect to the coordinate x l. So, to first order in the expansion, the vector force reads F = ( m ) B (ext) (9.124) Using the vector identity ( a ) b = ( a. b) a(. b) and the fact that B (ext) is divergence free, the expression of the force reduces to F = ( m. B (ext) ) r=0 = ( m. )B (ext) r=0 (9.125) where in obtaining the second equality we supposed that the applied magnetic field is due entirely to external sources so that its curl inside the current distribution vanishes. Note that the force has the form F = ( m. B (ext) ) U (9.126) Hence U = m. B (ext) can be interpreted as the potential energy of magnetic dipole moment m in the external field B (ext). This shows that the magnetic dipole tends to orient itself parallel to direction of the applied field to lower the potential energy. It is important to note that the expression of U in (9.126) does not represent the total energy of a magnetic dipole moment in an external magnetic field, since work must be done to keep m constant (through the current that produces the dipole) when bringing the dipole to its final position in the field. To calculate the torque with respect to the origin on the current distribution, we substitute the leading contribution in the expansion of the external magnetic field into the definition of the torque given in (??). Unlike the the magnetic force on the current distribution, the zeroth order term in B (ext) yields a non vanishing contribution given by (in cgs units) τ = 1 c = 1 c V V r ( J( r ) B (ext) (0) ) d 3 r (9.127) ( r.b (ext) (0) ) J( r )d 3 r B (ext) (0) ( r.j)d 3 r V 133

137 where we used the vector identity (9.54) to obtain the last equality. The second integral can be written as V ( r.j)d 3 r = 1 2 = 1 2 V V.(r 2 J)d 3 r 1 2 r 2 Jd s = 0 V r 2.Jd 3 r (9.128) where we used the divergence theorem to turn the first term into an integral over a surface at infinity at which the current vanishes. Hence, (9.127) can be written in component form as [ 1 τ i = c V [ 1 = 2c V x jj i d 3 r ] B (ext) j (0) (9.129) (x jj i x ij j )d 3 r ] B (ext) j (0) where Eq (9.104) has been used. Thus, the total torque on the current distribution reads 9.11 Gyromagnetic ratio τ = m B (ext) (0) (9.130) If the current is formed by a system of N particles of masses m i with electric charges q i moving with velocities v i << c, then we have J( r) = N q α v α δ( r r α ) (9.131) α=1 where r α is the position vector of the α th particle. In this case the magnetic dipole moment in the SI unit system is m = 1 2c or, equivalently, N q α ( r α v α ) (9.132) α=1 m = N α=1 q α 2m α Lα (9.133) 134

138 where L α = ( r α p α ) is the angular momentum of the α th particle 71. So the above equation shows that of a current distribution generated by a N charges can be expressed via their angular momenta. If all the particles have the same charge to mass ratio, i.e. then the expression of m reduces to q α = q, α = 1,..., N (9.134) m α m m = q 2m L γ L (9.135) where L = N α=1 L α is the total angular momentum of the N particles, and the quantity γ = q/2m is called the gyromagnetic ratio. It is often measured in units of γ e = e/m, the gyromagnetic ratio of the electron. For instance, if the current in a conductor is made of ions rather than electrons, then the gyromagnetic ratio will be less than the one of the electron by more than three order of magnitude. 71 We have also used that for v α << c, the particle momentum p α is given by m α v α. 135

139 10.1 Introduction 10 Magnetostatics in Matter There are three effects on magnetic materials: 1. Paramagnetism Every electron has intrinsic spin, and so it carries a little magnetic dipole moment with a dipole moment m = eh 2m e. In the absence of an external magnetic field, these dipoles are randomly oriented. However, in an applied a magnetic field, the dipole moments try to align themselves with the magnetic field. As a result of the aligned dipole moments the magnetic field is enhanced. This enhancement of a magnetic field in matter due to an applied magnetic field is called paramagnetism. It seems that since every material contains electrons paramagnetism should be a universal effect for all substances. However the Pauli exclusion principle (quantum mechanics) does not allow electrons in the same orbit to have their spins aligned. So as a rule of thumb only substances with an odd number of electrons exhibit paramagnetism due to spin alignment, while for substances with even number of electrons the effect cancels out. Even when a substance is paramagnetic the alignment of spins is usually (room temperatures) far from complete due to thermal fluctuations. A last point to remember is that as soon as the external magnetic field disappears, the paramagnetic effect disappears, and the dipole moments are randomly distributed once more. The orbital motion of electron around the nucleus giving rise to a current loop, which results in a magnetic dipole moment. However, this effect only gives a negligible contribution to the paramagnetism. 2. Diamagnetism 10.2 Magnetically permeable media Although, in principle one could apply the Maxwell equations of magnetostatics (9.62) and (9.65) at the microscopic level to the entire system of atoms and molecules, it is not practical. Furthermore, in the presence of an external magnetic field the atomic magnetic dipole moments can align theme selves giving rise to an average macroscopic dipole moment. Thus, one can give a macroscopic description of the system by averaging out the microscopic quantities over a volume V small compared to the object size but very large compared to the molecular scale. Let J (n) be the current from the n th molecule of the dielectric. Then microscopic vector potential field outside the n th molecule is A (n) J (n) ( ξ) ( r) = V M r r n ξ d3 ξ (10.1) 136

140 where V M represents the volume of the molecule. Considering the multipole expansion of (9.106), yields A (n) ( r) = m( r n) ( r r n ) r r n 3 (10.2) where m( r n ) is the total magnetic dipole moment of the n th molecule: m( r n ) = 1 2 V M ( r n + ξ) J (n) ( ξ) d 3 ξ = 1 2 V M ξ J (n) ( ξ) d 3 ξ (10.3) where we used the fact that for a localized current we have J (n) ( ξ) d 3 ξ = 0. The total microscopic vector potential, A (micro), is obtained by adding up the field of each molecule in the material, i.e. A (micro) ( r) = n m( r n ) ( r r n ) r r n 3 (10.4) which can also be written as an integral A (micro) m M ( r ) ( r r ) ( r) = d 3 r (10.5) V D r r 3 where now the integration is taken over the whole volume of the substance, and m M is the magnetic dipole moment density of the molecules in the substance, defined as m M ( r ) = n m( r n )δ( r r n ) (10.6) Now, by averaging out over a volume V, we obtain the macroscopic vector potential where M( r A (Macro) ) ( r r ) ( r) = d 3 r (10.7) V D r r 3 M( r ) = V f( y) m M ( r + y) d 3 y (10.8) = N ( r ) < m > M where N ( r ) is the number density of the molecules at a point r, and < m > M denotes the average magnetic dipole moment per molecule within V. 137

