(a) Product rule with $\vec\nabla(\vec d\cdot\vec x)=\vec d$ and $\vec\nabla r^{-3}=-3\,\vec x/r^5$:

(b) Use $\vec\nabla\times(\vec P\times\vec Q)=\vec P(\vec\nabla\cdot\vec Q)-\vec Q(\vec\nabla\cdot\vec P)+(\vec Q\cdot\vec\nabla)\vec P-(\vec P\cdot\vec\nabla)\vec Q$ with $\vec P=\vec d$ (constant: its divergence and directional derivatives vanish) and $\vec Q=\vec x/r^3$:

First piece: $\vec\nabla\cdot(\vec x/r^3)=3/r^3+\vec x\cdot(-3\vec x/r^5)=0$ for $r\neq0$ (this is $\nabla^2(−1/r)$, WT 2.55, minus its delta). Second piece, component $i$: $(\vec d\cdot\vec\nabla)(x_i r^{-3})=d_i r^{-3}+x_i\,\vec d\cdot(-3\vec x/r^5) =d_i/r^3-3(\vec d\cdot\vec x)x_i/r^5$. Hence

Both routes agree numerically to $\sim10^{-8}$ at random points (central differences, $h=10^{-5}$). A field that is simultaneously a gradient and a curl is harmonic component-wise — possible here because $r\neq0$ excludes the source.

(a) The unit cube is $\{\sum_j t_j\hat e_j,\ t_j\in[0,1]\}$. Linearity maps it to $\{\sum_j t_j\vec v_j\}$ with edge vectors $(\vec v_j)_i=M_{ij}$ — the columns of $M$. That set is the parallelepiped spanned by $\vec v_1,\vec v_2,\vec v_3$, whose volume is the scalar triple product

the Leibniz expansion of the determinant over columns. $\blacksquare$ (b) For a coordinate map $\vec x(u,v,w)$ the local linearization is $M=\partial(x,y,z)/\partial(u,v,w)$, whose columns are $\partial\vec x/\partial u$, $\partial\vec x/\partial v$, $\partial\vec x/\partial w$. Orthogonal system means these are mutually perpendicular, with norms $U,V,W$ (WT 2.29). For a right-handed orthogonal triple the triple product is the product of the norms: $|\det M|=|\vec v_3\cdot(\vec v_1\times\vec v_2)|=W\,(UV)=UVW$ (the cross product of two perpendicular vectors has magnitude $UV$ and points along $\vec v_3$). This is WT (2.28) — the infinitesimal volume $d^3x=UVW\,du\,dv\,dw$. Numerically, for spherical $(u,v,w)=(r,\cos\theta,\phi)$: $|\det J|=r^2=UVW$ to $10^{-7}$.

(a) Write $\varphi_\epsilon(x)=\epsilon^{-1}f(x/\epsilon)$ with $f(s)=1/[\pi(1+s^2)]$: indeed $\frac1\pi\frac{\epsilon}{x^2+\epsilon^2} =\frac1\epsilon\cdot\frac1{\pi(1+x^2/\epsilon^2)}$. $f$ is nonnegative and

so by WT (2.9)–(2.10) the family is a delta sequence. Concretely, for continuous bounded $g$: substituting $x=\epsilon s$, $\int\varphi_\epsilon g\,dx=\int f(s)g(\epsilon s)\,ds\to g(0)$ (dominated convergence: $|f g|\le\|g\|_\infty f$). Numerically with $g=\cos$, the integral is $e^{-\epsilon}\to1=g(0)$. (b) Radial symmetry: total mass with $\rho^2=x^2+y^2$,

for every $\epsilon$. And for fixed $\vec x\neq0$ the density $\le\epsilon/2\pi\rho^3\to0$. Scaling form: $\varphi_\epsilon(\vec x)=\epsilon^{-2}f(\vec x/\epsilon)$ with $f(\vec s)=1/[2\pi(s^2+1)^{3/2}]$, nonnegative with unit integral — the 2-D version of (2.9). Hence $\int\varphi_\epsilon g\,d^2x\to g(0)$: the family tends to $\delta^{(2)}(\vec x)$. $\blacksquare$ (Physically this $\varphi_\epsilon$ is the charge density a grounded plane "sees" from a point charge at height $\epsilon$ — the image-plane induced density of Ch. 3.)

(a) Write $\mu=\cos\theta$, $s=\sin\theta=\sqrt{1-\mu^2}$, so $\vec x=(rs\cos\phi,\ rs\sin\phi,\ r\mu)$. The Jacobian matrix has columns $\partial\vec x/\partial r=(s\cos\phi,s\sin\phi,\mu)$, $\partial\vec x/\partial\mu=(-r\mu\cos\phi/s,\ -r\mu\sin\phi/s,\ r)$ (using $\partial s/\partial\mu=-\mu/s$), and $\partial\vec x/\partial\phi=(-rs\sin\phi,\ rs\cos\phi,\ 0)$. Expanding along the third column (or multiplying out) gives

(one clean route: this map is spherical $(r,\theta,\phi)$ — Jacobian $r^2\sin\theta$ — composed with $\theta\mapsto\mu$, which contributes $|d\theta/d\mu|=1/\sin\theta$; the product is $r^2$). WT (2.25) then gives

(b) Scale factors: the three columns above are mutually orthogonal (spherical is an orthogonal system, and replacing $\theta$ by $\mu$ rescales one column without rotating it). Norms: $U=|\partial\vec x/\partial r|=1$; $V=|\partial\vec x/\partial\mu|=\sqrt{r^2\mu^2/s^2+r^2}=r/s$; $W=|\partial\vec x/\partial\phi|=rs$. Product $UVW=1\cdot(r/s)\cdot rs=r^2$ — same denominator, by P2(b). Note the $\mu=\cos\theta$ choice absorbs the usual $\sin\theta$: $d^3x=r^2\,dr\,d\mu\,d\phi$ with no angular weight, which is why the book prefers it.

