Applying the last two equation with a rectangular loop that crosses the interface,shows that i×(E2-E1)=0, 元×(H2-H1)=K (39) where K is the surface current density (units:amp/m).(If H2 points up and n points to the right,K points out of the paper.) Specializing to electrostatics,we have that the tangential components of the electric field are continuous,and the discontinuity of the normal components of D=eE are proportional to the surface charge density.Since the electric field vanishes inside a conductor this gives a direct connection between the field just outside a conductor and the surface charge density. The field is normal to the conductor and of magnitude o/eo. 2.3 Uniqueness of electrostatic solutions,Green's theorem Suppose 1,2 are two solutions of Poisson's equation-V2=p/eo.Then =2-is a solution of Laplace's equation V2=0.One can find lots of solutions of Laplace's equation but they all have growing behavior at infinity.To see this consider a5n®=人r-(®r)=人T电电 (40) J8R R If the left side vanishes (either because -0 fast enough at infinity or because Dirichlet of Neumann boundary conditions apply on the boundary of R),then must be a constant. This is because the integrand on the right is positive definite.If the charge density p is localized and has finite total charge,will vanish faster than 1/r at infinity. More generally one has 7.(070-07b)=必72中-72沙 (41) which,upon integrating over a region R,yields Green's Theorem dv(UV26-0V2v)=p dS(vi.Vo-on.Vv) (42) R In the context of electrostatics,suppose o is the potential associated with charge density p=-V2co and =is associated with charge density p=-V2'co,and furthermore that the boundary of R is a conducting surface with surface charge density o =-nEeo =n.Voco in the first case and a'=n.Vo'eo in the second case.Then Green's theorem becomes the reciprocity theorem: dV(-o'p+op)= dS(p'o-oo） dvop'+ dSoo'= dW'p+φ dsoo (43) a JaR 10 ©2010 by Charles Thorn
Applying the last two equation with a rectangular loop that crosses the interface, shows that nˆ × (E2 − E1) = 0, nˆ × (H2 − H1) = K (39) where K is the surface current density (units: amp/m). (If H2 points up and nˆ points to the right, K points out of the paper.) Specializing to electrostatics, we have that the tangential components of the electric field are continuous, and the discontinuity of the normal components of D = E are proportional to the surface charge density. Since the electric field vanishes inside a conductor this gives a direct connection between the field just outside a conductor and the surface charge density. The field is normal to the conductor and of magnitude σ/0. 2.3 Uniqueness of electrostatic solutions, Green’s theorem Suppose φ1, φ2 are two solutions of Poisson’s equation −∇2φ = ρ/0. Then Φ = φ2 − φ1 is a solution of Laplace’s equation ∇2Φ = 0. One can find lots of solutions of Laplace’s equation but they all have growing behavior at infinity. To see this consider Z ∂R dSΦnˆ · ∇Φ = Z R dV ∇ · (Φ∇Φ) = Z R dV ∇Φ · ∇Φ (40) If the left side vanishes (either because Φ → 0 fast enough at infinity or because Dirichlet of Neumann boundary conditions apply on the boundary of R), then Φ must be a constant. This is because the integrand on the right is positive definite. If the charge density ρ is localized and has finite total charge, Φ will vanish faster than 1/r at infinity. More generally one has ∇ · (ψ∇φ − φ∇ψ) = ψ∇2φ − φ∇2ψ (41) which, upon integrating over a region R, yields Green’s Theorem Z R dV (ψ∇2φ − φ∇2ψ) = I ∂R dS(ψnˆ · ∇φ − φnˆ · ∇ψ) (42) In the context of electrostatics, suppose φ is the potential associated with charge density ρ = −∇2φ0 and ψ = φ 0 is associated with charge density ρ 0 = −∇2φ 0 0, and furthermore that the boundary of R is a conducting surface with surface charge density σ = −nˆ·E0 = nˆ·∇φ0 in the first case and σ 0 = nˆ · ∇φ 0 0 in the second case. Then Green’s theorem becomes the reciprocity theorem: Z R dV (−φ 0 ρ + φρ0 ) = I ∂R dS(φ 0σ − φσ0 ) Z R dV φρ0 + I ∂R dSφσ0 = Z R dV φ 0 ρ + I ∂R dSφ 0σ (43) 10 c 2010 by Charles Thorn
