Wilcox Chapter1R

1.1 Chapter 1: Introduction to Electrostatics 1.1 Electric field definition We begin at a very basic level. Definition of E≥ in terms of a test charge, q: F≥ = q E≥. (1.1) Coulomb's law in the Gaussian system of units: F≥ 12 = q1q2 x≥ 1-x≥ 2 |x≥ 1-x≥ 2|3. (1.2) (force of 2 on 1) Picture: q2 x2≥>x≥ origin x≥>x≥qx-x2 >x-x≥≥ ≥ ≥' ' 11 1 E2≥(x≥ 1) = q2 x≥1-x≥ 2 |x≥ 1 - x≥ 2|3 or E≥(x≥) = q x≥- x≥' |x≥ - x≥'|3(1.3) For a charge density: E≥(x≥) = ∫ d3x'ρ(x≥') x≥-x≥' |x≥ - x≥'|3. (1.4) (Similarly for surface charge or line charge densities.) Of course, if we choose ρ = q δ (x≥-x≥'), we get back the previous discrete expression. 1.2 1.2 Dirac delta function properties One way to define Dirac delta functions: δ (x-x') ≡ α .0lim 1 α√⎯⎯2π exp[- 1 2α2 (x-x')2], | < >|~å x=x~1/ å ' We get ∫ -∞ ∞ δ(x-x') d(x-x') = 1 α√⎯⎯2π ∫ -∞∞ dx e-1/(2α2)x2 = 1. (Easy way of seeing the last integral: I ≡ ∫ -∞∞ dxe-ax2, I2 = ∫ dx ∫ dx'e-a(x2+x'2) = ∫ rdφdre-ar2, = 2π∫ 0∞ drr e-ar2 = 2π 1 2a = π a .) In 3-D space: δ (x≥-x≥') ≡ α .0lim ⎝⎜⎜⎛ ⎠⎟⎟⎞1 α√⎯⎯2π 3 exp[ -1 2α2 [(x-x')2+(y-y')2+(z-z')2]]. 1.3 Generalize it to any other orthogonal coordinate system: old: ds2 = dx2 + dy2 + dz2 new: ds2 = du2 U2 + dv2 V2 + dw2 W2 U,V,W are called "scale factors" (they are intrinsically positive) old: d3x = dxdydz new: d3x = du dv dw U.V.W ((UVW)-1 is just the Jacobian of the transformation, where U-1 = |∂x≥ ∂u|, etc.) General idea: δ(x≥-x≥') =α .0lim ⎝⎜⎜⎛ ⎠⎟⎟⎞1 √⎯⎯2π α 3 exp[-1 2α2(extendedinfinitismal length element)] δ(x≥-x≥') = α .0lim ⎝⎜⎜⎛ ⎠⎟⎟⎞1 √⎯⎯2π α 3 exp[-1 2α2((u-u') 2 U2 + (v-v') 2 V2 + (w-w') 2 W2)] => δ(x≥-x≥') = δ ⎝⎜⎛ ⎠⎟⎞u-u' U δ ⎝⎜⎛ ⎠⎟⎞v-v' V δ ⎝⎜⎛ ⎠⎟⎞w-w' W. Given (useful in many other situations) δ(f(x)) = ∑ i 1 |df dx(xi)| δ (x≥-x≥ i), so δ(ax) = 1 a δ(x), => δ(x≥-x≥') = U V W δ (u-u') δ (v-v') δ (w-w'). 1.4 Makes sense since this now gives ∫ δ(x≥-x≥')d3x = ∫ U V W δ (u-u') δ (v-v') δ (w-w') x dudvdw U V W, = 1. √ In particular, can show that the repn of the 3-D delta function in spherical coord. is δ(r≥-r≥') = 1 r2 δ ( r-r') δ(cosθ − cosθ ') δ (φ -φ'). These three dimensional delta functions can be "smeared" or integrated over some dimensions or objects, producing newlower dimensional delta functions. For example, the "linedelta function" seen in the introduction is nothing but a 2dimensional delta function integrated over the line inquestion. Then, an integration across the line producesunity. These delta functions can also be "directed"; that is,the result can depend on the direction of integration. Wewill see these types of objects repeatedly in Chs.5 and 6. 1.3 Gauss' law and solid angles The integral of the normal component of E≥ due to point charges is given as follows. n^ Vcontains charges q , =1,...,njj S 1.5 ∫o s da n^.E≥ = ∑ j qj ∫o s da n^.x≥-x≥ j |x≥ -xj ≥| 1 |x≥ -xj ≥|2(1.5) Here da n^ = ds≥, (1.6) is the outward directed element of surface area. Then da n^.x≥-x≥ j |x≥ -xj ≥| = ds≥.x≥-x≥ j |x≥ -xj ≥| = ds|| (1.7) unit vector along( away from)the line from the j th charge where ds || is the component || to the "line of sight" from the jth charge: ds≥= nda^ x-xj≥ ≥ qxj≥x≥ j Thus, since ds|| rds r2Ωd ||= => dΩj = da n^ . x≥-x≥ j |x≥ -xj≥|3 ()+ for charge "inside" - for charge "outside"(1.8) 1.6 Interpretation: solid angle subtended by ds≥ as seen from location of jth charge. There are now 2 possible situations: jV jV ∫o dΩj= 4π∫ o dΩj= 0 Reason (simple object): ds≥ jx-xj≥≥ overall neg.contributionoverall pos. contribution cancel Hence ∫o s da n^.E≥ = ∑ j in V 4π qj. (1.9) This is a statement about the global behavior of E≥. We need to get statements about the local behavior of E≥; that is, we need differential equations (supplemental by BC's). Canalways imagine a charge density as being due to point chargesin infinitismal volumes. Then we may replace ∑ j in V qj .∫ V d3xρ(x≥). (1.10) Now, our flux statement becomes (by Gauss) 1.7 4π ∫v d3x ρ(x≥) = ∫o s da n^.E≥ = ∫v d3x ∇≥.E≥ (x≥ ). (1.11) True for an arbitrary volume, V. Result: ∇≥.E≥ = 4πρ(x≥ ). (1.12) We notice that E≥ (x≥ ) = ∫ d3x'ρ(x≥ ') x≥-x≥' |x≥-x≥'|3 = -∇≥Φ (x≥ ), (1.13) where Φ(x≥ ) (the "scalar potential") is given by Φ(x≥ ) = ∫ d3x' ρ(x≥') |x≥-x≥'|. (1.14) Comparison with the above informs us that ∇2Φ (x≥ ) = -4πρ(x≥ ), (1.15) and therefore that ∇2 1 |x≥-x≥'| = -4πδ (x≥-x≥'). (1.16) (We also see that ∇≥×E≥= 0.) [There are also many direct ways of showing (1.16).] A more general statement is in fact (x≥'- x≥ .x≥ for convenience) ∇i∇j 1 r = 3xixj-r2δij r5 - 4π 3 δijδ (x≥ ) (1.17) (r = | x≥|) This equation will be justified later in the semester in Ch.5. 