141 If in addition there exists current, J free, generated by the motion of free charges, then the total vector potential reads J free( r ) A( r) = V D r r M( r d 3 r ) ( r r ) + d 3 r (10.9) V D r r 3 where, to ease the notation, we denoted the macroscopic vector potential simply by A. The second integral in the above equation can be written as: V D M( r ) ( r r ) r r 3 d 3 r = V D M( r ) ( 1 ) r r d 3 r (10.10) where denotes the gradient with respect to the vector r. Using the identity (F C) = ( F C + F C), we can write V D M( r ) ( r r ) r r 3 d 3 r = M( r ) V D r r M( r = ) V D r r ( ) M( r d 3 r ) d 3 r V D r r ( ) M( r d 3 r ) + r r V D ˆn ds (10.11) where the second integral in the last step was obtained as a result of the identity 72 C( r) d 3 r = ˆn C d s (10.12) V where V is surface bounding the volume V, and ˆn is the outward normal vector at d s. Substituting these result into (10.9), we get J free( r ) + A( r) = M( r ) d 3 r + V D r r V V D M( r ) ˆn r r ds (10.13) This equation has the form of (9.84), for the vector potential of appropriate volume and surface current densities if we identify in terms of which j b ( r ) = M( r ) (10.14) K b = M( r ) ˆn 72 This identity can be derived by applying Gauss s divergence theorem to ( α C), where α is an arbitrary constant vector, and using the identity ˆn.( α C) = α.(ˆn C). 138

142 A( r) = V D J free ( r ) r r d 3 r + V D J b ( r ) r r d 3 r + V D K b ( r ) r r ds (10.15) Thus, J b ( r ) can be interpreted as the volume density bound current at r in the magnetized body, and K b ( r ) is the surface density bound current at r on its surface. So bound currents are induced by magnetization of the medium due to the orbital motion and spin of charges (electrons) bound within the atoms or molecules in the substance. If the integration is performed over all space, then the second integral vanishes since M is localized to a finite region V D. Thus, in general the effective current in a magnetized medium is a sum of a free currents and the volume bound density current. Consequently the macroscopic equation for the curl of B is given by B( r) = µ 0 [J free ( r) + M( r) ] (10.16) We can combine the term M with the field B and define a new macroscopic field such that we can re-write(10.16) as H = ( 1 µ 0 B M) (10.17) H = J free (10.18) Many authors call the field H the magnetic field, whereas B is called the induction field. The field H is the analogue of the displacement field D in electrostatics. Since H is directly given in terms of J, the field B is usually regarded as a function of H, with the later being regarded as the independent parameter, i.e. B = µ 0 [H + M(H) ] (10.19) In particular, for some materials, there is an approximate linear relation between the magnetization and the field H: M(H) = χ m H (10.20) where χ m is known as the magnetic susceptibility of the material. For this type of substances, the relation between B and H reads 139

143 Material χ m [ 10 5 ] (SI units) Alluminuim +2.3 Copper 0.98 Diamond 2.2 Gold 3.4 Hydrogen (at 1 atm) Mercury 2.85 Silicon 0.4 Tungsten +6.8 B = µ 0 (1 + χ m ) H µh (10.21) The constant µ is called the magnetic permeability. The equation relating B and H is, sometimes, called constitutive equation. Materials with µ > 1 are said to be paramagnetic whereas those with µ < 1 are diamagnetic. Note that for most of paramagnetic and diamagnetic materials, the magnetic susceptibility is very small, typically of order For ferromagnetic materials 73, which are strongly affected by the applied magnetic field, the magnetization is usually a non linear, multivalued function of H. Hence in a ferromagnet, B is not a single value function of H. It is found that for a given value of the induction B In the table below, we list the magnetic susceptibility of some paramagnetic and diamagnetic substances at room temperature. Example 10.1: Induction field inside a very long bar magnet We would like to calculate the induction field inside an infinitely long bar magnet with uniform magnetization. Integrating (??) over some surface S, we have S ( B). d s = µ 0 which after using Stoke s theorem, we obtain C B. d l = µ 0 S C ( M). d s (10.22) M. d l (10.23) where C is the loop that limits the surface S. We chose the surface S such that C is a rectangular loop with one short side of length L inside the magnet and its other 73 A ferromagnet can have a finite magnetic induction, B, with zero magnetic field as well as the converse. 140

144 short side sufficiently far removed so that both B and the magnetization vanish on it. The remaining two sides are assumed to be infinitely long. The integral along these long sides vanishes since they are perpendicular to B. Now, Eq (10.23) reads So, the field B inside the infinitely long bar magnet is B (inside) z L = µ 0 M z L (10.24) B (inside) = µ 0 M (10.25) If we repeat the same argument above but with the far short side of the loop brought close to the magnet we get B (ouside) = 0 (10.26) 10.3 Boundary conditions for B and H at mediums interface The boundary conditions on the magnetic field at the interface between two media, 1 and 2, can be determined by performing appropriate integrals of Maxwell s equations of magnetostatics: H = 4π c J (10.27). B = 0 (10.28) We will first find the boundary conditions for the induction field. The Integration of the second equation over a Gaussian surface S, yields S B. d s = 0 (10.29) If we chose the surface S to be a small pill box with the surfaces above and below the interface each has an area S, and by breaking up the surface integral into integrals over each of the three surfaces of the pillbox, we obtain B 2. n 21 S + B 1. n 12 S + 0 = 0 (10.30) where n 21 = n 12 is a unit vector normal to the interface and pointing from medium 2 to 1. Hence we obtain 141

145 (B 2 B 1 ). n = 0 (10.31) The boundary condition on the magnetic field H can be obtained by integrating the first equation in (10.27) over a surface S of a closed rectangular loop, C, where the line above and below the interface each has length l, and the width of the side through the interface is h. We chose the interface on the xy plane such that l = l ˆx, and h = h ẑ, with the medium 1 in the negative z-axis. Using Stokes theorem, we have C H. d l = 4π c S J. n ds (10.32) where n is a unit vector normal to the surface S (pointing out of the page). For small rectangular loop,the integral on the left-hand side of the equation above reads C H. d l = l (H 2 H 1 ). ( n n) (10.33) = [ n (H 2 H 1 )]. n Let us assume that the current density consists of a body current density J vol ( r) and a surface current density J s ( r) = K δ(z) parallel to the surface layer between the two mediums. Then, the integral on the right-hand side of (10.32) reads J. n ds = h l J vol ( r). n + l K. n (10.34) S So, in the limit h 0, the first term on the right-hand side of the above equation vanishes. lim J. n ds = l K. n (10.35) h 0 S Using the results of equations (10.33) and (10.35), we get from which it follows that (H 2 H 1 ). ( n n) = 4π c K. n (10.36) n (H 2 H 1 ) = 4π c K (10.37) This equation can be used to determine the surface current density if H 1 and H 2 are known at the interface between the two mediums. 142

146 10.4 Boundary-value problem in magnetostatics for J = 0 When there is no free current in the system, the differential equation for the magnetic field reads H = 0 (10.38) This implies that H can be expressed as H = Φ m, where φ m is a scalar field, called the magnetic scalar potential. This is similar to the expression that we encountered in chapter 1 for the electrostatic field in terms of the scalar potential. The second equation of magnetostatics. B = 0 gives µ (H + M) = µ 2 Φ m + µ. M = 0 (10.39) where we have assumed that the permeability of the system is constant. Thus the above equation implies that Φ m satisfies Poisson s equation where the effective magnetic charge density is given by 2 Φ m = ρ M (10.40) The solution to equation (??) can be writing as Φ m ( r) = 1 4π = 1 4π V V ρ M =. M (10.41). M( r ) d 3 r (10.42) r r ( ) M( r ). d 3 r + 1 ( ) 1 M( r ). d 3 r r r 4π V r r where the volume V contains all the magnetization. Using the divergence theorem, the first term in the above expression can be converted to a surface, which vanishes since the magnetization is localized within the material volume. Hence, the magnetic potential reads Φ m ( r) = 1 4π V = 1. 4π M( r ). V M( r ) r r d3 r ( 1 ) r r d 3 r (10.43) Far from region of non-vanishing magnetization, we can approximate the potential by 143