2.4 Green functions The typical electrostatic problem specifies a fixed charge distribution p(r)and a number of surfaces on which the potential satisfies boundary conditions.Commonly these surfaces are the boundaries of conductors which are held at various fixed potentials,e.g.by batteries Disregarding the surfaces a solution of Poisson's equation is just (34).This solution will not satisfy the boundary conditions:it is necessary to add a solution of Laplace's equation to it. The Green function G(r,r)is the solution of a boundary value problem on these surfaces where p/eo is replaced by that of a point charge at r':(r-r').If there were no surfaces at all,G would just be the Coulomb potential 1/4r-r'l of a unit (g=eo)point charge.The Dirichlet Green function Gp vanishes on all the surfaces,whereas the Neumann function has normal derivative a constant on all surfaces.Since the Neumann function is the potential for a unit point charge,the total electric flux leaving the region must be 1,this constant can be taken zero only if some flux can escape to infinity.Knowledge of the Green function for a given set of surfaces allows the explicit construction of the solution for an arbitrary charge distribution and any specified boundary values on the given surfaces.Unless the surfaces have very special geometries,though,it is virtually impossible to actually find the Green function! This construction is a simple application of Green's theorem,with o the potential we seek,and G the Green function: -oV2(r)=p(r), -72G(r,r)=6(r-r) (44) 1 dvp(r)G(r,r')+(r')=dSG(r,r)i.Vo- dso(r)i.VG(r,r'X45) EO R If the value of o is specified on the boundary,then we use the Dirichlet function which vanishes on the boundary: dvp(r)GD(r,r')-dSo(r)i.VGD(r,r') (46) EO R JaR On the other hand if the normal derivative of o is specified on the boundary,we need the Neumann function.If the problem is such that the flux can escape to infinity,we can take the normal derivative of Gy to vanish on the finite surfaces and obtain d(r)fdvp(r)Gx(r,)+f dsGx(r.r)n.Vo(r) (47) E0 R JOR These formulas summarize a profound physics point:the solution of an electrostatics problem is uniquely determined by the charge distribution and either Dirichlet or Neumann boundary conditions.On the other hand they betray no hint of how to find that solution,beyond reducing the problem to finding the Green function.Developing techniques and tricks for doing that in various simple situations is the subject of the first three chapters of Jackson and the next three or four weeks of our course. 11 ©2010 by Charles Thorn
2.4 Green functions The typical electrostatic problem specifies a fixed charge distribution ρ(r) and a number of surfaces on which the potential satisfies boundary conditions. Commonly these surfaces are the boundaries of conductors which are held at various fixed potentials, e.g. by batteries. Disregarding the surfaces a solution of Poisson’s equation is just (34). This solution will not satisfy the boundary conditions: it is necessary to add a solution of Laplace’s equation to it. The Green function G(r, r 0 ) is the solution of a boundary value problem on these surfaces where ρ/0 is replaced by that of a point charge at r 0 : δ(r − r 0 ). If there were no surfaces at all, G would just be the Coulomb potential 1/4π|r − r 0 | of a unit (q = 0) point charge. The Dirichlet Green function GD vanishes on all the surfaces, whereas the Neumann function has normal derivative a constant on all surfaces. Since the Neumann function is the potential for a unit point charge, the total electric flux leaving the region must be 1, this constant can be taken zero only if some flux can escape to infinity. Knowledge of the Green function for a given set of surfaces allows the explicit construction of the solution for an arbitrary charge distribution and any specified boundary values on the given surfaces. Unless the surfaces have very special geometries, though, it is virtually impossible to actually find the Green function! This construction is a simple application of Green’s theorem, with φ the potential we seek, and G the Green