1.8 1.4 Deviations from inverse square law Now that we have an expression for Φ (x≥ ), we can talk about searching for deviations from the inverse square law. Assume F≥ 12 = q1q2 n^12 r122+δ,from 2 to 1, δ<<1 (1.18) Regard q 1 as a test charge. Then E≥ 2 (x≥ 1) = q2 n^12 r122+δ ,( 1.19) => Φ 2 (x≥ 1) = q2 (1+δ)r121+δ. (1.20) Verify last statement: E≥ 2 = -∇ ≥ 1Φ2 = - q2 (1+δ) ∇≥ 1 1 |x≥ 1-x≥ 2|1+δ (1.21) = − q2 (1+δ) {- x≥ 1-x≥ 2 |x≥-x≥'|3+δ (1+δ)}. (1.22) Picture: θ dsrr'σsphere radius=R dΦ = σds (1+δ)r'1+δ(q = σS, S = 4 πR2) (1.23) 1.9 Imagine rΦ(r)ab What one can show is that Φ(r) = Φin(r) + Φout(r), (1.24) where Φin = qa (1-δ2) 1 2ar [(a+r)1-δ - (a-r)1-δ], (1.25) Φout = qb (1-δ2) 1 2br [(r+b)1-δ - (r-b)1-δ]. (1.26) Now imagine connecting a wire between the spheres: => Φ(a) = Φ(b). (1.27) Can show that this requires qb ~~ qaδ 2(b-a) [a ln ⎝⎜⎛ ⎠⎟⎞a-b a+b + b ln ⎝⎜⎛ ⎠⎟⎞4a2 a2-b2], (1.28) so that q b vanishes when δ =0 (inverse square law.) 1.5 Surface charge and dipole layers We learned in the introduction that (D≥ 2-D≥ 1).^n21 = 4π σ, (1.29) (E≥ 2-E≥ 1)×^n21 = 0. (1.30) 1.10 If both volumes are vacuum, then (E≥ 2-E≥ 1).^n21 = 4π σ ( (E2-E1)n = 4π σ), (1.31) and the potential is Φ (x≥ ) = ∫ s σ(x≥')da' |x≥-x≥'|. (1.32) This eqn is consistent with the statement about the discontinuity in E≥. Can see it as follows. Imagine a flat surface sheet of charge: E≥ = -∇≥Φ = -∫ s da'σ (x≥ ')∇≥ 1 |x≥-x≥'|, (1.33) => En = -∫ s da'σ (x≥ ')n^21.∇≥ 1 |x≥-x≥'|. (1.34) ↑↑↑↑constant vector But remember da' ^n21.∇≥ 1 |x≥-x≥'| = ds≥'.(x≥'-x≥) |x≥-x≥'|3 = dΩ'. (1.35) Observation position is now from x≥ (field position) rather than x' (source position as it was before). Picture: 1.11 x'-x≥≥n^ 21 ≥≥x' xorigin can be "inside" or "outside" relative to n21^.12 Also, remember that d Ω' >0 "inside", d Ω '<0 "outside." => En(x≥ ) = -∫ s σ(x≥')dΩ' , flat surfaces (all x≥). (1.36) Now imagine we are very close to an arbitrary surface. We can imagine partitioning the surface into an infinitismaldisk and a remainder: 2 1 ≥x' x≥n21^ disk very close to surface(2-D cutout) Because the fields are linear E≥ = E≥ remainder + E≥ disk. (1.37) But because we are very close to the disk's surface, it is clear that σ(x≥') and n^21 are nearly constant over the integration over the disk, so that (E2,disk)n ≈ 2πσ, (1.38) (E1,disk)n ≈ -2πσ. (1.39) 1.12 very near the surface. This means there is a discontinuity in En near the surface given by (E2-E1)n = 4πσ, (1.40) as stated above. (The E n due to the remainder is obviously continuous.) Since the contribution of the remainder toE 1n & E2n is not known, E 1n & E2n are separately unknown. Another type of surface distribution is the dipole layer. 2 1n21^ σ≥(x') σ-≥d (x') Define D(x≥') = lim d(x≥') .0 σ(x≥')d(x≥'). (1.41) One can show (see the derivation preceeding prob. 1.7; the surface S = S 2) |not a const.vector Φ (x≥ ) .d .0 ∫ s D (x≥') n^21.∇≥' 1 |x≥-x≥'| da' -dΩ(from field point)' => Φ (x≥ ) = -∫ s D (x≥')dΩ'. (1.42) For constant D this means that the potential is dependent only on the perimeter of an open surface. (For a closed 1.13 surface see for prob. 1.19(b)). This implies (as before take a small disk) Φd 2 = 2πD,Φd 1 = -2πD . (1.43) Rest is continuous => ( Φ2-Φ1) = 4πD. Really a local statement at every point on surface. An easier way of seeing this last result is 12n21^ σ Previous result: E1d n = -2πσ (single layer). Therefore, we have Etotal n = -4πσ between the two layers, so that ΔΦ = -∫ E≥.d l≥ = 4πσd ⇒ d .0 4πD. (1.44) 1.6 Boundary conditions and uniqueness of solutions General question: When are the solutions to the eqn s of electrostatics, ∇≥.E≥ = 4πρ, ∇≥×E≥ = 0, (or ∇2Φ = -4πρ, E≥ = -∇≥Φ) unique? An application of Gauss ("Green's first identity"): ∫ v ( ƒ∇2ψ + ∇≥ψ.