147 Φ m ( r) = 1 4π. ( ) 1 r V M( r ) d 3 r = m. r 4πr 3 (10.44) which has similar form as the expression we derived for the scalar potential of an electric dipole moment. An alternative way to calculate Φ m is take the integration over the volume of the material, V material, instead of the volume of the whole space. However, in this case there is a surface boundary where the magnetization changes suddenly, and so the solution is given by Φ m ( r) = 1 4π V. M( r ) material r r d 3 r + 1 4π V material σ M r r ds (10.45) where σ M is the effective surface magnetic density. Integrating (10.41) over a small pillbox at the interface, and using the divergence theorem, we find σ M = ( M 2 M 2 ). n (10.46) where n is a unit vector normal to the interface and directed from medium 1 to medium 2. In the special case of uniform magnetization throughout the material volume, the first term vanishes and only the surface integral over σ M contributes. Example 10.2: Induction field of a magnetized sphere Consider a sphere of radius R with a uniform permanent magnetization M = M 0 ẑ. There are no external free currents, and so we can use the magnetic potential method of the previous section. Since. M = 0 (uniform magnetization), using equation (10.47), we have Φ m ( r) = M 0 cos θ 4π Vmaterial R2 r r dω (10.47) Now we expand r r 1 in terms of spherical harmonics 1 r r = l l=0 m= l 4π (2l + 1) Y lm(ω )Y lm (Ω) rl < r l+1 > (10.48) where r < (r > ) is the smaller (larger) of r and R. Noting that only the term with (l, m) = (1, 0) survives the integration over Ω, and writing Y10(Ω) 3 = P 4π 1(cos θ), we get 144

148 Φ m ( r) = 1 4π M 0R 2 P 1 (cos θ) r < r> 2 P 1 (cos θ )P 1 (cos θ )2πd cos θ (10.49) In the above expression we rewrote cos θ as P 1 (cos θ). Since the Legendre polynomials satisfy the orthonormality condition we obtain 1 1 P n (x)p m (x)dx = 2π 2n + 1 δ nm (10.50) Φ m ( r) = 1 3 M 0R 2 cos θ) r < r 2 > (10.51) Therefore, the magnetic potential is given by Φ m ( r) = 1 M 3 0R 2 cos θ) r, R 2 r < R 1 M 3 0R 2 cos θ) R, r R r 2 (10.52) Note that we will obtain the same result by solving the Laplace equation 2 Φ m = 0 Using azimuthal symmetry of the system, the fact that Φ m (r ) = 0, and requiring that potential to be finite when r approaches zero, we can write the solution of the form Φ in m = Φ m (r < R) = Φ out m = Φ m (r > R) = ( Al r l + ) P l (cos θ) (10.53) l=0 l=0 B l r l+1 P l(cos θ) The boundary conditions on the magnetic field at r = R, read 1 R Φ out θ r=r = 1 Φ in R Φ out r r=r = 1 R θ r=r (10.54) Φ in θ r=r + M 0 cos θ 145

149 The first condition yields A l = the second equation in (10.53) implies that l=0 from which it follows that B l R 2l+1 (10.55) [ l + 1 ] R B 2l+1 l A l lr l 1 P l (cos θ) = M 0 cos θ (10.56) A 1 = M 0 2B 1 (10.57) R 3 A l = l + 1 B l l R ; l 1 2l+1 The equations (10.55) and (10.58)can be satisfied only if A l 1 = B l 1 = 0, and A 1 = M 0 3 B 1 = M 0 3 R3 (10.58) Substituting these results in (10.53) gives the same expression for the magnetic potential in equation (10.51). So, now that we determined the magnetic potential, we deduce the magnetic induction field using the relation H = Φ m. Outside the sphere, B out = µ 0 H out = µ 0 Φ out m. Moreover, we see that Φ out m is the potential of a dipole with dipole moment Thus, m = 4πR3 3 M 0 (10.59) B out = µ [ 0 m ] 3( m. ˆr)ˆr + 4π r3 r 3 (10.60) Inside the sphere we have H in = µ 0 Φ in m, and B = µ 0 (H + M), which give Hence, both H and B are uniform inside the sphere. B in = 2 3 µ 0 M 0 (10.61) 146

150 10.5 Magnetic shielding Consider a spherical shell of material with magnetic permeability µ, with inner and outer radii R 1 and R 2, respectively. The regions inside and outside the shell are assumed to be empty. The shell is placed initially in a uniform magnetic induction field B 0. Since there are no free external currents, we can use the method of potential to determine the induction field. So, we need to solve Laplace s equation 2 Φ m in the three regions : r > R 2, R 1 < r < R 2, and r < R 1. Let us chose the z-axis along the external induction field B 0. Expanding Φ m in spherical polar coordinates in each of the three regions, we get Φ m (r > R 2 ) = B 0 µ 0 r cos θ + Φ m (R 1 < r < R 2 ) = Φ m (r < R 1 ) = l=0 l=0 C l r l + D l r P l(cos θ) l+1 E l r l P l (cos θ) l=0 A l r l+1 P l(cos θ) (10.62) Note that the first term in the expansion of Φ m (r > R 2 ) guarantees that at sufficiently large distance from the shell, Φ m B 0. The boundary conditions at r = R 1 and r = R 2 requires that the radial components of B and the tangential components of H to be continuous. In terms of the magnetic potential, we have µ Φ m r R + = µ 0 1 Φ m θ R + 1 Φ m r R, µ Φ m 0 1 = Φ m θ R 1, r R + 2 Φ m θ R + 2 = µ Φ m r R 2 ; (10.63) = Φ m θ R 2 where R + i and R i are approached from above (plus sign) and below (negative sign), respectively. Substituting the expansions of Φ m into the above equations, and using the linear independence of the Legendre polynomials, we find µlc l R1 l 1 µ(l + 1)D l R (l+2) 1 µ 0 le l R1 l 1 = 0, (10.64) µ lc l R2 l 1 µ(l + 1)D l R (l+2) 2 + µ 0 (l + 1)A l R (l+2) 2 = B 0 δ l,1 C l R1 l + D l R (l+1) 1 E l R1 l = 0 C l R l 2 + D l R (l+1) 2 A l R (l+1) 2 = R 2 B 0 µ 0 δ l,1 It is straightforward to show that all the coefficients with l 1 vanish, whereas for l = 1 the solution is given by 147