Write $\mu=\cos\theta$, $s=\sin\theta$, $A=\sqrt{\xi^2+1}$. Scale factors (the system is orthogonal in $(\xi,\theta,\phi)$, hence in $(\xi,\mu,\phi)$): - $\partial\vec x/\partial\xi=\bigl(R\xi s\cos\phi/A,\ R\xi s\sin\phi/A,\ R\mu\bigr)$, so

using $s^2+\mu^2=1$. - $\partial\vec x/\partial\theta=(RA\mu\cos\phi,\ RA\mu\sin\phi,\ -R\xi s)$, so $h_\theta=R\sqrt{A^2\mu^2+\xi^2s^2}=R\sqrt{\xi^2+\mu^2}$; converting to $v=\mu$ costs $|d\theta/d\mu|=1/s$: $h_\mu=R\sqrt{\xi^2+\mu^2}/s$. - $h_\phi=RA\,s=R\sqrt{\xi^2+1}\,s$.

Product (= $|\det J|$ by P2(b)):

Hence by WT (2.25)

Sanity limits: $\xi\gg1$ gives $R^3\xi^2\to r^2/R\cdot R^2$, matching spherical with $r=R\xi$ (P4). The factor vanishes at $\xi=0,\ \theta=\pi/2$ — the focal ring $\rho=R$, where the coordinates degenerate; that ring is exactly the rim of the charged disk in P12 (Exercise 2.6.1), which is why the disk problem is natural in these coordinates.

(a) For a unit charge at $\vec x_j$, the flux element through arc $d\ell$ of $C$ is $d\ell\,\hat n\cdot(\vec x-\vec x_j)/|\vec x-\vec x_j|^2=d\theta_j$ — the plane angle subtended by $d\ell$ at $\vec x_j$ (the 2-D copy of WT 2.42: $d\ell_\parallel/\rho$ instead of $ds_\parallel/r^2$). Closing the curve, $\oint d\theta_j=2\pi$ if $\vec x_j$ is enclosed (full horizon) and $0$ otherwise (front/back cancellation) — the same in/out dichotomy as WT Figs. 2.7–2.8, one dimension down. Superposing, $\oint d\ell\,\hat n\cdot\vec E=2\pi\sum_{\rm in}q_j$; smearing $q_j\to\sigma\,da$ and applying the 2-D divergence theorem to arbitrary regions equates integrands: $\vec\nabla\cdot\vec E=2\pi\sigma$. $\blacksquare$ (b) For $q=1$ at $\vec x'$: $\vec E=(\vec x-\vec x')/|\vec x-\vec x'|^2 =\vec\nabla\ln|\vec x-\vec x'|=\vec\nabla\ln(|\vec x-\vec x'|/K)$ (the constant $K$ only fixes the zero of the log and drops under $\vec\nabla$ — needed because a pure $\ln|\vec x|$ has dimensions inside the log). By (a), $\vec\nabla\cdot\vec E=2\pi\delta^{(2)}(\vec x-\vec x')$, i.e. $\nabla^2\ln(|\vec x-\vec x'|/K)=2\pi\delta^{(2)}(\vec x-\vec x')$. Note also directly $\nabla^2\ln\rho=\rho^{-1}\partial_\rho(\rho\,\partial_\rho\ln\rho)=\rho^{-1}\partial_\rho(1)=0$ for $\rho\neq0$. (c) Put $\vec x'=0$, integrate over the disk $\rho\le R$: right side gives $2\pi$ by the sampling property. Left side, by the 2-D divergence theorem: $\int\nabla^2\ln(\rho/K)\,da=\oint_{\rho=R}\partial_\rho\ln(\rho/K)\,R\,d\phi =\frac1R\cdot R\cdot2\pi=2\pi$. Equal, for every $R$ — consistent. $\blacksquare$ (d) The two "unit charges" are normalized by different Gauss laws. 2-D: $E\cdot2\pi\rho=2\pi q\Rightarrow E=1/\rho$, potential $-\ln(\rho/K)$. A unit-density line in 3-D: $E\cdot2\pi\rho=4\pi\lambda\Rightarrow E=2\lambda/\rho$, potential $-2\lambda\ln(\rho/K)$ (also obtainable by integrating $dz'/\sqrt{\rho^2+z'^2}$ with a cutoff, P7). The extra 2 is the ratio $4\pi/2\pi$ of the solid angle of a sphere to the angle of a circle — a bookkeeping constant between dimensions, not new physics.

The single-line building block, for a line along $x$ from $x_1$ to $x_2$ seen from $(x,y,z)$ with $b^2=y^2+z^2$:

since $\int du/\sqrt{u^2+b^2}=\sinh^{-1}(u/b)=\ln(u+\sqrt{u^2+b^2})-\ln b$. (a) Superpose the $x$-line ($x_{1,2}=\mp L$, $b_1^2=y^2+z^2$) and the $y$-line ($b_2^2=x^2+z^2$):

For $L\gg$ everything: numerator $\to2L$; denominator $\sqrt{(L+x)^2+b^2}-(L+x)=\dfrac{b^2}{\sqrt{(L+x)^2+b^2}+(L+x)}\to\dfrac{b^2}{2L}$. Each log $\to\ln(4L^2/b^2)$, so

This is exactly two copies of the 3-D line-charge potential $-2\lambda\ln(b/K)$ of P6(d), one per line, with the cutoff scale $K=2L$ set by the line length — the finite ends regulate the log divergence. (b) For $L\ll r$: $\sinh^{-1}\frac{L\mp\zeta}{b}$ expands as $\int dx'/\sqrt{\cdots}\approx2L/r$ per line (the integrand is $\approx1/r$ across the whole segment), so $\Phi\approx\lambda(2L/r)+\lambda(2L/r)=4\lambda L/r=Q_{\rm tot}/r$ with $Q_{\rm tot}=2\times2L\lambda$ — the expected monopole. (c) Half-lines: $+\lambda$ on $0\le y'\le L$, $-\lambda$ on $0\le x'\le L$:

— more transparently, using the antiderivative directly,

where the lower limits produced $\sqrt{y^2+x^2+z^2}-y=r-y$ and $r-x$. For $L\gg|x|,|y|,|z|$ both numerators $\to2L$ and cancel between the two logs:

The $L$-dependence drops because the far ends of the $\pm\lambda$ lines cancel; what survives is the near-origin imbalance. (The equipotentials $r-y=c(r-x)$ are the cones of the classic "bent-wire" problem.)