∇≥ƒ) d3x = ∫o s da ƒ∂ψ ∂n . (1.45) 1.14 Suppose we have two solutions ( Φ1,2) whose B.C.'s are: Dirichlet: Φ1|s= Φ2|s = f(x,y,z) (1.46) Neumann:∂Φ1 ∂n|s = ∂Φ2 ∂n|s = g(x,y,z) (1.47) Mixed: ⎭⎪⎪⎬⎪⎪⎫ Φ1|s' = Φ2|s' ∂Φ1 ∂n|s" = ∂Φ2 ∂n|s" S'U S" = S (1.48) ↑"union" ↑whole surface (no overlap) Let U = Φ1-Φ2. Then ∇2U = 0 in V and U|s = 0 , Dirichlet (1.49) ∂U ∂n|s = 0 , Neumann (1.50) U|s' = 0, ∂U ∂n|s" = 0, Mixed. (1.51) Let ψ = ƒ = U in Green's first identity. Result: ∫ v (U∇2U + (∇≥.U)2) d3x = ∫o s daU ∂U ∂n |s . (1.52) For all three cases, we have the rhs vanishing. =>∫ v d3x(∇≥U)2 = 0, (1.53) => ∇≥U = 0 in V, => U = const. (For Dirichlet & Mixed cases we know that U| s = 0, so that the constant is zero there.) 1.7 Dirichlet and Neumann Green functions Rather than trying to solve for each charge distribution as a special case, there is a more powerful method whichallows us to solve a class of problems with a given set of 1.15 BC's. Called Green functions. Let us say we are trying to solve ∇' 2Φ (x≥') = -4 πρ (x≥'), (1.54) in V. (Can have boundaries at infinity or in finite regions.) The trick is first to solve ∇' 2G (x≥,x≥') = -4 πδ (x≥-x≥'), (1.55) for the same volume. G (x≥,x≥') is then the potential of a point charge at x≥, the BC of which are as yet totally unspecified. (I probably should have introduced (1.55) as ∇2G (x≥ ',x≥) = -4πδ (x≥-x≥') to conform with the usage that x≥' is regarded as the source and x≥ as the field coordinate. In any case you can regard the coordinates inside as G(source, field).) We willuse this freedom of choice of BC to simplify our expressions. "Green's second identity": ∫ v d3x'( ƒ∇' 2ψ - ψ ∇' 2 ƒ) = ∫o s da'[ ƒ∂ψ ∂n' - ψ ∂ ƒ ∂n']. (1.56) Let ƒ = G (x≥,x≥'), ψ = Φ(x≥'). Then => ∫ d3x'[G(x≥,x≥') ∇' 2Φ(x≥') - Φ(x≥')∇' 2G(x≥,x≥')] = ∫o s da'[G(x≥,x≥') ∂Φ ∂n' - Φ(x≥') ∂G(x≥,x≥') ∂n'], (1.57) LHS = -4 π∫ d3x'G (x≥,x≥')ρ (x≥') + 4πΦ (x≥). (1.58) (n^ = outward normal) We now use the freedom of choice of BC for G (x≥,x≥') to specifiy GD(x≥,x≥ ') = 0, x≥ ' on S (Dirichlet). (1.59) 1.16 Then Φ(x≥ ) = ∫ d3x' GD(x≥,x≥') ρ (x≥') - 1 4π ∫o s da'Φ(x≥') ∂ GD(x≥,x≥') ∂ n'. (1.60) Physically, this Green function is the solution for the potential of a +unit point charge at x≥ ' in the presence of grounded conductors. For Neumann BC, we must be more careful. One must realize that since ∇' 2GN(x≥,x≥') = -4 πδ (x≥-x≥'), (1.61) => ∫ v d3x' ∇≥'.∇≥'GN(x≥,x≥') = -4 π, (1.62) => ∫o s da'∂ GN(x≥,x≥') ∂ n' = -4π. (1.63) One can not take ∂GN ∂n' = 0 on S in general. Easiest thing to do: ∂GN ∂n' = - 4π S for x≥' on S. (1.64) S = total surface area. Then Φ(x≥ ) = 1 S ∫s da'Φ(x≥') + ∫ d3x' GN(x≥,x'≥ )ρ (x≥') ≡ <Φ>s+ 1 4π ∫o da'GN(x≥,x≥')∂Φ ∂n'. (1.65) A possible situation is to find the fields outside of: ("exterior problem") 1.17 take to ∞n^' imaginary thin "Neumann" tuben^' n^'BC's specified If we add the imaginary surfaces shown, then we have a single surface, S, which has infinite area as we take theouter part to ∞. Clearly, we then have <Φ>s = 0, (1.66) and we may consistently take ∂GN ∂n' = 0, (1.67) on the finite volumes. (Doesn't contradict ∫o s ∂GN ∂n' da' = -4 π.) Thus, G N(x≥,x≥') is the potential for a point charge for zero normal electric field on the boundaries for the exterior problem. Can think of Dirichlet as (conductor has surface charge density which excludes E≥ norm.) 1.18 conductor .∞ε()x'≥ x≥ Φ(x≥,x≥') = GD(x≥,x≥') Can think of Neumann as ("Neumann conductor" has dipole density which excludes E≥ tang. Has properties of a "dual superconductor".) x'≥ x≥ (dual superconductor)ε .0 "Neumann conductor" ( ) Φ(x≥,x≥') = GN(x≥,x≥') For electrostatics then, Neumann is not as physical as Dirchlet. It makes more sense, however, in fluid flow problems. Note the ε .∞,0 remarks above. ε is the "dielectric constant" to be introduced in Ch.4. I am saying that the ε .∞ limit of a point charge near a dielectric gives you the correct Dirichlet Green's function for this geometry (except that the conductor will be uncharged) and the ε . 