151 (2µ + µ 0 )(µ µ 0 )(R2 3 R 3 A 1 = 1) B 0 (10.65) (µ + 2µ 0 )(2µ + µ 0 ) 2(µ µ 0 ) 2 R1/R µ 0 (2µ + µ 0 )R1 3 C 1 = B (µ + 2)(2µ + µ 0 ) 2(µ µ 0 ) 2 R1/R (µ µ 0 )R1 3 D 1 = B (µ + 2µ 0 )(2µ + µ 0 ) 2(µ µ 0 ) 2 R1/R µ E 1 = B (µ + 2µ 0 )(2µ + µ 0 ) 2(µ µ 0 ) 2 R1/R So, outside the spherical shell the magnetic induction field is the sum of of the applied uniform field B 0 and the field of a dipole with a dipole moment 4πA 1 parallel to B 0. Inside the cavity there is a uniform magnetic field of magnitude B(r < R 1 ) = µ 0 E 1 and parallel to the applied field B 0. When the permeability of the shell µ is much greater than that of the vacuum, we have B(r < R 1 ) = 9 µ/µ 0 2 (1 R1/R 3 2) B 3 0 (10.66) which shows that for high permeability shells (i.e µ in the range 10 3 µ 0 to 10 9 µ 0 ) will result in a great reduction of the magnetic field in the interior of the shell. For example, taking µ = 10 6 µ 0, and R 1 /R 2 = 0.95, gives B(r < R 1 ) B

152 Exercise 1 11 Exercices on Magnetostatics A conductor in the shape of an n-sided polygon of side d carries current I. Calculate the magnitude of the magnetic induction at the centre of the polygon. Exercise 2 A thin plastic disk of radius R has a uniform surface charge density σ. The disk is rotating about its own axis with an angular velocity ω. Find the magnetic induction at a distance z along the axis from the centre of the disk. Exercise 3 Consider an infinitely long, cylindrical conductor of radius R carrying a current I with a non-uniform current density: J = αr ẑ where ẑ is along the symmetry axis of the cylinder, and α is a constant. Find the magnetic induction at points inside and outside the cylinder. Exercise 4 Calculate the vector potential at distance r from an infinitely long straight wire carrying a current I. Exercise 5 A circular loop of wire having a radius R and carrying a current I is located in vacuum with its center a distance d away from the a semi-infinite slab of permeability µ. Find the force acting on the loop when (a) The plane of the loop is parallel to the face of the slab. (b) The plane of the loop is perpendicular to the face of the slab. (c) Determine the limiting form of the answer to parts (a) and (b) when d >> R. Exercise 6 A particle of mass M and magnetic dipole moment m is placed on the axis of a circular loop of radius R and carrying a current I, at a distance z 0 from the center. (a) Find the force of attraction between the m and the loop. 149

153 (b) When m is released, it moves toward the loop. What is the kinetic energy of the particle when it arrives at the center (assumes m is constrained to the z-axis). (c) If the particle is originally placed at the center of the loop, find the frequency of small oscillations about this position for motion along the z-axis. Exercise 7 Consider a conducting sphere of radius R carrying a surface charge density σ which rotates about its axis at an angular frequency ω. In this case the current density is given by J = σωr sin θδ(r R) ˆφ (a) Calculate the vector potential inside and outside the sphere, and deduce the the magnetic dipole moment of the charged sphere. (b) Determine the magnetic induction B inside and outside the sphere. Exercise 8 A right circular solenoid of finite length L and radius R has N turns per unit length and carries a current I. Find the magnetic induction on the cylinder axis in the limit NL. Exercise 9 A compact circular coil of radius R, carrying a current I and has N turns, each with current I/N, lies in the x-y plane with its center at the origin. (a) Using Biot-Savart law, find the magnetic induction at any point on the z-axis. (b) An identical coil with same magnitude and sense of the current is located on the same axis, parallel to, and a distance h above the first coil. With the coordinate origin relocated at the point midway between the centers of the two coils, determine the magnetic induction on the axis near the origin as an expansion in powers of z, up to order z 2. (c) Find up to second order in the coordinates the axial and radial components of the magnetic induction at point off-axis near the origin. Exercise 10 A circular loop wire carrying a current I is located with its center at the origin of coordinates and the normal to its plane spherical angles θ 0, φ 0. There is an applied induction field B = B 0 (1 + βy) ˆx + B 0 (1 + βx) ŷ 150

154 (a) Calculate the force acting on the loop without making any approximations. (b) Calculate the torque in lowest order. Can you deduce anything about higher order contributions? Do they vanish for circular loop? What about the other shapes? Exercise 11 A magnetic dipole m is embedded at the center of a sphere of radius R of a linear magnetic material with permeability µ. Find the induction field inside and outside the sphere. Exercise 12 Consider a sphere of radius R which carries a uniform charge distribution σ, and is rotating about its axis with a constant angular velocity. (a) Find the vector potential and the induction magnetic fields inside the sphere. (b) Find the vector potential and the induction magnetic fields inside the sphere. Exercise 13 A circular wire of radius b carries a current I. A solid sphere of radius a << b made of paramagnetic material with permeability µ is placed at the center of the circular wire. Find the magnetization and the total magnetic dipole moment of the sphere. Exercise 14 Find the magnetic potential of a magnetized sphere with magnetization M( r ) = M 0 z ˆx. 151

155 12 Answers to the Exercices on Magnetostatics Exercise 1 The magnitude of the magnetic induction at the centre of the polygon is Exercise 2 B = µ 0In πa sin ( π d the magnetic induction at a distance z along the axis from the centre of the disk is ( ) R 2 + 2z 2 B = µ 0 σω 2z. (R 2 + z 2 ) 3/2 Exercise 3 the magnetic field inside and outside the cylinder are given by ). B(r < R) = αµ 0 3 r2 ˆφ, B(r > R) = αµ 0R 2 ˆφ. 3r Exercise 4 The vector potential at distance r ( ) µo I A = 2π ln r ẑ + λ (12.1) where λ is an arbitrary function. Exercise 5 (a) the force F = ẑ πµ 0 R 2 II n=0 [ ] ( 1) n (2n + 1)!! R 2n+1 R 2 n (R 2 + 4d 2 ) n+2 n! P d 2n+1(cos θ) (n + 1)! P1 2n+1(cos θ) where cos θ = 2d/ R 2 + 4d 2, and I is an image given by ( ) µ I µ0 = µ + µ 0 152

156 Exercise 6 (a) The force of attraction between the magnetic dipole and the loop is F = 6πIR2 m c(r 2 + z0) 2 5/2 ẑ. (b) The kinetic energy of the particle when it arrives at the center of the loop is ( ) K = 2πIR2 m 1 c R 1. 3 (R 2 + z0) 2 5/2 (c) The frequency of small oscillations along the z axis about the center of the loop is f = 1 2πIm 2π cmr. 3 Exercise 7 (a) The vector potential inside and outside the sphere are given by A(r < R) = µ 0σωR r sin θ 3 ˆφ, A(r > R) = µ 0σωR 4 sin θ 3r ˆφ. 2 From the expression of the vector potential at r > R, we can deduce that m = qr2 3 ω. (b) The magnetic induction inside and outside the sphere are given by Exercise 8 B(r < R) = 2µ 0σωR (cos θ ˆr sin θ 3 ˆθ), B(r > R) = µ 0σωR 4 (2 cos θ ˆr + sin θ 3r ˆθ). 3 The magnetic induction on the cylinder axis in the limit NL is B = µ 0NI (cos θ 1 + cos θ 2 )