(a) The on-axis field of a uniform disk (radius $b$, surface density $\sigma$, at $z'=0$) follows from the ring formula (P9a) integrated over radius: $E_z^{\rm disk}(z)=2\pi\sigma[\mathrm{sgn}(z)-z/\sqrt{z^2+b^2}]$. Slice the cylinder into disks $\sigma=\rho\,dz'$ at height $z'$ and superpose:

Both pieces integrate exactly. With $\int\mathrm{sgn}(z-z')dz'=a$, $2z$, $-a$ for $z>a/2$, $|z|\le a/2$, $z<-a/2$, and $\frac{d}{dz'}\sqrt{(z-z')^2+b^2}=-\frac{z-z'}{\sqrt{(z-z')^2+b^2}}$:

(b) Inside, set $f(z)=\sqrt{(z+\frac a2)^2+b^2}-\sqrt{(z-\frac a2)^2+b^2}$; $f(0)=0$ and

So $E_z\approx2\pi\rho[2z-f'(0)z]=4\pi\rho\Bigl[1-\dfrac{a}{\sqrt{a^2+4b^2}}\Bigr]z$:

For $b\gg a$ the bracket $\to1$ and $E_z\to4\pi\rho z$ — predictable without calculation: an infinite slab has, by Gauss on a pillbox straddling $|z|$, $2E_z=4\pi\rho\cdot2z$, i.e. $E_z=4\pi\rho z$; the edge correction $-a/\sqrt{a^2+4b^2}\approx-a/2b$ is the finite-radius leak. (Both linear coefficients are checked numerically against a symmetric-difference slope.)

(a) Every element $dq=\lambda R\,d\phi$ sits at distance $s=\sqrt{R^2+z^2}$ from the axis point $(0,0,z)$. Transverse components cancel by symmetry; the axial projection is $z/s$:

(Limit $z\gg R$: $E_z\to2\pi R\lambda/z^2=q_{\rm ring}/z^2$ ✓.) (b) Parametrize the cone surface by the axial coordinate $z'\in[L_1,L_2]$: the ring at height $z'$ has radius $\rho'=z'\tan\alpha$ and slant width $d\ell=dz'/\cos\alpha$, hence charge $dq=\sigma\,2\pi\rho'\,d\ell=2\pi\sigma\tan\alpha\,z'\,dz'/\cos\alpha$. By (a) with the field point at the origin ($z=0$, i.e. axial offset $-z'$):

using $1+\tan^2\alpha=1/\cos^2\alpha$. So

Every ring contributes $\propto dz'/z'$ — the cone is the classic logarithmic geometry (self-similar: each octave of distance contributes equally). The sign says the field at the tip points away from the (positive) cone, down the $-z$ axis; the magnitude $\pi\sigma\sin2\alpha\ln(L_2/L_1)$ is extremal at $\alpha=45°$.

(a) For a uniform shell of radius $s$, charge $q$, surface density $\sigma=q/4\pi s^2$, put the field point at distance $r$ and use the axial angle $\theta$ of the source point: $u^2\equiv|\vec x-\vec x'|^2=r^2+s^2-2rs\cos\theta$, so $u\,du=rs\sin\theta\,d\theta$ and $da'=s^2\sin\theta\,d\theta\,d\phi$:

i.e. $\Phi_{\rm shell}=\dfrac{q}{1-\epsilon^2}\dfrac{(r+s)^{1-\epsilon}-|r-s|^{1-\epsilon}}{2rs}$. With $s=a\le r$ this is $\Phi_{\rm out}$ (2.62) ($|r-a|=r-a$); with $s=b\ge r$ it is $\Phi_{\rm in}$ (2.63) ($|r-b|=b-r$). Superposition gives $\Phi=\Phi_{\rm out}+\Phi_{\rm in}$ in the gap. (Conductors keep each shell uniform by symmetry, so this is exact.) $\blacksquare$ (b) Impose $\Phi(a)=\Phi(b)$ and multiply through by $2ab(1-\epsilon^2)$:

Expand $x^{1-\epsilon}\approx x(1-\epsilon\ln x)$. The $O(1)$ parts: $\mathcal A_0=2b-(a+b)+(b-a)=2(b-a)$, while $\mathcal B_0=2a-(a+b)+(b-a)=0$ — the whole right bracket is first order, which is the Cavendish point: $q_a=O(\epsilon)$. The $O(\epsilon)$ part of $\mathcal B$:

Hence $q_a\approx q_b\,\mathcal B_1/\mathcal A_0=-\dfrac{q_b\,\epsilon}{2(b-a)}\,N$ with $N=2a\ln2b-(a+b)\ln(a+b)+(b-a)\ln(b-a)$. Regroup $N$ by collecting the $a$- and $b$-coefficients of the logs:

so, flipping the denominator sign,

(Numerically, $a=1,b=2$: $N=2\ln4-3\ln3=-0.5232$, so $q_a=-\epsilon q_bN/[2(b-a)]=+0.2616\,\epsilon\,q_b$; the exact linear solve at $\epsilon=10^{-4}$ matches to $\sim10^{-4}$ relative.) (c) Expand the outer shell's interior potential to first order: $\Phi_{\rm in}(r)\approx q_b\bigl[\frac1b-\epsilon\bigl(\frac{\ln b+1}{b}-\frac{r^2}{6b^3}\bigr)\bigr]$, which increases with $r$ when $\epsilon>0$ (faster-than-Coulomb fall-off makes the near wall dominate). So with $q_a=0$ the inner sphere would sit below the outer's potential ($q_b>0$); since a charge $q_a$ raises $\Phi(a)$ by $\approx q_a/a$ but $\Phi(b)$ only by $\approx q_a/b$, equalization needs $q_a>0$: same sign as $q_b$. Flipping $\epsilon$ or $q_b$ flips the argument: $\mathrm{sgn}\,q_a=\mathrm{sgn}(\epsilon\,q_b)$ — matching (b) since the bracket $N/(a-b)>0$.