0 limit in the same situation gives the correct Neumann one. Note that the inside fields in these cases may or may not be physical. An important aspect of Green functions is the symmetry property, G (x≥,x≥') = G (x≥',x≥ ). We can show it as follows: ∫v d3x( ƒ∇2ψ - ψ∇2 ƒ) = ∫o s da[ ƒ∂ψ ∂n - ψ∂ ƒ ∂n ]. (1.68) Choose (Dirichlet case) 1.19 ƒ = GD(x≥",x≥ ) , ψ = GD(x≥',x≥ ) => ∫ d3x [G(x≥",x≥ ) ∇2G(x≥',x≥ ) - G (x≥',x≥ ) ∇2G(x≥",x≥ )] = ∫o da[G(x≥",x≥ ) ∂ ∂n G (x≥',x≥ ) - G (x≥',x≥ ) ∂ ∂n G (x≥",x≥ )]. (1.69) But of course since G D(x≥',x≥) = 0 for x≥ on S, we have GD(x≥",x≥') = GD(x≥',x≥"). (1.70) If we do the same thing for Neumann, the RHS does not automatically vanish, and we get: (GN(x≥",x≥') - GN(x≥',x≥")) = <GN(x≥",x≥ ) - GN(x≥',x≥ )>s(x≥)(1.71) However, each of these terms vanishes separately if S includes the surface at infinity since <GN(x≥',x≥ )>s(x≥) = 1 S ∫o sda GN(x≥',x≥ ) .R . ∞0 (1.72) GN(x≥',x≥ ) ~ 1 R (|x≥ | >> |x≥'| ) da ~ R2dΩ S = 4πR2 (If we choose G N = const. on the surface at infinity, the RHS of (1.71) still vanishes.) Therefore, G N(x≥",x≥') = GN(x≥',x≥") also for the exterior Neumann problem. Is not necessarily true for interior Neumann problems. Let's look at this insome more detail. Have shown (G ≡ G N ; x' ≡ x≥') G(x", x') - G (x', x") = <G(x",x) - G(x',x) >s(x)(1.73) 1.20 =><G(x",x')-G(x',x") >s(x')=<G(x",x)>s(x )-<G(x',x)>s(x),s(x') source | | field =><G(x', x")>s(x') ≡ const. (1.74) Note that (1.74) implies ∂ ∂n' <G(x, x')>s(x) = <∂ ∂n' G(x, x')>s(x) = 0. (1.75) But we also have that ∂ ∂n' G(x, x') = -4π S, (1.76) for x' on S. Therefore, it would seem from (1.76) that <∂ ∂n' G(x, x')>s(x) = -4π S, (1.77) contradicting (1.75) above. We can clear this up with the hypothesis that there is a surface delta function lurking in (1.76). The hypothesis is that ∂ ∂n' G(x, x') = -4π S +4πδ(s)(x-x'). (1.78) [What the last statement really means is lim x' .s ∂ ∂n' G(x, x') |x on s = -4π S + 4πδ(s)(x-x'). ] This is now consistent with (1.75) since ∂ ∂n' <G(x, x')>s(x) = <∂ ∂n' G(x, x')>s(x) = -4π S + 4π S = 0.(1.79) We can confirm this hypothesis from our earlier expression for Φ: 1.21 Φ(x≥ ) = <Φ>s + ∫ d3x' G(x,x') ρ(x') + 1 4π ∫o s da'G(x,x') ∂Φ ∂n',(1.80) which implies ∂Φ ∂n |s =∫ d3x' ∂ ∂n G(x,x') ρ(x') + 1 4π ∫o s da' ∂ ∂n G(x,x') ∂Φ ∂n' . (1.81) (Note an indescretion here: the normal derivatives are on the first, not second arguments of G(x,x'). We will clear this upin prob. 1.7.3.) This must reduce to an identity: ∂Φ ∂n|s = ∫ d3x'{-4π S + 4πδ (s)(x-x')}ρ(x') + 1 4π ∫o s da'{-4π S + 4πδ (s)(x-x')}∂Φ ∂n', (1.82) => ∂Φ ∂n|s = -4πQ S + 4π ∫ d3x'δ(s)(x-x') ρ(x') - 1 S ∫o s da'∂Φ ∂n' + ∂Φ ∂n|s. (1.83) But -1 S ∫o s da' ∂Φ ∂n' = -1 S ∫ d3x'∇' 2Φ(x') = 4πQ S. (1.84) -4ρx'()π Therefore ∂Φ ∂n|s = ∂Φ ∂n|s + 4π∫ d3x'δ(s)(x-x') ρ(x'), (1.85) and consistency is maintained as long as we take ∫ d3x'δ(s)(x-x')f(x') = 0, (1.86) for any f(x'). (This last fact can be seen from other, more mathematical, points of view as well. So the surface is notcontained in the volume!) 1.22 Actually, surface delta functions appear using Dirichlet B.C. as well. Remember Φ(x) = ∫ d3x' G(x,x') ρ(x') - 1 4π ∫o s da'Φ(x')∂G ∂n' . (1.87) Take x≥ on S. Then Φ|s = - 1 4π ∫o s da'Φ(x') ∂G ∂n', (1.88) which means that - 1 4π ∂G ∂n' |s = δ(s)(x-x'). (1.89) [This really means lim x .s - 1 4π ∂G ∂n' (x,x')|x'on S = δ(s)(x-x').] We will illustrate the presence of surface delta functions in some explicit problems later. I claimed earlier that G N(x',x") is not necessarily symmetric in it's arguments for interior BC's. Can see thisas follows. First, we will show that, in general, theDirichlet Green function is unique, while the Neumann one isnot. Both cases at once: ∇ 2 G1(x',x) = -4 πδ(x-x') ∇2 G2(x',x) = -4 πδ(x-x')either Dirichlet or Neumann (1.90) Form U(x',x) = G 1(x',x) - G 2(x',x), (1.91) 1.23 =>∇2U(x',x) = 0. (1.92) Then U(x',x)|s(x) = 0 Dirichlet (1.93) ∂ ∂n U(x',x)|s(x) = 0 Neumann (1.94) As before, look at Green's first identity: ∫v d3x(U(x',x) ∇2 U(x',x) + [ ∇≥U(x',x)]2) = ∫o s da U(x',x) ∂ ∂n U(x',x), (1.95) vanishes in either case => ∫v d3x(∇≥U(x',x))2 = 0, (1.96) => ∇≥U(x',x) = 0 in V. (1.97) In the Dirichlet case, we proved that G D(x',x") = G D(x",x') Combined with U |s = 0, we have that U(x',x) = 0, (1.98) and GD(x',x") is unique. On the other hand, we have proven the symmetry G N(x',x") = G N(x",x') only for the exterior problem (in which case G N(x',x") is unique up to a constant.) Thus, in general for the interior problem we only have that ∇≥U(x',x) = 0 implies U(x',x) = F(x'), (1.99) where F(x') is some function of x'. It is also easy to see that the most general solution is then G N(x',x) = G Nsym(x',x) + F(x') (1.100) 1.24 where GNsym(x',x) = G