157 Exercise 9 (a) The the magnetic induction at any point on the z-axis is given by B = µ 0 IR 2 2(R 2 + z 2 ) 3/2. (b) The magnetic induction up to order z 2 is given by B µ ] 0IR [ (h2 R 2 )z 2 ẑ. h 3 2d 4 (c) The axial and radial components have the form where σ 0 and σ 2 are constants. B z = σ 0 + σ 2 (z 2 ρ2 2 ); B ρ = σ 2 z ρ. Exercise 10 (a) The force acting on the loop is F = πr 2 Iβ (sin θ 0 sin φ 0 ˆx + sin θ 0 cos φ 0 ŷ). (b) The torque to lowest order is given by τ = RIB 0 π [ cos θ 0 ˆx + cos θ 0 ŷ + sin θ 0 (cos φ 0 sin φ 0 )]. Exercise 11 Exercise 12 B(r R) = µ 4π B(r > R) = µ 0 4π (a) Inside the sphere, the fields are given by: [ 1 r [3( m. ˆr) ˆr m] + 2(µ 0 µ) m 3 (2µ 0 + µ) R ( ) 3 3µ 1 [3( m. ˆr) ˆr m]. 2µ 0 + µ r3 A(r < R) = 1 3 µ 0σωR sin θr ˆφ, B(r < R) = 2 3 µ 0σωR(cos θ ˆr sin θ ˆθ). ], 154

158 (b) Outside the sphere, the fields are given by: Exercise 13 A(r < R) = 1 3 µ 0σωR 4 sin θ r 2 The magnetization of the sphere is B(r < R) = 1 3 µ 0σω R4 (2 cos θ ˆr + sin θ ˆθ). r3 ˆφ, M = B 0 µ 0 µ µ 0 ( 2 3 µ µ) where B 0 = µ 0 I/2b ẑ is the induction field at the center of the circular wire. The total magnetic dipole moment is Exercise 14 m = 4πa3 3 M ẑ = 4πa3 3 I µ µ 0 2b ( 2µ µ) ẑ 3 The magnetic potential inside and outside the magnetized sphere is Φ m (r < R) = M 0zx, 5 Φ m (r > R) = M 0zxR 5. 5r 5 155

159 13 Time-Dependent Phenomena and Maxwell s Equations In this chapter we will study the effect of time dependent magnetic field Faraday s Law of Induction When an electric current flow between two points A and B in a wire if B A Ed l 0, it means that there is a potential difference between between them. However, for a closed circuit, this seems in contradiction with Ed l = 0 which follows from E = 0 for static fields. Therefore, the force driving the charges around the circuit must be non electrostatics in nature. If f s is the force per unit charge responsible for doing of moving the charges in the wire, then the work done by f s along the circuit C is E = C f s. d l (13.1) E is called the electromotive force, or emf for short, of the circuit C. The name is rather misleading since the emf is not a force, rather, it is the work performed on a unit charge in moving around the loop. Around 1830, Faraday observed that when a current-carrying circuit either was moved or had its current varied as a function of time, there will be an induced current in a near-by circuit. He also found that a current is generated in a circuit when a permanent magnet is in motion through its surface. He concluded that if the magnetic flux through a closed loop changes with time, there will be an induced emf in the circuit. To write this in a mathematical form, let us denote by Φ the flux of a magnetic field B through a closed loop C: Φ = S B. nd s (13.2) Here S is an open surface that ends on the loop C, and n is a unit vector normal to the surface and its direction is found by the right-hand rule. The emf is defined as E = C E. d l (13.3) with E is the electric field that causes the motion of the electrons around the circuit as seen by an observed in the rest frame of the loop. So, Faraday s law reads E = α d Φ B dt (13.4) This is known as Faraday s law of induction. Note that the minus sign is required by Lenz s rule, which states that the current in the loop will flow in such a direction as to produce a magnetic flux through S that will be opposite in sign to the change in Φ B 156

160 that gave rise to the emf. The factor α has the dimension of length over time in the cgs system of units, that is the inverse of a speed. In the appendix, we will show that α = 1/c in the Gaussian unit system. So in these units, Faraday s law of induction reads 74 E = 1 c d Φ B dt (13.5) As an application of Faraday s law, is the motional emf, where a wire loop moves through a magnetic field B such that the effective area of the loop crossed by B changes with time. This will give rise to an emf in the loop, called motional emf. This the basic principle of the electric generator produces electric currents. Example 11.1: Bar sliding on two parallel rail in magnetic field Consider a metal bar that can slide along two parallel wires that are at distance h. At the end the wires are connected by a resistor R. The entire system is in a uniform magnetic field B, pointing into the page. At instant t, the bar is at distance x(t) from the resistor, and so the magnetic flux is Thus, the induced emf is Φ B = Bhx(t) (13.6) This implies that the induced current is given by E = Bhυ (13.7) I = E R = BLυ R (13.8) As a result of the induced current there will be a magnetic force on the rod to the left according to F m = I h B. An external force, F ext, equal in magnitude, but opposite in direction, to F m and directed to the right is needed to maintain the constant velocity motion of the rod. Furthermore, the input energy from this external force is dissipated in the resistor from the induced current, and it is given by 74 To transfer to the SI units, we just make replacement B/c by B in the expression (13.5), which amounts to setting α (SI) = 1 and we obtain E = d Φ B dt 157

161 P dissipated = RI 2 = B2 h 2 υ 2 = E 2 R R = IhBυ = F ext υ = P ext (13.9) Thus the mechanical energy get converted to electrical energy and then to thermal energy. The velocity υ is given by so, υ = P ext = B2 h 2 υ 2 /R F ext IhB υ = IR Bh (13.10) (13.11) 13.2 Displacement current and Maxwell s equations 158

162 14 Appendices 14.1 Some Physical Quantities in SI and CGS system of units Symbol Dimension Units conversion factor Physical quantity SI Guassian SI unit Gaussian Force F ML T 2 ML Energy U 2 ML 2 T 2 Capacitance C ML T 2 Newton Dyne 10 5 T 2 Joule erg 10 7 T 2 Q 2 ML 2 L Farad cm M Electric charge Q Q 1/2 L 3/2 Coulomb StatCoulomb T Q M Charge density ρ 1/2 Coulombm 3 StatCoulomb/cm L 3 L 3 T T Q Conductivity σ 2 1 Siemensm sec ML 3 T Q M Electric current I 1/2 L 3/2 Ampere StatAmpere T T 2 Q M Current density J 1/2 Amperem 2 StatAmpere/cm L 2 T L 1/2 T 2 ML M Electric field E 1/2 1 Volt/m StatVolt/cm T 2 Q L 1/2 T T Q Resistance R 2 T 1 Ohm sec/cm LM L ML Electromotive force E 2 M 1/2 L 1/2 1 Volt StatVolt T 2 Q T QL M Magnetic field H 1/2 Ampere-turn/m Oersted 4π 10 3 T L 1/2 T M M Magnetic induction B 1/2 Tesla Gauss 10 4 T Q L 1/2 T 14.2 Useful Formulas in Vector Caculus 1. Gradient of scalar field Let φ( r) be a scalar field at a point r. At a neighbouring point ( r + δ r), we have φ( r + δ r) = φ( r) + φ x Thus, the variation of φ reads where δx + φ y φ δy + δz (14.1) z δφ( r) = φ. δ r (14.2) φ = ( φ x, φ y, φ z ) (14.3) is a vector field called the gradient of φ at point r. it can be interpreted as follows. Consider lines of the scalar field along which φ( r) = const, also known as the equipotential lines 75. If we take δ r eq to be an infinitesimal change of the vector r along an equipotential line, then 75 The scalar field φ is often called scalar potential. δf along the equipotential line = 0 = φ. δ r eq (14.4) 159