function: −0∇2φ(r) = ρ(r), −∇2G(r, r 0 ) = δ(r − r 0 ) (44) − 1 0 Z R dV ρ(r)G(r, r 0 ) + φ(r 0 ) = Z ∂R dSG(r, r 0 )nˆ · ∇φ − Z ∂R dSφ(r)nˆ · ∇G(r, r 0 )(45) If the value of φ is specified on the boundary, then we use the Dirichlet function which vanishes on the boundary: φ(r 0 ) = 1 0 Z R dV ρ(r)GD(r, r 0 ) − Z ∂R dSφ(r)nˆ · ∇GD(r, r 0 ) (46) On the other hand if the normal derivative of φ is specified on the boundary, we need the Neumann function. If the problem is such that the flux can escape to infinity, we can take the normal derivative of GN to vanish on the finite surfaces and obtain φ(r 0 ) = 1 0 Z R dV ρ(r)GN (r, r 0 ) + Z ∂R dSGN (r, r 0 )nˆ · ∇φ(r) (47) These formulas summarize a profound physics point: the solution of an electrostatics problem is uniquely determined by the charge distribution and either Dirichlet or Neumann boundary conditions. On the other hand they betray no hint of how to find that solution, beyond reducing the problem to finding the Green function. Developing techniques and tricks for doing that in various simple situations is the subject of the first three chapters of Jackson and the next three or four weeks of our course. 11 c 2010 by Charles Thorn
2.5 Electrostatic energy A static configuration of charges stores energy:the charges are held in place by external forces.If those forces were released the charges would begin moving converting that energy to kinetic energy.We can evaluate the stored energy by computing the work required to bring all the charges in from infinite separation.As each charge is successively brought in work is done against the forces due to all the charges already in their final positions,AW =AQ: n-1 Wn gnqi (48) 4TeO Tn -Ti =1 Wtot ∑W= 1 gnqi 1 1 (49) n= 8r01 in lro-rdl 1 2-1∑ 8πe0 FryP()( (50) x-y The last term subtracts off the "self-energy"of the charges.It is more convenient to include these self-energies in the total energy: 1 Etot三Wtot+Esef p(c)p(y） = 8元e0 x-y 2 dxp(x)o(x) 2 dz(-V2o)o(x)= dxE2 (51) This last form makes no reference to the location and sizes of whatever charges produce the field.The energy can be attributed solely to the electric field itself.A nonzero electric field at a point adds w=eoE2/2 to the energy density at that point. 2.6 Capacitance A very typical problem in electrostatics involves a set of conductors of various shapes and locations each held at some potential Vi.The potential everywhere outside these conductors is uniquely determined,which uniquely determines the field near the surface of each conduc- tor,which uniquely determines the surface charge density on each conductor,so the total charge Qi on each conductor is determined.Because of the linear relation between potential and charge density we can assert that Q:=∑Cy (52) The coefficients Cii(which have dimension eoxLength)depend on the geometry of the various conductors and are usually impossible to compute,though they can clearly be measured.Ca is called the capacitance of conductor i and Cii is the coefficient of induction ofj on i.The 12 ©2010 by Charles Thorn
2.5 Electrostatic energy A static configuration of charges stores energy: the charges are held in place by external forces. If those forces were released the charges would begin moving converting that energy to kinetic energy. We can evaluate the stored energy by computing the work required to bring all the charges in from infinite separation. As each charge is successively brought in work is done against the forces due to all the charges already in their final positions, ∆W = φ∆Q: Wn = Xn−1 i=1 qnqi 4π0|rn − ri | (48) Wtot = X N n=2 Wn = 1 8π0 X i6=n qnqi |rn − ri | = 1 8π0 X i,n qnqi |rn − ri | − 1 8π0 X i q 2 i |0| (49) → 1 8π0 Z d 3xd3 y ρ(x)ρ(y) |x − y| − 1 8π0 X i q 2 i |0| (50) The last term subtracts off the “self-energy” of the charges. It is more convenient to include these self-energies in the total energy: Etot ≡ Wtot + Eself = 1 8π0 Z d 3 xd3 y ρ(x)ρ(y) |x − y| = 1 2 Z d 3 xρ(x)φ(x) = 0 2 Z d 3 x(−∇2φ)φ(x) = 0 2 Z d 3 xE 2 (51) This last form makes no reference to the location and sizes of whatever charges produce the field. The energy can be attributed solely to the electric field itself. A nonzero electric field at a point adds w = 0E 2 /2 to the energy density at that point. 