(a) Same substitution as P10(a) — the "perfect differential" hint: with $u=|\vec x-\vec x'|$, $\int da'\,f(u)=\frac{2\pi s}{r}\int_{|r-s|}^{r+s}u f(u)\,du$, and here $uf(u)=e^{-u/R}$ integrates immediately:

Evaluate the four combinations:

The wire enforces $\varphi_a(a)+\varphi_b(a)=\varphi_a(b)+\varphi_b(b)$; multiplying by $2ab/R$ and collecting the $q_a$ and $q_b$ terms:

(the sign of the exponent flips to $e^{(a-b)/R}$ because $e^{-(b-a)/R}$ moved across with the same value: $b>a$). Solving for $q_a$ gives the displayed ratio. $\blacksquare$ (b) Expand each bracket in $1/R$ (write $x=1/R$): $1-e^{-2bx}=2bx-2b^2x^2+\tfrac43b^3x^3+\ldots$, so $\tfrac ab(1-e^{-2bx})=2ax-2abx^2+\tfrac43ab^2x^3$; and

(the odd/even split of the two exponentials leaves $-[(a-b)+(a+b)]x=-2ax$, $+\frac{(a+b)^2-(a-b)^2}{2}x^2=+2abx^2$, $-\frac{(a+b)^3+(a-b)^3}{6}x^3=-\frac{a^3+3ab^2}{3}x^3$). Numerator: the $x$ and $x^2$ terms cancel exactly, leaving $\bigl[\tfrac43ab^2-ab^2-\tfrac13a^3\bigr]x^3=\tfrac13a(b^2-a^2)x^3$. Denominator: swap $a\leftrightarrow b$ in the first bracket only: $2bx-2abx^2+\tfrac43a^2bx^3-2ax+2abx^2-\ldots=2(b-a)x+O(x^3)$. Hence

Positive for $q_b>0$: unlike P10, the Yukawa deviation is even in the sense that the induced charge appears only at second order in $b/R$ — which is why Cavendish null results translate into such strong photon-mass limits ($\epsilon_{\rm eff}\sim(b/R)^2$; Williams–Fowler–Hill 1971 reach $\epsilon\sim3\times10^{-16}$, i.e. $R\gtrsim10^{7}\,$m).

(a) From P5's map: $\rho^2\equiv x^2+y^2=R^2(\xi^2+1)\sin^2\theta$ and $z=R\xi\cos\theta$, so $r^2=\rho^2+z^2=R^2\xi^2+R^2\sin^2\theta$, giving $\sin^2\theta=(r^2-R^2\xi^2)/R^2$ and

The positive root of this quadratic in $\xi^2$:

which substituted in $\Phi=A\tan^{-1}(1/\xi)$ is the displayed form. $\blacksquare$ (b) On the disk ($z\to0$, $r=\rho<R$): $r^2-R^2<0$, and $\sqrt{(r^2-R^2)^2+4R^2z^2}\to R^2-r^2$, so the inner sum $\to0^+$ and $1/\xi\to\infty$: $\tan^{-1}(1/\xi)\to\pi/2$. Consistency with the boundary value $\Phi|_{\rm disk}=V$ forces $A=2V/\pi$. (Check the other limits: $r\to\infty$ gives $\xi\to r/R$, $\Phi\to AR/r\to0$ ✓; the potential is continuous across the plane outside the disk ✓.) (c) Near the disk $\xi\to0$: $\Phi=A\tan^{-1}(1/\xi)=A(\pi/2-\xi+O(\xi^3))$, and at fixed $\rho<R$, $z=R\xi\cos\theta$ with $\cos\theta=\sqrt{1-\rho^2/R^2}$ at $\xi=0$, so $\partial\xi/\partial z=1/(R\cos\theta)=1/\sqrt{R^2-\rho^2}$ and

By symmetry $E_z(0^-)=-E_z(0^+)$, so the total (both faces) density from the jump condition (WT 2.68), $E_z(0^+)-E_z(0^-)=4\pi\sigma$:

(The inverse-square-root rim divergence is the conductor's edge singularity.) (d) Total charge:

so

— the $C_{\rm cir}$ quoted and reused in P37–P39 (Exercises 2.12.7–2.12.9). Far field check: $\Phi\to A R/r=Q/r$ ✓ (numerically the ratio $\to1$ at $r/R\sim10^3$).

The geometric content: the negative layer sits at $\vec x''=\vec x'-\hat n_{21}d$, so $|\vec x-\vec x''|=|(\vec x-\vec x')+\hat n_{21}d|$. Expand in the small shift $\vec a=\hat n_{21}d$ using $f(\vec v+\vec a)\approx f(\vec v)+\vec a\cdot\vec\nabla_v f$ with $f=1/|\cdot|$, $\vec v=\vec x-\vec x'$:

But $\vec\nabla'\,|\vec x-\vec x'|^{-1}=+(\vec x-\vec x')/|\vec x-\vec x'|^3$ (derivative with respect to the source point — the sign that makes the book's hint come out right), so

Insert into the two-layer superposition: the leading $1/|\vec x-\vec x'|$ pieces cancel (equal and opposite charge at corresponding points), leaving

with $D\equiv\sigma d$ held finite as $d\to0$, $\sigma\to\infty$ (WT 2.76). This is (2.77); combined with the solid-angle identity (2.71) it is $\Phi=-\int D\,d\Omega'$ (2.78). $\blacksquare$ Numerically: two disks of radius $1$ carrying $\pm\sigma=\pm D/d$ at separation $d$ reproduce the $d\to0$ formula with error $\propto d^2$ (the next, quadrupole, order — the linear term is the answer).

(a) Let $\Phi^{(2)}(\vec x)=\int_{S_2}\sigma(\vec x')/|\vec x-\vec x'|\,da'$ be the single-layer potential. The negative copy lives at source points $\vec x'+\hat xd$ (displaced rigidly by $\hat xd$; Fig. 2.19's orientation makes the pair's dipole moment point along $+\hat x$), so its potential at $\vec x$ is

So the pair potential is a finite difference of the same function at two field points:

The overall sign tracks which copy is positive: with the positive layer displaced by $+\hat xd$ relative to the negative one (equivalently the negative copy at $\vec x'-\hat xd$), the same two lines give $\Phi_d(\vec x)=\Phi^{(2)}(\vec x)-\Phi^{(2)}(\vec x+\hat xd)=+\,\hat x\cdot\vec E^{(2)}(\vec x)\,d$, which is the book's orientation (Fig. 2.19). Either way: the pair potential is the $d$-weighted component of the single-layer field along the offset direction, valid in both regions since only $\Phi^{(2)}$, defined everywhere, enters. $\blacksquare$ (b) $\Phi_d$ inherits its discontinuity from $\vec E^{(2)}$. Crossing $S_2$ along $\hat n_{21}$, the tangential part of $\vec E^{(2)}$ is continuous (WT 2.67) and the normal part jumps by (WT 2.68): $\vec E^{(2)}_2-\vec E^{(2)}_1=4\pi\sigma\,\hat n_{21}$. Hence

This generalizes (2.80): the standard normal-offset layer is $\hat x=\hat n_{21}$, jump $4\pi D$; a layer sheared parallel to itself ($\hat x\perp\hat n_{21}$) produces no potential jump — numerically the two-disk potential with in-plane offset is continuous across the disk to $O(d^2)$.