163 Hence, φ is a vector perpendicular to the equipotential lines. In general, we can express the variation of φ as δφ = φ δ r cos θ (14.5) with θ being the angle between r eq and φ. This shows that δφ can be maximum along the direction of δ r perpendicular to the equipotentials pointing to larger values of of the scalar field. To find the change of φ between two points A and B that are not necessarily close to each other, we divide the line joining them into N infinitesimal intervals so that φ( r B ) φ( r B ) = (φ( r B ) φ( r N )) + (φ( r N ) φ( r N 1 )) (φ( r 1 ) φ( r A )) N =. δ r i i=1 Taking N, we obtain 76 φ( r B ) φ( r B ) = B A φ. d r (14.6) Note that the path from A to B is arbitrary. In particular,for a closed path, i.e. A = B, we get B A φ. d r = 0 (14.7) 2. Vector operator identities Here bold characters denote vector fields and φ a scalar field. following identities: We have the. (φ A) = φ. A + A. φ (14.8) (14.9) (φ A) = φ ( A) + ( φ) 76 This result could be also derived using the concept of directional derivative of a field φ along the direction δ l = δlˆl, defined as dφ dl = lim φ( r + δlˆl) φ( r) δl φ. ˆl = φ( r) + φ. ˆlδl φ( r) δl 160

164 3. Divergence (or Gauss s) theorem Let S be a positively-oriented closed surface with interior D, and let F be a vector field continuously differentiable in a domain containing D. Then 4. More on the distribution δ (3) ( r) S F. d s =. D. F dv (14.10) ( ) ˆr = δ (3) ( r) (14.11) 4πr 2 ( ) 1 2 = δ (3) ( r) 4πr (14.12) Note that the second equality is just a consequence of the first one since the gradient. (1/r) is equal to ˆr/r The delta-distribution in the spherical and the cylindrical system of coordinates [Spherical] : δ (3) ( r r ) = δ(r r ) δ(θ θ ) δ(φ φ ) (14.13) rr sin θ = δ(r r ) δ(cos θ cos θ )δ(φ φ ) rr = δ(r r ) δ(ω Ω ) rr [Cylindrical] : δ (3) ( r r ) = δ(ρ ρ ) δ(φ φ )δ(z z ) (14.14) ρ 14.3 Cylindrical Coordinates When a system has an axial symmetry along some axis, say the z-axis, it is useful to describe it in terms of the cylindrical coordinates (ρ, φ, z), with ρ = x 2 + y 2 ; (14.15) ( y ) φ = tan 1 x 161

165 with φ is allowed to vary in the range [0, 2π]. We denote by ˆρ and ˆφ the unit vectors along the direction of ρ and the perpendicular direction to it, respectively, with ˆρ ˆφ = ẑ. The element of surface at constant ρ is given by A volume element in cylindrical coordinate reads The vector position r can be written as d S ρ=const. = ρdφdz ˆρ (14.16) ( d r = ρdφ ˆφ ) dzẑ.dρˆρ = ρdρdφdz (14.17) The infinitesimal vector element d r can be written as r = ρ cos φ ˆx + ρ sin φ ŷ + z ẑ (14.18) d r = r ρ dρ + r φ dφ + r z = dρ ˆρ + ρdφ ˆφ + dz ẑ dz (14.19) Each term in the second equation above represent a change in length along each direction. Hence, we have ˆρ = r ρ ˆφ = r ρ = cos φ ˆx + sin φ ŷ (14.20) = sin φ ˆx + cos φ ŷ In this system of coordinates the gradient of a field fi can obtained by writing the differential of φ as df = f ρ from which we deduce that dρ + f φ dφ + f z dz := f. d r (14.21) = ( f. ˆρ)dρ + ( f. ˆφ)ρdφ + ( f. ẑ)dz = ˆρ ρ + ˆφ 1 ρ φ + ẑ z For a vector field A = with components (A ρ, A φ, A z ) the divergence is Using the fact that 77 (.A = ˆρ ρ + ˆφ 1 ρ φ + ẑ ) ( ). A ρ ˆρ + A φ ˆφ + Az ẑ) z (14.22) (14.23) 77 These can be shown by taking the derivative of the expressions of ˆρ, ˆφ given in Eq (14.20). 162

166 ˆρ ρ = 0, ˆρ φ = ˆφ, ˆφ ρ = 0, ˆφ φ = ˆρ (14.24) we obtain that.a = 1 (ρa ρ ) + 1 A φ ρ ρ ρ φ + A z z (14.25) 14.4 Spherical Coordinates For a system with spherical symmetry it is convenient to describe it with the spherical coordinates (r, θ, φ) defined in terms of the cartesian coordinates as x = r sin θ cos φ (14.26) y = r sin θ sin φ z = r cos θ with 0 0θ π, 0 0φ 2π, and r > 0. In terms of the spherical coordinates, the volume reads d 3 r = r 2 sin θdrdθdφ (14.27) We denote the units vectors in the spherical coordinate system by ˆr = r/r, ˆθ and ˆφ, with the following orthonormality properties ˆr.ˆθ = ˆθ. ˆφ = ˆφ.ˆr = 0 (14.28) ˆr ˆθ = ˆφ, ˆθ ˆφ = ˆr, ˆφ ˆr = ˆθ The infinitesimal vector position d r can be written as Hence, we have d r = r r dr + r θ dθ + r φ = dr ˆr + rdθˆθ θ + r sin θdφ ˆφ dφ (14.29) 163

167 ˆr = r r = (14.30) ˆθ = 1 r r θ = ˆφ = 1 r sin θ r φ = The gradient of a scalar field f can be obtained from the definition of the differential of f, i.e. from which it follows that df =... (14.31) = = ˆr r + ˆθ 1 r θ + ˆφ 1 r sin θ φ (14.32) Now the divergence of a vector field A is.a = ( ˆr r + ˆθ 1 r θ + ˆφ 1 r sin θ ) ( ). A r ˆr + A θ ˆθ + Aφ ˆφ φ (14.33) The partial derivatives of ˆr, ˆθ and ˆφ with respect to the spherical coordinates can be computed using the expression of r, and we find r θ = ˆθ; ˆθ r = 0, φ r = 0, r φ = sin θ ˆφ (14.34) ˆθ θ = ˆr; φ θ = 0; ˆθ = cos θ ˆφ φ φ φ = ˆρ Then, the divergence of A in the spherical coordinate system is given by.a = 1 ( ) r 2 1 A r 2 r + r r sin θ θ (sin θa θ) + 1 A φ r sin θ φ (14.35) 164