Nsym(x,x'). (1.101) It is easy to see that ∇2F(x) = 0 and that G Nsym(x',x) is unique up to a constant. We can get a physical interpretation of F(x') from anti-symmetrizing (1.100). We can write F(x') - F(x") = G N(x',x") - G N(x",x'). (1.102) We now compare to our earlier (1.73): <GN(x',x) - G N(x",x)>s(x) = GN(x',x") - G N(x",x'). Averaging over the s(x) surface there gives F(x') = <GN(x',x)>s(x) + Const. (1.103) Thus, the arbitrary function F(x') represents the average potential of the surfaces (which vanishes when the surface at ∞ is included) up to a constant. The fact that this potential can take an arbitrary value for each source position is due to the fact that the system is isolated. The sensible thing to do would be to refer it to a constantoutside potential (via a thin wire, say), which would thenremove the arbitrariness. Thus, we can always require the Neumann Green function, in any case, to be symmetric in itsarguments. 1.8 Electrostatic energy Enough of this! Let's investigate some aspects of electrostatic energy. Imagine sources in infinite space.Work done on a test charge: 1.25 Wi = qiΦ(x≥ i) (brought in from ∞ ) (1.104) If Φ is due to point charges, [(n-1) of them ] Φ(x≥ i) = ∑ j=1n-1 qj |x≥ i - x≥ j|. (1.105) Total energy: (now n of them) W = ∑ i=1n ∑ j<i qiqj |x≥ i - x≥ j|, (1.106) or W = 1 2 ∑ i≠j qiqj |x≥ i - x≥ j|. (1.107) Continuous distributions (now has self-energy contributions): W = 1 2 ∫ ∫ ρ(x≥ ) ρ(x≥') |x≥ - x≥'| d3xd3x'. (1.108) In contrast to the discrete sum, this expression is in general finite (because of the d3x, d3x'elements) when self- energies are included as long as ρ(x'≥ ) is not too singular. Another expression: W = 1 2 ∫ ρ(x≥ )Φ(x≥ ) d3x. (1.109) But now ρ(x≥ ) = -1 4π ∇2Φ (x≥ ), (1.110) => W = 1 8π ∫ d3x [∇2Φ]Φ, (1.111) 1.26 = -1 8π ∫ d3x [∇≥.(Φ∇≥ Φ) - ∇≥ Φ .∇≥ Φ], (1.112) = -1 8π ∫o s da Φ ∂ ∂n Φ + 1 8π ∫ d3x |E≥|2(1.113) because da ~ R2dΩ, Φ ~ 1 R => ∂ ∂n Φ ~ 1 R2 => W = ∫ d3x w, w = 1 8π |E≥|2. (1.114) “w” is energy density. What if we try to calculate self- energy of a stationary point charge: |E≥|2 = q2 |x≥ - x≥'|4. (1.115) Then (take x≥' = 0; x ≡ 1 r ) W self = q2 ∫ d3x r4 = 4πq2 ∫ .0∞ dr r2 ~ 4π q2∫ 0.∞ dx. (1.116) Divergence is at small distances, which means high energies. Called a linear divergence. Actually, divergent in the quantum theory also, but only logarithmically. This permits renormalization to take place. (Ask me what this means.) The equation W i = qiΦ (x≥ i) also allows a physical interpretation of the fact that we may always choose G(x',x") = G(x",x') for Green functions. We learned that G(x',x")represents the potential of a (+) unit charge in given BC's.Because of this connection between interaction energy andpotential, we see that the Dirichlet or Neumann Greenfunction can also be considered the interaction energy of two positive (or two negative) point charges (at x',x") with given BC's. Since the two charges are equivalent, the Greenfunction must be symmetric in its arguments. This argumentfails in the case of interior Neumann boundary conditions, 1.27 again because one can imagine defining the zero of potential for each new source position. This is not possible in theDirichlet case since the potentials are specified by theBC's. 1.9 Capacitance Let’s talk about capacitance. To emphasize the fact that the Green function contains all the information to solve anelectrostatic problem in a given geometry, let’s connect thisto the Green function. I learned the following formulas fromSchwinger. Adopting Dirichlet BC's, the potential is Φ (x ≥) = ∫ d3x GD(x≥ ,x≥')ρ(x≥') - 1 4π ∫o s da'Φ (x≥')∂GD(x≥ ,x≥') ∂n'. (1.117) Let us apply this to a volume, V, bounded by a number of conducting surfaces. Let's let ρ(x≥') = 0 (1.118) in V, and Φ (x≥')|si .Vi (const.) (1.119) on the ith surface. Then, using superposition, the potential in the volume V is Φ (x≥) = -1 4π ∑ j Vj ∫o sjdaj'∂GD(x≥ ,x≥') ∂n'j. (1.120) Therefore, the surface charge density, given by the discontinuity of the normal electric field, is (n^i is an outward normal to V on the ith conductor, not necessarily to the conductor volume) σi (x≥) = 1 4π n^i.