2.6 Capacitance A very typical problem in electrostatics involves a set of conductors of various shapes and locations each held at some potential Vi . The potential everywhere outside these conductors is uniquely determined, which uniquely determines the field near the surface of each conductor, which uniquely determines the surface charge density on each conductor, so the total charge Qi on each conductor is determined. Because of the linear relation between potential and charge density we can assert that Qi = X j CijVj . (52) The coefficients Cij (which have dimension 0×Length) depend on the geometry of the various conductors and are usually impossible to compute, though they can clearly be measured. Cii is called the capacitance of conductor i and Cij is the coefficient of induction of j on i. The 12 c 2010 by Charles Thorn
SI unit of capacitance is the farad (F).Thus in SI units we have o8.854 x 10-12 F/m. Applying the reciprocity theorem for the two systems Q,V and Q,V we find 0=∑(QY-Qw)=∑(CyYW-Cy%=∑(C,-C)ygg (53) which implies that Cii Cii.Thus a system of N conductors has at most N(N+1)/2 independent Cii.If the system has symmetry there may be even fewer independent com- ponents.For example,three identical conducting spheres at the vertices of an equilateral triangle would have C1=C22=C33 and Cii =C12 for all ij.Reciprocity and symmetry have reduced the 9 Ci;to only two independent ones in this case! The electric energy of the system of conductors is E=∑Q%=∑c%=∑c写Q.Q, (54) Here C-1 is the inverse of the matrix C,CC-1=C-1C=I.Suppose we want to find the force on one of the conductors,say 1,exerted by all the others.Then after disconnecting the batteries,so that the chareges Qi remain fixed,we move conductor 1 a small distance OZk in the k direction.Then =9盟。=-∑品co.,=+号∑品cy=+ E (55) The subscripts V indicate which quantities are held fixed in taking the derivative with respect to r.Note the sign difference! Finally,a very common special case is a simple capacitor,which has only two conductors, and further is electrically neutral (Q2=-Q1=Q).Then the capacitor relations reduce to the single equation =CV where V=V2-Vi.In terms of the Ci of the general two conductor system,one finds C11C22-Ci2 C=Cu+Ca+2C1 (56) 3 Electrostatic Boundary-Value problems 3.1 Method of Images This method exploits the fact that the potential of any system of charges satisfies Laplace's equation at all points in space where the charge density p=0.Thus we can try to make an educated guess for a solution of a boundary value problem by placing various "image" charges behind the boundary,i.e.outside the region where we need to find o.Frequently the symmetry of the boundaries helps guide the choice for the location of the image charges. 13 ©2010 by Charles Thorn
SI unit of capacitance is the farad (F). Thus in SI units we have 0 ≈ 8.854 × 10−12 F/m. Applying the reciprocity theorem for the two systems Q, V and Q0 , V 0 we find 0 = X i (QiV 0 i − Q 0 iVi) = X ij (CijVjV 0 i − CijV 0 j Vi = X ij (Cij − Cji)VjV 0 i (53) which implies that Cij = Cji . Thus a system of N conductors has at most N(N + 1)/2 independent Cij . If the system has symmetry there may be even fewer independent components. For example, three identical conducting spheres at the vertices of an equilateral triangle would have C11 = C22 = C33 and Cij = C12 for all i 6= j. Reciprocity and symmetry have reduced the 9 Cij to only two independent ones in this case! The electric energy of the system of conductors is E = 1 2 X i QiVi = 1 2 X ij CijViVj = 1 2 X ij C −1 ij QiQj (54) Here C −1 is the inverse of the matrix C, CC −1 = C −1C = I. Suppose we want to find the force on one of the conductors, say 1, exerted by all the others. Then after disconnecting the batteries, so that the chareges Qi remain fixed, we move conductor 1 a small distance δxk in the k direction. Then Fi = − ∂E ∂xi Q = − 1 2 X ij ∂ ∂xk