168 14.5 Helmholtz theorem (1855) This is one of the fundamental theorems of vector analysis, derived by Helmholtz ( ). It states as follows: " Let F ( r) be any continuous vector field with continuous first partial derivatives. Then F ( r) can be uniquely expressed as the sum of a gradient and a curl". Mathematically, this means that F = φ + ω (14.36) where φ and ω are called the scalar and the ω potential fields, respectively. Further more, the decomposition is unique if F vanishes at infinity." Let us first prove that the decomposition (14.36) exists. So let us suppose that we do not know F, but we do know its divergence and its curl, i.e. We define the scalar field. F ( r) = D( r) (14.37) F ( r) = C( r) (14.38) φ( r) = 1 4π V D( r ) [ r r [ d3 r (14.39) where the volume V is taken as the whole space. approaches zero faster than 1/r 2 at infinity. Then So, we must assume that D( r) 2 φ( r) = 1 4π = 1 4π V V =. F ( r) Similarly, we define the vector field ω( r) = 1 4π V ( ) 1 D( r ) 2 d 3 r (14.40) [ r r D( r ) 2 ( 4πδ( r r )) d 3 r C( r ) [ r r d3 r (14.41) with C( r) goes to zero faster than 1/r 2 at infinity. As in (14.40), we find that 165

169 The divergence of ω is. ω( r) = 1 4π V = 1 4π = 1 4π V 2 ω( r) = C( r) (14.42) ( ) C( r ) 1 [ r r V C( r ) ( 1 [ r r 1 [ r r. C( r )d 3 r 1 4π d 3 r (14.43) ) d 3 r V C( r ) [ r r d s The first integrand is zero because. C( r ) = 0 (divergence of a curl), and the second integral also vanishes since we C( r) goes to zero faster than 1/r 2. So, we get Using the vector identity and equations (14.42) and (14.44), we obtain. ω( r) = 0 (14.44) ( ω) = 2 ω (. ω) (14.45) ( ω) = F (14.46) Thus, the results (14.40) and (14.46) suggest that we can write F as a sum of a gradient of a scalar field and a curl of a vector field. However, the question is :"Is this decomposition unique? ". in general, the answer is no. Because we can always add an arbitrary vector field who divergence and curl are both zero. However, if we require that F vanishes at infinity (which is usually the case for the electric and magnetic field), then, as we will show below, the decomposition is unique. So, let λ a vector field that goes to zero as infinity and has both zero divergence and curl. Then, we can write λ = ψ, where the scalar field ψ satisfies Lapalce s equation: 2 ψ = 0 (14.47) If λ vanishes at infinity, which can be represented by a sphere of radius R, then, in particular, its normal (radial) component λ r ψ/ r is zero there. By inspection, the solution to (14.47) with the boundary condition λ r ψ/ r(r = R) = 0 is ψ = constant. However, according to the uniqueness theorem (see subsection 1.12 of chapter one in these notes), the solution to equation (14.47) is uniquely determined 166

170 up to an additive constant if the normal derivative of ψ is specified at the boundary. Hence, ψ =constant is the unique solution, and it follows that λ = 0 everywhere. Therefore, if the divergence and the curl of a vector field F are specified and both go to zero faster than 1/r 2 at infinity, and if F ( r) 0 as r, then F is uniquely decomposed as in (14.36), in which the scalar function φ( r) is given by Eq (14.39) and the vector potential ω( r) by Eq (14.41) Legendre Polynomials The second order differential equation given as (1 x 2 )y 2xy + λy =; λ > 0, x < 1 (14.48) is known as Legendre equation. Here the prime sign in the y(x) denotes the derivative with respect to x. This equation is frequently encountered in Physics and Engineering. In particular, it occurs when solving Laplace s equation in spherical coordinates. It can also be written as where y + P (x)y + Q(x)y = 0 (14.49) P (x) = 2x 1 x ; Q(x) = λ (14.50) 2 1 x 2 The functions P (x) and Q(x) are analytic at x = 0, and so it is natural to look for a solution of the form y(x) = c n x n (14.51) n=0 Substituting the above series into (14.48), yields (2c 2 + λc 0 ) + (6c 3 6c 1 + λc 1 )x (14.52) + [(k + 2)(k11)c k+2 (k(k + 1) λ) c k ] x k = 0 k=2 This requires that the coefficients of in front of x l have to vanish, i.e. c 2 = λ 2 c 0 (14.53) (λ 2) c 3 = c 1 6 k(k + 1) λ c k+2 = (k + 2)(k + 1) c k 167

171 For the solution to be physically feasible, the series must terminates after a finite number of steps, i.e. the solutions have to be polynomials. We note that when λ is of the form n(n + 1), where n is a non negative integer, then requiring that c n+2 = 0 implies that all coefficients c l+2k = 0, whereas c n+(2k+1) may still be non-zero. However, by choosing one of c 0 or c 1, we can make all the c 2k or all the c 2k+1 equal to zero. So the solutions we are interested in will be polynomials with even powers for n being even or polynomials with odd powers for λ = n(n + 1) and n odd. So, with λ = n(n11), the recurrence relation in (??) can be written as For even numbered coefficients we find k(k + 1) n(n + 1) c k+2 = c k (14.54) (k + 2)(k + 1) (n 2)(n1k + 1) = (k12)(k + 1) n(n + 1) c 2 = c 0 (14.55) 2 (n 2)(n + 3) n(n 2).(n + 3)(n + 1) c 4 = = c ! (n 4)(n + 5) n(n 2)(n 4).(n + k)(n + 3)(n + 1) c 6 = = c !. m m(m 2)... (l 2(m 1)).(l + 2m 1)(l + 2m 3)... (m + 1) c 2m = ( 1) (2m)! which can be re-arranged in a closed formula c 0 [( m (l + 2m)! l ] 2 c 2m = ( 1) 2)! (2m)! l! ( l + m)! ( l m)!c 0 (14.56) 2 2 from which we can deduce that [( c l 2k = ( 1) k (2l 2k)! l ] 2 2l 2)! 2 l ( 1)l/2 c 0 (14.57) (l 2k)! (l k)! k! l! Choosing c 0 = ( 1) l/2 2l [( 2)!] l 2, the even-numbered solution to the Legendre equation l! is given by P l (x) = [ l 2] k=0 ( 1)k (2l 2k)! 2 l k!(l k)! (l 2k)! xl 2k (14.58) 168

172 This known as the Legendre polynomials. Below I list the first six Legendre polynomials: P 0 (x) = 1 (14.59) P 1 (x) = x (14.60) P 2 (x) = 1 2 (3x2 1) (14.61) P 3 (x) = 1 2 (5x3 3x) (14.62) P 4 (x) = 1 8 (35x4 30x 2 + 3) (14.63) P 5 (x) = 1 8 (63x5 70x x) (14.64) 14.7 Spherical Harmonics These are set of functions, Y lm (θ, φ), with l = 0, 1,..; m = l, l + 1,..., l, and obey the equation where the operator Ω is defined as Ω Y lm (Ω) = l(l + 1)Y lm (Ω) (14.65) Ω F = 1 ( sin θ F ) F sin θ θ θ sin 2 (14.66) θ φ 2 which acts only on θ and φ and does not depend on r (sometimes it is called the Laplacian on the sphere). The general solution of (14.65) is given by This shows that Y lm (θ, φ) = ( 1) m (2l+1) (l m)! 4π (l+m)! Pl m (cos θ) e imφ (14.67) Y l, m (θ, φ) = ( 1) m (Y lm ) (14.68) The spherical harmonics satisfy the orthonormality condition dωy lm(ω)y l m (Ω) := dφ sin θdθy lm(ω)y l m (Ω) = δ ll δ mm (14.69) 169