Work from the layer potential in solid-angle form (P13 / WT 2.77):

(a) On the axis, $\vec x=(0,0,z)$; hemisphere points $\vec x'=a(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$, $\hat n'=\vec x'/a$, $\mu=\cos\theta\in[0,1]$: $\hat n'\cdot(\vec x-\vec x')=(z\mu a-a^2)/a=z\mu-a$ and $|\vec x-\vec x'|^2=z^2+a^2-2az\mu$, so

Substitute $Q=z^2+a^2-2az\mu$ (so $z\mu-a=(z^2-a^2-Q)/2a$, $d\mu=-dQ/2az$):

evaluated between $\mu=0$ ($Q=z^2+a^2$, bracket $=2z^2/\sqrt{z^2+a^2}$) and $\mu=1$ ($Q=(z-a)^2$, bracket $=(z^2-a^2)/|z-a|+|z-a|=\pm2z$ — $+$ for $z>a$, $-$ for $z<a$). Hence

upper sign for $z>a$ (outside the dome), lower for $z<a$ (below/inside — including all $z<0$). Limits: $\Phi\to0$ as $z\to\pm\infty$ ✓; crossing the dome at $z=a$, $\Phi(a^+)-\Phi(a^-)=4\pi D$ ✓ (WT 2.81). The constant $\pm$ never differentiates:

a single smooth formula on the whole axis, pointing along $+\hat z$ for $D>0$ (value $2\pi D/a$ at O). Far field: the layer's dipole moment is $\vec p=\int D\,\hat n'\,da'=D\times(\text{vector area of the dome})=D\,\pi a^2\hat z$ (the vector area of any cap equals that of its rim disk). A point dipole $p\hat z$ has $E_r=2p\cos\theta/r^3$, $E_\theta=p\sin\theta/r^3$, i.e. exactly $E_r=2\pi Da^2\cos\theta/r^3$, $E_\theta=\pi Da^2\sin\theta/r^3$; on the axis $E_z\to2\pi Da^2/z^3$, matching the exact form. $\blacksquare$ (b) Disk of radius $a$ in the $z=0$ plane, $\hat n'=\hat z$: $\hat n'\cdot(\vec x-\vec x')=z$, $|\vec x-\vec x'|^2=z^2+\rho'^2$:

so again $E_z=2\pi Da^2/(z^2+a^2)^{3/2}$ everywhere off the layer, and $\vec p=D\pi a^2\hat z$: identical axis field and far field. No accident: the dome and the disk share the same rim, and a constant-$D$ layer's potential is $-D\,\Omega(\vec x)$ with $\Omega$ the (signed) solid angle of the surface — which depends only on the rim, up to the $4\pi$ jump bookkeeping of which side you're on. (c) Full sphere: from (a) plus the complementary dome ($\mu\in[-1,0]$, bracket at $\mu=-1$: $(z^2-a^2)/(z+a)+(z+a)=2z$ for $z>-a$), the lower dome contributes $2\pi D[z/\sqrt{z^2+a^2}-1]$, so

both constants (the axis calculation suffices — by symmetry $\Phi$ depends only on $r$, and the solid-angle form gives $\Omega=4\pi$ inside, $0$ outside). Therefore

the pure gauge-jump of a closed uniform double layer — the "Neumann conductor" field configuration of §2.7. $\blacksquare$

The product rule for divergences gives, for any $C^2$ pair,

Integrate over $V$ and apply the divergence theorem (WT 1.9) to the left side:

$\hat n$ the outward normal. Hence

(Antisymmetrizing in $\phi\leftrightarrow\psi$ kills the symmetric $\vec\nabla\phi\cdot\vec\nabla\psi$ term and yields Green's second identity (2.95) — the engine of §2.8.) Closed-form check used by the tests: on the ball $r\le R$ with $\phi=r^2,\ \psi=r^4$: $\nabla^2r^4=20r^2$, $\vec\nabla r^2\cdot\vec\nabla r^4=8r^4$, so the volume side is $\int_0^R28r^4\,4\pi r^2dr=16\pi R^7$; the surface side is $R^2\cdot4R^3\cdot4\pi R^2=16\pi R^7$. Equal. (analytic — the numeric check is of the verifying-instance kind.)

(a) Both states' potentials are Coulomb superpositions of all their charges (volume + surface): $\Phi'(\vec x)=\int d^3y\,\rho'(\vec y)/|\vec x-\vec y|+\oint da_y\,\sigma'(\vec y)/|\vec x-\vec y|$. Insert into the left side:

writing surface charge as a surface-delta volume density (P–§2.3). The double integral is symmetric under exchanging primed and unprimed sources (the kernel $1/|\vec x-\vec y|$ is symmetric), and the right side is the same double integral with the roles swapped. Hence LHS = RHS. $\blacksquare$ (Equivalently: both sides equal twice the mutual interaction energy of the two charge systems.) (b) Choose the auxiliary state to convert unknown induced charges into knowns. State 1 (actual): $q$ at radius $r$; both shells grounded, $\Phi=0$ on $S_a$ and $S_b$; unknown totals $Q_a$, $Q_b$. State 2 (auxiliary): no interior charge, inner shell at potential $V'$, outer grounded; between the shells the auxiliary potential is the standard capacitor form

Reciprocation: the RHS vanishes ($\rho'=0$; $\sigma'$ lives on shells where the actual $\Phi=0$). The LHS: the actual surface charge on $S_a$ sees $\Phi'=V'$, on $S_b$ sees $0$; the point charge sees $\Phi'(r)$:

Repeating with the auxiliary roles swapped (outer at $V'$, inner grounded, $\Phi'(s)=V'(1/a-1/s)/(1/a-1/b)$):

Sum: $Q_a+Q_b=-q\,[a(b-r)+b(r-a)]/[r(b-a)]=-q$ — all field lines from $q$ end on the grounded pair. Limits: $r\to a$ gives $(Q_a,Q_b)\to(-q,0)$; $r\to b$ gives $(0,-q)$ ✓. Independent verification (used by the test): solve the $\ell=0$ radial problem directly — $\Phi=\alpha(1/s-1/a)$ for $s<r$, $\beta(1/s-1/b)$ for $s>r$; Gauss gives $\alpha=Q_a$, $\beta=q+Q_a$; continuity at $s=r$ reproduces exactly $Q_a=-qa(b-r)/[r(b-a)]$. $\blacksquare$