∇≥ Φ (x≥) = 1 4π ∂ ∂ni Φ, (1.121) 1.28 = − 1 16π2 ∑ j Vj ∫o sj da'j∂ ∂ni ∂ ∂n'j GD(x≥ ,x≥'). (1.122) The total charge on the ith surface is therefore Qi = ∫o si dai ßi = -1 16π2 ∑ j Vj ∫o si,sj dai da'j ∂ ∂ni ∂ ∂n'j GD(x≥ ,x≥'). (1.123) We can write this more simply as Qi = ∑ j Cij Vj, (1.124) where the capacitances (C ii) or coefficients of induction (Cij,i≠j) are given by Cij = - 1 16π2 ∫o si,sj dai da'j ∂ ∂ni ∂ ∂n'j GD(x≥ ,x≥'). (1.125) The capacitance of a conductor is therefore the charge on the conductor when it is maintained at unit potential, all otherconductors being grounded . A similar statement holds for the coefficients of capacitance, except the total charge ismeasured on a conductor other than the one being held at unitpotential. The coefficients are symmetric in their indices, C ij = Cji. (1.126) because G D(x≥ ,x≥') = G D(x≥',x≥) . Given the energy expression, W = 1 2 ∫ d3xd3x' ®(x≥) ®(x≥') |x≥ - x≥'| .1 2 ∫ dada' ß(x≥) ß(x≥') |x≥ - x≥'|, (1.127) => W = 1 2 ∫ da Φ(x≥) ß(x≥), (1.128) we have, in this case 1.29 W = 1 2 ∑ i Vi ∫o si dai ßi(x≥) = 1 2 ∑ i ViQi (1.129) or W = 1 2 ∑ ij Cij ViVj (gives diff. perspective (1.130) on the C ii's.) We have to be careful when the surface at ∞ is included in our analysis since it's capacitance and coeff.'s of induction can only be defined in a limiting sense in a given problem.With this understanding then, we can now show that ∑ i Cij = 0, (1.131) for an interior or “exterior” problem as follows: ∑ i Cij = -1 16π2 ∫o sj da'j ∂ ∂n'j (∑ i ∫o si dai ∂ ∂ni GD(x≥',x≥)). (1.132) But 1 4π ∑ i ∫o si dai ∂ ∂ni GD(x≥',x≥) = 1 4π ∫ d3x ∇2 GD(x≥',x≥) = -1. (1.133) Since ∂ ∂n'j (const.) = 0, the top line says ∑ i Cij = 0. This means that Q ≡ ∑ i Qi = ∑ ij CijVj = 0, (1.134) as it must since we are imagining a closed system with no volume charge. (Can you see why this line of reasoning failsin the exterior problem, when the surface at infinity is notincluded?) In addition, only relative values of thepotential have significance here. We can see this from ∑ j Cij(Vj + const.) = ∑ j CijVj. (1.135) Simple example: parallel conductor geometry 1.30 aVS S2 1 Since the C ij are symmetric in i & j and must sum to zero on one index, this means C11 = -C21 = -C12 = C22 ≡ C. (1.136) Also Q 1 = -Q2 = C(V 1-V2) = C ΔV. (1.137) The energy is E = 1 2 ∑ i,j CijViVj = 1 2 C ΔV2. (1.138) These are the usual expressions for a parallel plate capacitor. C can be shown from Gauss' law to be, C ≈ A 4πa. (finite surface, A. Can be directly evaluated; see prob. 2.4.) (1.139) Note: Usually it is difficult to get exact solutions for the Green function for a given geometry, so approximationmethods must be used in computing the C ij. 1.31 Problems 1.1.1 (a) A vector field is defined by E≥ = -∇≥Φ, with Φ = d≥.r≥ r3 . (d≥ = constant vector) Derive an expression for E≥. (b) Prove that E≥ in part (a) can be written as follows: E≥ = ∇≥×A≥, with A≥ = d≥×r≥ r3. (c) Calculate ∇≥.E≥ and ∇≥×E≥. [Note: This problem neglects certain subtleties that occur at r≥= 0. Adapted from DiBartolo's "Classical Theory of Electromagnetism".] 1.2.1 Given the scale factor definitions, U-1 ≠ |∂x≥ ∂u|, V-1 ≠ |∂x≥ ∂v|, W-1 ≠ |∂x≥ ∂w|, and δ(x≥-x≥') = δ(u-u')δ(v-v')δ(w-w')UVW, find the form of the delta function for: 1.32 (a) cylindrical coordinates x= ρ cos φ, y= ρ sin φ, where u= ρ, v=φ, w=z; (b) spherical coordinates x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, where u=r, v=cos θ, w=φ. 1.2.2 Find the form of the Dirac delta function in oblate spheriodal coordinates. Use (x) x1 = R√⎯⎯⎯⎯Z2+1sinθ cosφ, (y) x2 = R√⎯⎯⎯⎯Z2+1sinθ sinφ, (z) x3 = RZ cos θ. Take u=Z, v=cos θ, w=φ as the new coordinates. (The θ and φ corrdinates correspond to the usual polar and azimuthal angles, respectively, in spherical coordinates.] 1.3.1 (a) Given in two spatial dimensions the form E≥(x≥) = ∫ da'σ(x≥') x≥-x≥' |x≥ - x≥'|2, for the electric field, argue as in the notes that one has ∫o c d l n^.E≥ = ∑ j in S 2πqj, for enclosed charge within an surface area S bounded by C, and ∇≥.E≥ = 2πσ(x≥ ), for the electric field due to charge density σ(x≥ ). 