173 and the completeness relation δ (2) (Ω Ω ) = l,m Y lm (Ω)[Y lm (Ω )] (14.70) Below I list some of the spherical harmonics Y 0,0 = 1 (14.71) 4π 3 Y 1,±1 = 8π sin θe±iφ (14.72) 3 Y 1,0 = cos θ (14.73) 4π 15 Y 2,±2 = 32π sin2 θe ±2iφ (14.74) 15 Y 2,±1 = 8π sin θ cos θe±iφ (14.75) 15 Y 2,0 = 16π (3 cos2 θ 1) (14.76) 14.8 Addition Theorem The aim of this subsection is to show that if γ is the angle between two position vectors r = (r, θ, φ) and r = (r, θ, φ ), then 78 P l (cos γ) = 4π 2l + 1 m=l m= l Y l,m(θ, φ )Y l,m(θ, φ) (14.77) known as the addition theorem. To prove it, we first note that by choosing a new coordinate system with axis x, ỹ and z in which r lies along the z-axis, the angle γ is the polar coordinate and we have 2 P l (cos γ) = l(l + 1) r 2 P l (cos γ) (14.78) where is the gradient operator in the new coordinate system. Hence, we have 78 Note that cos γ := ˆr. ˆr = cos θ cos θ + sin θ sin θ cos φ φ 170

174 4π P l (cos γ) = 2l + 1 Y l,0(γ, β) (14.79) with β is the corresponding angle in the new system of axis. Since 2 is invariant under rotation, Eq (14.78) also hold in the original coordinate system. Therefore, we can expand P l (cos γ) as where l P l (cos γ) = A m (θ, φ ) Y l,m (θ, φ) (14.80) m= l A m (θ, φ ) = P l (cos γ) Y l,m(θ, φ) dω (14.81) It is tempting to substitute the expression of cos γ in terms of θ, φ, θ and θ in the above integrant and try to evaluate the integral. However, this way is too difficult. Instead, we use the fact that θ and φ are functions of γ and β, and so we can express Y l,m (θ, φ) as Y l,m(θ, φ) = l m = l B m,m Y l,m (β, γ) (14.82) where the coefficients B l,m are given by B m,m = Y l,m(θ, φ)y l,m (γ, β) dω γ (14.83) where dω γ represents the infinitesimal element of solid angle in the new coordinate system. In particular, for m = 0 we have B m,0 = Y l,m(θ, φ)y l,0(γ, β) dω γ = 2l + 1 4π Y l,m(θ, φ)p l (cos γ) dω γ (14.84) Since under rotation the element of solid angle remains unchanged 79, we have 2l + 1 B m,0 = 4π Y l,m(θ, φ)p l (cos γ) dω = 2l + 1 4π A m (14.85) 79 This can be seen from the fact that an infinitesimal element of surface on a sphere, r 2 dω, subtended by the infinitesimal solid angle dω is invariant under rotation. 171

175 Now, from Eq (14.82) we note that Y l,m (θ, φ) γ=0 = l m = l B lm Y l,m (0, β) = B l0 2l + 1 4π (14.86) where we used the fact that Y l,m (0, β) = δ m 0 2l + 1 4π (14.87) However in the limit γ 0, the vectors r and r coincide, and so θ θ and φ φ. Thus, A m = 4π 2l + 1 Y l,m(θ, φ) (14.88) which after substituting into Eq(14.80) yields the addition theorem given in Eq(14.77) The constant α in Faraday s law of induction To determine the constant α in (13.4), we consider the rate change of the magnetic flux through a loop C moving with a constant velocity υ relative to an observer in the laboratory frame whose space-time coordinates are labeled by ( r, t). So, in the laboratory frame, we have dφ dt = d B( r, t). n ds (14.89) dt = S S B ( r, t). n ds + lim t t 0 [S (B( r, t) B( r + υ t, t)). n ds] t where υ is the displacement of the loop in an infinitesimal time dt. Expanding B( r + υ t, t) and neglecting terms of order (dt) 2 and higher, we obtain dφ dt = B ( r, t). n ds + ( υ. )(B. t n) ds (14.90) S Using the identity ( υ B) = υ(. B) ( υ. )B, and the fact that. B = 0, we get dφ dt = B [ ] ( r, t). n ds ( υ B). n) ds (14.91) S t S B = ( r, t). n ds ( υ B). d t l S C S 172

176 Now, Faradays law of induction reads [E α( υ B)]. d l = α C S B ( r, t). n ds (14.92) t where E is the electric field in the loop rest frame. 173

177 15 Who is Who in Electromagnetism 15.1 Charles Augustin de Coulomb ( ; French). Figure 7. Charles de Coulomb. Between 1785 and 1790, he wrote seven important treatises on electricity and magnetism. Using a torsion balance experiment, he demonstrated that the force between two point- bodies electrically charged is inversely proportional to the square of the distance between them. His name is one of the 72 names inscribed on the Eiffel Tower Andre Marie Ampere ( ; French) Figure 8. Andre Ampere. 80 Among the 72 inscribed on the Eiffel tower are Simeon Poisson. 174

178 Andre Marie Ampere was mathematician and physicist. In 1820, shortly after the announcement of the discovery by the Danish physicist Hans Christian Oersted that an electric current produces a magnetic effect, Ampere repeated Oersted s experiment, and gave a mathematical explanation of it. In addition he discovered that electric currents in parallel wires exert a magnetic force on each other. He also formulated many of the laws of electricity and magnetism and was the father of electrodynamics. The unit of electric current, the ampere, was chosen in his honor in Johann Carl Friedrich Gauss ( ; German) Figure 9. Johann Gauss. Gauss was a mathematician and scientist who contributed significantly to many fields, including number theory, statistics, analysis, differential geometry, geodesy, electrostatics, astronomy and optics. His work has had an immense influence in many areas Jean-Baptiste Biot ( ; French) and Felix Savart ( ; French). Together they discovered the law governing the force between a long, straight conductor carrying current. They showed that the (magnetic) force is proportional to the current in the wire and inversely proportional to the perpendicular distance from the point of measurement to the conductor 15.5 Simeon Denis Poisson ( ; French) Simeon Denis Poisson was a French mathematician, geometer, and physicist. He published about 300 to 400 mathematical works in the areas of integration, potential theory, differential equations, physics, and electricity. For example, he is known for Poisson process, Poisson equation, Poisson kernel, Poisson distribution, Poisson summation formula. 175

179 Figure 10. Biot (left) and Savart (right). Figure 11. Simeon Poisson 15.6 Michael Faraday ( ; English) Figure 12. Michael Faraday 176