(a) Define the average over the sphere of radius $R$ about $\vec x_0$,

Differentiate under the integral with respect to $R$:

by the divergence theorem and $\nabla^2\Phi=0$ (charge-free). So $\langle\Phi\rangle_R$ is independent of $R$; letting $R\to0$, continuity gives $\langle\Phi\rangle_R=\Phi(\vec x_0)$ for every admissible $R$. $\blacksquare$ (b) Let $\Phi_1,\Phi_2$ solve Poisson's equation in $V$ with the same $\rho$ and the same Dirichlet values on $S$; $U=\Phi_1-\Phi_2$ is harmonic with $U|_S=0$. Suppose $U>0$ somewhere. $U$ is continuous on the compact $\overline V$, so it attains its maximum $M>0$ at some interior $\vec x_$ (not on $S$, where $U=0$). Take any small sphere around $\vec x_$ inside $V$: the MVT says $M=U(\vec x_)=\langle U\rangle_R\le M$, with equality only if $U\equiv M$ on the whole sphere (an average of values $\le M$ equals $M$ only when all equal $M$). Hence $U=M$ on every sphere around $\vec x_$ up to the boundary of its neighborhood — the set $\{U=M\}$ is open and closed in $V$, so $U\equiv M$ on the connected component, contradicting $U\to0$ at $S$. So $U\le0$; symmetrically $U\ge0$; therefore $U\equiv0$ and $\Phi_1=\Phi_2$. (This is the max-principle route — same squeeze as Griffiths' proof, with the MVT supplying the "no interior extrema" step; cf. ~EM-04 notes §2.)

Green's second identity in the primed variable:

Insert $\phi(\vec x')=G_D(\vec x',\vec x)$ and $\psi(\vec x')=G_N(\vec x',\vec x'')$, both Green functions in their first argument: $\nabla'^2G_D(\vec x',\vec x)=-4\pi\delta(\vec x'-\vec x)$, $\nabla'^2G_N(\vec x',\vec x'')=-4\pi\delta(\vec x'-\vec x'')$. The volume side collapses onto the two deltas:

The surface side: $G_D(\vec x',\vec x)=0$ for $\vec x'\in S$ (2.100) kills the first term, leaving $-\oint da'\,G_N(\vec x',\vec x'')\,\partial_{n'}G_D(\vec x',\vec x)$. Divide by $4\pi$:

Reading: the Neumann function is the Dirichlet one plus a boundary correction whose kernel $-\partial_{n'}G_D/4\pi$ is exactly the Dirichlet "boundary influence" weight of (2.98) — the correction re-distributes the $-4\pi/S$ wall field. 1-D version (what the test checks, WT 1-D conventions $G''=-\delta$, no $4\pi$): $G_N(x,x'')=G_D(x'',x)-\sum_{x'\in\{0,L\}}G_N(x',x'')\,\partial_{n'}G_D(x',x)$, with $\partial_{n'}=\mp d/dx'$ at $0,L$. With $G_D=x_<(1-x_>/L)$, $\partial_{n'}G_D|_{x'=0}=-(1-x/L)$, $\partial_{n'}G_D|_{x'=L}=-x/L$, and the unsymmetrized $G_N$ of P23 ($G_N(0,x'')=0$, $G_N(L,x'')=x''-L/2$), the right side is $G_D(x'',x)+(x''-L/2)\,x/L$, which equals $x/2$ ($x<x''$) or $x''-x/2$ ($x>x''$) — precisely $G_N(x,x'')$. Exact.

First, $G^{\rm symm}_N$ is itself a legitimate Neumann Green function: (2.111) subtracts $g(\vec x'')\equiv\langle G_N(\vec x,\vec x'')\rangle_{S(\vec x)}$, which depends only on the second slot, so in the first slot $\nabla'^2G^{\rm symm}_N=-4\pi\delta$ and $\partial_{n'}G^{\rm symm}_N|_S=-4\pi/S$ still hold. Now substitute $G^{\rm symm}_N(\vec x',\vec x)=G_N(\vec x',\vec x)-g(\vec x)$ into the two integrals of (2.103):

But Gauss (Green's first identity with $\phi\equiv1$) gives $\oint\partial_{n'}\Phi\,da'=\int\nabla'^2\Phi\,d^3x'=-4\pi Q$, so $\Delta_{\rm surf}=+g(\vec x)\,Q$ and the two shifts cancel identically: the integral part of the representation is the same function of $\vec x$ for both kernels. The only freedom left is the additive constant each convention appends — (2.103) adds $\langle\Phi\rangle$; calling $\Phi_{\rm symm}$ the symmetric-kernel representation with its average constant, the two answers differ by the constant $\langle\Phi\rangle-\langle\Phi_{\rm symm}\rangle$:

Exactly the expected Neumann non-uniqueness: the physical field $-\vec\nabla\Phi$ is untouched. (1-D numeric check: kernels $G_N=\{x/2;\,x''-x/2\}$ vs $G^{\rm symm}_N=-|x-x''|/2+L/4$ in (2.138) give $\Phi$'s differing by a constant $\sim$ machine-flat across the grid.)

With $\rho=0$ inside, (2.98) reduces to the boundary term

since $\Phi=V_i$ is constant on face $i$. Two observations at $\vec x=\vec x_c$ (the center): 1. Symmetry. The rotation group of the regular polyhedron acts transitively on the faces and fixes $\vec x_c$. $G_D$ is determined uniquely by the geometry (2.99)–(2.100), so it is invariant: $G_D(g\vec x',g\vec x)=G_D(\vec x',\vec x)$ for any symmetry $g$. Mapping face $i$ onto face $j$ therefore maps the integral $F_i(\vec x_c)$ onto $F_j(\vec x_c)$: all the $F_i(\vec x_c)$ are equal. 2. Normalization. Summing over all faces,

(Gauss, then the defining delta of $G_D$ with the source $\vec x_c$ inside). Equal weights summing to 1 must each be $1/n$:

(Physical reading: $-\partial_{n'}G_D/4\pi\,da'$ is the probability-like "influence measure" of boundary patch $da'$ — nonnegative, total mass 1 — and the center sees every face identically. The cube relaxation check lands at $0.1667$.)