C −1 ij (x)QiQj = + 1 2 X ij ∂ ∂xk Cij (x)ViVj = + ∂E ∂xi V (55) The subscripts Q, V indicate which quantities are held fixed in taking the derivative with respect to x. Note the sign difference! Finally, a very common special case is a simple capacitor, which has only two conductors, and further is electrically neutral (Q2 = −Q1 ≡ Q). Then the capacitor relations reduce to the single equation Q = CV where V = V2 − V1. In terms of the Cij of the general two conductor system, one finds C = C11C22 − C 2 12 C11 + C22 + 2C12 (56) 3 Electrostatic Boundary-Value problems 3.1 Method of Images This method exploits the fact that the potential of any system of charges satisfies Laplace’s equation at all points in space where the charge density ρ = 0. Thus we can try to make an educated guess for a solution of a boundary value problem by placing various “image” charges behind the boundary, i.e. outside the region where we need to find φ. Frequently the symmetry of the boundaries helps guide the choice for the location of the image charges. 13 c 2010 by Charles Thorn
Infinite Plane An elementary example,where the choice is easy,is a charge above a conducting plane. The Dirichlet Green function for the region above the xy-plane is,writing r=p+22,and similarly for r': 1 GD(r,） 1 4玩 p-p)2+(a-2元-√p-pP+z+元 (57) To solve the general Dirichlet problem we need i.7G=- 、aG 0z1z=0 1 2m(c-x')2+(y-)2+z2)3/2 (58) Here since the region of interest is the upper half plane,the outward normal to the bounding xy-plane points in the negative z-direction,i.e.n=-2.The general Dirichlet problem is to impose o(x,y,0)=V(r,y),Then,applying the Green formula,the potential above the plane is (r）= ddwV(,纱c-rP+g-P+2丽 (59) As a simple example,take V=0 outside a disk of radius R and V(,y)=Vo inside the disk: (r'）=(p,）= 2元0 pdp2pp co (60) The integral over p could be done leaving a complicated integral to do.However,when the observation point r'is on the z-axis,i.e.p'=0,the integral simplifies dramatically 60,0= 62m R 2元J0 %r du(u+z2)-32=61 2 2+元 (61) Also,for general V(r,y),it is easy to evaluate the behavior as r'=vp2+2R: Voz'R2 forrR 2Tr3 dxdyV(x,y）→ 28, (62) for the simple example. Sphere We next discuss a less trivial example:a point charge outside a spherical conductor held at 0 potential.(We assume the potential also vanishes at infinity.)Axial symmetry dictates 14 ©2010 by Charles Thorn
Infinite Plane An elementary example, where the choice is easy, is a charge above a conducting plane. The Dirichlet Green function for the region above the xy-plane is, writing r = ρ + zzˆ, and similarly for r 0 : GD(r, r 0 ) = 1 4π h 1 p (ρ − ρ0 ) 2 + (z − z 0 ) 2 − 1 p (ρ − ρ0 ) 2 + (z + z 0 ) 2 i (57) To solve the general Dirichlet problem we need nˆ · ∇G = −zˆ ∂G ∂z z=0 = − 1 2π z 0 ((x − x 0 ) 2 + (y − y 0 ) 2 + z 02 ) 3/2 (58) Here since the region of interest is the upper half plane, the outward normal to the bounding xy-plane points in the negative z- direction, i.e. nˆ = −zˆ. The general Dirichlet problem is to impose φ(x, y, 0) = V (x, y), Then, applying the Green formula, the potential above the plane is φ(r 0 ) = 1 2π Z dxdyV (x, y) z 0 ((x − x 0) 2 + (y − y 0) 2 + z 02) 3/2 (59) As a simple example, take V = 0 outside a disk of radius R and V (x, y) = V0 inside the disk: φ(r 0 ) = φ(ρ 0 , z 0 ) = V0 2π Z 2π 0 dϕ Z R 0 ρdρ z 0 (ρ 2 + z 02 + ρ 02 − 2ρρ0 cos ϕ) 3/2 (60) The integral over ρ could be done leaving a complicated ϕ integral to do. However, when the observation point r 0 is on the z-axis, i.e. ρ 0 = 0, the integral simplifies dramatically φ(0, z 0 ) = V0 2π Z 2π 0 dϕ Z R 0 ρdρ z 0 (ρ 2 + z 02) 3/2 = V0z 0 2 Z R2 0 du(u + z 02 ) −3/2 = V0 1 − z 0 √ R2 + z 02 (61) Also, for general V (x, y), it is easy to evaluate the behavior as r 0 = p ρ 02 + z 02 R: φ ∼ z 0 2πr 03 Z dxdyV (x, y) → V0z 0R2 2r 03 , for r 0 R (62) for the simple example. Sphere We next discuss a less trivial example: a point charge outside a spherical conductor held at 0 potential. (We assume the potential also vanishes at infinity.) Axial symmetry dictates 14 c 2010 by Charles Thorn