1.33 (b) For a positive point charge in two dimensions, show that (a) gives ∇2 ln|x≥-x≥'| K = 2πδ (2)(x≥-x≥'), where K is an arbitrary positive constant. (c) Confirm this delta function representation by puttingx≥'=0 and integrating both sides of (b) about an arbitrary area surrounding the origin. [Note: This identifies the potenial of a point charge in 2 dimensions as - ln|x≥-x≥'| K . However, the potential of a line charge in 3 dimensions with unit line density of charge is - 2 ln|x≥-x≥'| K. Do you know where the extra 2 comes from?] 1.3.2 (a) Find the potential everywhere for a straight uniform line charge density, λ, which makes a X-figure at the origin in the xy plane extending from L to -L on both axes (no conductors present): y xλλ L -L -LL Get the exact Φ(x,y,z), then show for L >> |x|,|y|,|z| that Φ(x,y,z) ~~ - λ ln ⎝⎜⎛ ⎠⎟⎞ (x2+z2)(y2+z2) 16L4. 1.34 Why is this answer obvious? For the opposite extreme, L << |x|,|y|,|z| show that you get the expected answer for aneffective point charge.(b) y xλλ -L L The straight uniform line charges in the xy plane are now given by one line charge, λ, situated along the y-axis from the origin to L, and another, - λ, situated along the x-axis. Again, get the exact expression, then show for L >> |x|,|y|,|z| that Φ(x,y,z) ~~ - λ ln ⎝⎜⎛ ⎠⎟⎞r-y r-x, where r = √⎯⎯⎯⎯⎯⎯⎯x2+y2+z2. 1.4.1 (a) For a potential of the form ( δ<<1) φ = q (1+δ)r121+δ , where r 12 is the distance between source and field points, show that the potential difference between two concentricspheres with uniform surface charge (radii a and b with b<a)is given by φ(r) = φ in(r) + φout(r) where 1.35 φin(r) = qa 1-δ2 1 2ar [(a+r)1-δ -(a-r)1-δ] , φ out(r) = qb 1-δ2 1 2br [(r+b)1-δ − (r-b)1-δ]. (b) After the spheres are connected by a wire, show that the charge on the inner sphere is given by qb ≈ qaδ 2(b-a) [a ln ⎝⎜⎛ ⎠⎟⎞a-b a+b + b ln ⎝⎜⎛ ⎠⎟⎞4a2 a2-b2]. 1.5.1 In deriving Eq.(1.42) of the text, consider two close surfaces, 1 and 2, with associated surface charge densities σ1(x≥') and σ(x≥"), respectively; see the figure on p. 1.12. The separation between the surfaces is such that the points at which the charge densities are equal and opposite areseparated a uniform distance, d, along the common perpendicular; ie, σ 1(x≥'-n^21d) = - σ(x≥'). We may characterize the potential in the limit of close separation of the two surfaces as, Φ(x≥) = ∫ S2 σ(x≥') |x≥-x≥'| da' - ∫ S1 σ1(x≥") |x≥-x≥"| da". With the connection x≥" = x≥' - n^21d, and the expansion, to lowest order in a≥, 1 |x≥-a≥| = 1 |x≥| + a≥.∇≥ ⎝⎜⎜⎛ ⎠⎟⎟⎞ 1 |x≥|, one may derive Eq.(1.42) of the text. Now consider the similar situation: 1.36 21n21^ σ≥(x')σ-≥d +xsurface 2 surface 1(x'+dx)^ Two regions in free space (3 dimensions) are separated by a type of dipole surface. The dipole surface is constucted byplacing equal and opposite surface charge densities oncomplementary surfaces which are then brought together suchthat a small constant separation, d, measured along one axis,remains (these are open surfaces). (a) Show that the potential can be written as Φ d(x≥) = dx^.E≥2(x≥), where x^ is a unit vector in the separation direction and E≥2 is the electric field due to surface 2. (b) Show that the discontinuity of the potential in the normal direction (to surface 2) across the surfaces is given by Φd2 − Φd1 = 4π D(x≥) x^.n^21, where D(x≥) ≡ σ(x≥)d. 1.5.2 (a) A hemisphere of radius "a" has a constant dipole surface density, D. D is defined relative to the unit vector,n^, pointing "out" as shown. Using the coordinates shown with the origin O at the hemisphere's enter, find the E≥ field everywhere along the z-axis (- ∞ <_ z <_ ∞), Then, using spherical coordinates, find the electric field components, E r and E θ, far from the region of the sphere, r>>R. 1.37 +z On^ constant D (b) A sphere of radius "a" is given a constant dipole surface charge density, D. D is defined relative to the sphere's outernormal, n ^. n^ D=constant Find the electric field both inside and outside the sphere. 1.5.3 A thin isolated metallic disk of radius R is charged until it has aquired a potential V. R V The solution for the electrostatic potential is found most easily in oblate spheroidal coordinates, Ζ and ξ, given by 1.38 z = R Ζ ξ, ρ = R[(1+ Ζ2)(1-ξ2)]1/2, (ρ2 = x2 + y2 where x, y and z are the usual Cartesian coordinates) and is given by Φ = A tan-1 ⎝⎜⎛ ⎠⎟⎞1 Ζ, where A is an unknown constant. (The third coordinate is the usual azimuthal angle φ; take your the origin of coordinates at the center of the disk and the z azis | _ to the plane of the disk.)