The two branches of $G_D$ as a function of the field coordinate $x$:

(a) The self-field average at the particle (the §2.11 prescription — the mean of the two one-sided fields removes the particle's own symmetric contribution):

$F<0$ (toward $0$) for $x'<L/2$, $F>0$ (toward $L$) for $x'>L/2$: the charge is attracted to its nearer grounded wall by its induced charge, with the (1-D) force growing linearly, vanishing by symmetry at midbox. (b) In these 1-D conventions ($\Phi''=-\lambda$) Gauss reads $dE/dx=\lambda$, so a wall's surface charge is the jump of $E$ across it (field vanishes outside the box):

Both negative (they attract the unit charge — consistent with (a)); total induced $\sigma_0+\sigma_L=-1$, screening the unit source completely. Note the proximity rule: the wall nearer the charge takes the larger share, exactly like the 3-D parallel-plate image problem. (c) With $\lambda=0$, (2.127) keeps only the boundary term $\Phi(x)=-[\Phi(x')\,\partial_{x'}G_D(x,x')]_{x'=0}^{x'=L}$. From the branches (now differentiating in $x'$): $\partial_{x'}G_D|_{x'=L}=-x/L$ and $\partial_{x'}G_D|_{x'=0}=1-x/L$, so

the straight line from $V$ to $0$ — which indeed solves $\Phi''=0$ with the given endpoint data. $\blacksquare$

(a) Write $G_N=C_1x+C_2$ ($x>x'$), $C_3x+C_4$ ($x<x'$). The endpoint conditions land on one branch each: $C_3=\tfrac12$ (at $x=0$), $C_1=-\tfrac12$ (at $x=L$). The unit-source jump $\partial_xG_N|^{x'^+}_{x'^-}=C_1-C_3=-1$ is then automatically satisfied — that is the 1-D image of the $-4\pi/S$ consistency (2.102): total wall flux must swallow the unit charge, and $\tfrac12+\tfrac12=1$ splits it evenly. Continuity at $x'$: $-\tfrac{x'}2+C_2=\tfrac{x'}2+C_4$, i.e. $C_2=x'+C_4$ with $C_4=c(x')$ free for each source position (Neumann non-uniqueness). So

Both branches combine as $G_N=-\tfrac12|x-x'|+\tfrac{x'}2+c(x')$. Choosing $c(x')=C-\tfrac{x'}2$ — allowed, it's a function of $x'$ only — gives the symmetric

The (2.111) prescription picks the constant: subtract the endpoint average of the $c=0$ function, $\tfrac12[G_N(0,x')+G_N(L,x')]=\tfrac12[0+(x'-\tfrac L2)]=\tfrac{x'}2-\tfrac L4$, which yields exactly $-\tfrac12|x-x'|+\tfrac L4$: $C=L/4$. $\blacksquare$ (b) $E(x)=-\partial_xG^{\rm symm}_N=\tfrac12\,\mathrm{sgn}(x-x')$: $E(x'^\pm)=\pm\tfrac12$, so $E_{\rm avg}=0$ and $F=0$ for every $x'$. Physical: constant-field Neumann walls exert no position-dependent pull — the "image pressure" is the same from both sides (contrast P22a). (c) In (2.138) with $\lambda=0$ and $\partial_{x'}\Phi(0)=\partial_{x'}\Phi(L)=-V/L$ (remember $E=-\Phi'$):

$G(x,L)-G(x,0)=-\tfrac12(L-x)+\tfrac12x=x-\tfrac L2$, so

the uniform field $E=V/L$, determined up to the arbitrary average — and consistent: the Neumann data $E(L)-E(0)=0=Q$ matches the empty box ($\lambda=0$), which is the solvability condition the book's note flags. $\blacksquare$

(2.127) splits the answer into a source part and a boundary part:

Volume term. Split at $x'=x$:

after the cubic terms cancel — the textbook parabola (zero at both walls, curvature $-\lambda$). Boundary term. From P22(c), $\partial_{x'}G_D|_{x'=0}=1-x/L$ and $\partial_{x'}G_D|_{x'=L}=-x/L$:

the straight line from $+V$ to $-V$. Total:

Verification: $\Phi''=-\lambda$ ✓; $\Phi(0)=V$, $\Phi(L)=-V$ ✓ — the GF machinery assembled superposition (particular + harmonic) automatically, which is its point.

(a) Multiply the $\Phi$ equation (in $x'$) by $G(x',x)$, the $G$ equation (source at $x$, variable $x'$) by $\Phi(x')$, subtract, and integrate:

where the $k_0^2\,G\Phi$ cross terms cancelled identically — that's why the whole (2.127) structure survives the Helmholtz shift. The left side is a total derivative, $\bigl[G\Phi'-\Phi\,\partial_{x'}G\bigr]_0^L$; with homogeneous Dirichlet data for $G$ the $G\Phi'$ endpoint piece dies, so

and the symmetry $G(x',x)=G(x,x')$ (proved as in (2.108), or read off the explicit form in (b)) rewrites it in the stated field-point form. $\blacksquare$ (b) Away from $x'$ the function solves $G''+k_0^2G=0$, so it is a sinusoid on each branch; the boundary conditions pick $\sin k_0x$ (vanishes at $0$) below the source and $\sin k_0(L-x)$ (vanishes at $L$) above:

(automatically continuous and symmetric). The unit source fixes $A$ via the slope jump $\partial_xG|^{x'^+}_{x'^-}=-1$:

by the sine addition formula — the $x'$-dependence collapses (this is the constancy of the Wronskian of the two homogeneous solutions), leaving $A=1/(k_0\sin k_0L)$:

At resonance $\sin k_0L=0$ the construction fails — the homogeneous problem has a nontrivial mode and no Green function exists (the 1-D cavity resonance). Limits: $k_0\to0$ gives $x_<(L-x_>)/L=x_<(1-x_>/L)$, recovering (2.136) ✓. Worked check: for $\lambda=$ const the GF quadrature reproduces $\Phi=\frac{\lambda}{k_0^2}\bigl[\cos k_0(x-L/2)/\cos(k_0L/2)-1\bigr]$, the direct particular-plus-homogeneous solution. $\blacksquare$