(a) Change variables to show that the potential may be written as (r 2 = ρ2 + z2) Φ = A tan-1 ⎝⎜⎜⎜⎛ ⎠⎟⎟⎟⎞ √⎯2R √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯(r2 - R2)+√⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯(r2 - R2)2 + 4R2z2 (b) By considering the limit z § 0 when r < R, argue from consistency that A = 2V π. (c) Show that the surface charge density on the disk, σ(ρ), is given by σ(ρ) = V π2 1 √⎯⎯⎯⎯⎯⎯R2 - ρ2. [Note: d dxtan-1 y = 1 1+y2 dy dx.] 1.6.1 (a) Prove Green's reciprocation theorem: For two problems with the exact same geometry including conductingsurfaces, one has 1.39 ∫ da σ Φ' + ∫ d3x ρ Φ' = ∫ da σ 'Φ + ∫ d3x ρ 'Φ, where Φ, ρ and σ are the potential, volume charge density and surface charge density of problem 1, while Φ', ρ' and σ' are similar quantities for problem 2. (b) A point charge, q, is located somewhere between twogrounded concentric spherical conducting shells (a < r < b) asshown. a br q0V0V Find the charge induced on the inner or outer conducting shells by use of the theorem of (a). 1.7.1 Given a certain geometry, show that the Neumann and Dirichlet Green functions are related by GN(x≥",x≥) = GD(x≥,x≥") - 1 4π ∫o Sda' GN(x≥",x≥')∂GD(x≥,x≥') ∂n'. 1.7.2 Show that a solution to the Poisson equation with both Φ and ∂Φ ∂n specified arbitrarily on a closed boundary (Cauchy boundary conditions) does not exist. [Hint: Show that if one is known, the other is determined .] 1.7.3 The Neumann Green's function in 3 dimensions satisfies ∇2GN(x≥',x≥) = -4πδ(x≥ - x≥'). 1.40 Assume that the normal derivative is ∂GN(x≥',x≥) ∂n|x≥on S = - 4π S where S is the total surface area. We saw that according to (1.100), G N(x≥',x≥) = Gsym(x≥',x≥) + F(x≥'). Considering the interior problem, show that F(x≥') gives no contribution to the potential, Φ(x≥). When considering the normal derivative, ∂Φ ∂n|s, show that the general form for GN(x≥',x≥) above and Eq.(1.78) remove the "indescretion" in Eq.(1.81). 1.8.1 Prove a simplified Thompson's theorem for a single closed surface: If a surface is fixed in position and a giventotal charge is placed on it, then the electrostatic energy inthe region bounded by the surface and infinity is an absoluteminimum when the charges are placed so that the surface is anequipotential. 1.8.2 Prove the following theorem: If a number of conducting surfaces are fixed in position with a given total charge oneach, the introduction of an uncharged, insulated conductorinto the region bounded by the surfaces lowers theelectrostatic energy. 1.8.3 Investigate the self-energy of a one-dimensional straight string of charge with linear charge density λ = Q a and length a. Is this quantity divergent, and if so, is it guadratically, linearly or logarithmically divergent? 1.41 1.8.4 Consider an infinitely thin circular disk of radius R with constant surface charge density σ. (This is not a conductor!) Give an argument that the self-energy is finite. [Hint: The origins of the primed and unprimed coordinatesystems do not have to coincide. I used cylindricalcoordinates and got an upper limit.] 1.9.1 Two conductors, A and B, form an isolated system. Show, based upon the positivity of the field energy W, that thecoefficient of capacitance, C AB = CBA, and the two capacitances, C AA and C BB, obey the inequality, CAB2 < CAACBB. 1.9.2 (a) Use Green's reciprocity theorem (above) to prove Cij = Cji. (b) Show the same result from (1.125) of the script. (c) Show C ii > 0 from first principles. 1.9.3 Using Q i = ∑ jCijVj and the definition of the C ij, find the six independent coefficients of capacitance for three co-centered spherical conducting shells having radii a,b and c(a<b<c). [ Call these C aa, Cbb, Ccc, Cab, Cac and Cbc.] 1.9.4 Given a system of two conductors of arbitraryshape, with conductor Bbeing hollow and containingan interior conductor A,show that the capitancessatisfy AB 1.42 CBB ≥ CAA. 1.9.5 Consider two arbitrary, closed conductors, labelled "a" and "b" in free space, with total charge Q and -Q present.Define, C ≠ Q V a - Vb, where V a and V b are the potentials of conductors A and B. Show that C = det()Caa Cab Cab Cbb ∑ i,jCij. 1.9.6 Consider an isolated conductor, a, with capacitance Ca1. Show that the introduction of a second conductor, b, raises the capacitance of the first conductor: Ca2 > Ca1.