Let $\vec E_0=-\vec\nabla\Phi_0$ be the equilibrium field before, and $\vec E_1=-\vec\nabla\Phi_1$ after the neutral conductor (occupying $v$, surface $\partial v$) is inserted; all old conductors keep their totals $Q_i$, the new one has total $0$. Compare energies over the field region $\mathcal V$ (all space outside the old conductors; note $\vec E_1=0$ inside $v$ while $\vec E_0$ is not):

the algebraic identity $a^2-b^2=(a-b)^2+2b(a-b)$. Kill the cross term: with $\vec E_1=-\vec\nabla\Phi_1$,

The second integral vanishes: both fields obey Gauss's law with the same volume charge (none, or the same fixed $\rho$) in $\mathcal V\setminus v$, and inside $v$ $\vec\nabla\cdot\vec E_0=0$ (no charge sat there before) while $\vec E_1\equiv0$. The first becomes boundary terms over the old conductors, $\partial v$, and infinity: on each old conductor $\Phi_1=$ const and $\oint(\vec E_0-\vec E_1)\cdot d\vec a=4\pi(Q_i-Q_i)=0$; on $\partial v$, $\Phi_1=$ const (it is a conductor in state 1) and $\oint\vec E_0\cdot d\vec a=0$ (no enclosed charge), $\oint\vec E_1\cdot d\vec a=4\pi\cdot0$ (neutral); at infinity the product falls as $1/r^3$. So the cross term is zero and

strictly, because $\vec E_1\equiv\vec E_0$ is impossible: $\vec E_0\neq0$ in $v$ generically while $\vec E_1=0$ there. $\blacksquare$ Instance with closed forms (the numeric check): sphere $a$ with $Q$, neutral shell $(b,c)$: $W_0=Q^2/2a$; after insertion $E=Q/r^2$ except for the dead zone $b<r<c$, so $W_1=\frac{Q^2}2(\frac1a-\frac1b+\frac1c)$ and $W_1-W_0=\frac{Q^2}2(\frac1c-\frac1b)<0$ — energy drops by exactly the energy the shell's metal expelled. (Physically: the induced surface charges rearrange to screen; any rearrangement allowed at fixed totals that changes the field can only have been adopted because it lowers $W$ — Thompson's principle in action.)

The (2.144) double integral on the segment:

which diverges on the diagonal. Regulate by cutting out $|x-x'|<\delta$:

using the $x\leftrightarrow x'$ symmetry. Substituting $u=a-x$:

So $W\to\infty$ as $\delta\to0$, but only logarithmically — each decade of resolution adds the same $\lambda^2a\ln10$. $\blacksquare$ Placement in the §2.10 hierarchy: point charge $\sim1/\delta$ (linear, WT 2.149); line $\sim\ln(1/\delta)$; surface — finite (P28 computes one). The dimensional rule of thumb: $W\sim\int\rho\Phi$, and the potential a $d$-dimensional uniform charge produces on itself scales as $\int d^dx\,/r$ near a point — convergent for $d\ge2$, log-marginal at $d=1$, divergent at $d=0$. (Same log that made the physical "wire" capacitance depend on its radius.)

With $\sigma=Q/R^2$, scale lengths by $R$ ($\vec u,\vec v$ in the unit square):

Difference-variable reduction (the hint): for each Cartesian component, $\int_0^1\!\int_0^1f(|u-v|)\,du\,dv=2\int_0^1(1-w)f(w)\,dw$ (the density of $w=|u-v|$ on $[0,1]$ is $2(1-w)$). Applying it in $x$ and $y$ independently:

Expand the numerator; each piece has an elementary primitive ($L\equiv\ln(1+\sqrt2)=\sinh^{-1}1$): 1. $\displaystyle\int_0^1\!\!\int_0^1\frac{dw_xdw_y}{\sqrt{w_x^2+w_y^2}} =2L$ — integrate $\sinh^{-1}(1/w)$-type terms; standard square integral. 2. $\displaystyle\int_0^1\!\!\int_0^1\frac{w_x\,dw_xdw_y}{\sqrt{w_x^2+w_y^2}} =\int_0^1\bigl[\sqrt{1+w_y^2}-w_y\bigr]dw_y =\frac{\sqrt2+L}2-\frac12$ (and the same with $x\leftrightarrow y$). 3. $\displaystyle\int_0^1\!\!\int_0^1\frac{w_xw_y\,dw_xdw_y}{\sqrt{w_x^2+w_y^2}} =\int_0^1w_y\bigl[\sqrt{1+w_y^2}-w_y\bigr]dw_y =\frac{2\sqrt2-1}3-\frac13=\frac{2\sqrt2-2}3 .$ So

Numerically $A=0.8813736-0.1380712=0.7433023$, $I=4A=2.9732093$, and

(The book quotes the value from numerics; the reduction above gives it in closed form, agreeing to all quoted digits. Compare the disk of equal area: a disk of radius $R_d$, $W_{\rm disk}=\frac{8}{3\pi}Q^2/R_d$; at equal area $R_d=R/\sqrt\pi$, $W=1.5045\,Q^2/R$ — the square, being more spread out into its corners, stores slightly less.)

(a) Enclosed charge: $Q(r)=\int_0^r4\pi C r'^{2+\alpha}dr'=\dfrac{4\pi C}{3+\alpha}r^{3+\alpha}$ (needs $\alpha>-3$), and $Q=Q(a)$ fixes $C$. Gauss:

Field energy (2.147):

provided the interior integral converges at $r=0$: exponent $4+2\alpha>-1$, i.e. $\boxed{\alpha>-\tfrac52}$. Combine:

Sanity: $\alpha=0$ (uniform) $\to\frac35Q^2/a$; $\alpha\to\infty$ (charge pushed to the surface) $\to\frac12Q^2/a$, the conducting-shell value; $\alpha\to-\frac52^+$ diverges — too much charge concentrated at the center. (b) Uniform ball: $\frac35\,e^2/a=m_ec^2$, so

Gaussian numbers: $e=4.80321\times10^{-10}$ esu, $m_e=9.10938\times10^{-28}$ g, $c=2.99792\times10^{10}$ cm/s give $r_e=e^2/m_ec^2=2.81794\times10^{-13}$ cm and

— the classical electron radius picture whose tensions (what holds it together? cf. §14.1) previewed the need for QED, where the divergence softens to a log (footnote to (2.149)).

(a) Spherical symmetry ⟹ radial field, Gauss (2.44) on a sphere of radius $r$: - $r<R$: enclosed charge $\frac43\pi r^3\rho$, so $4\pi r^2E=4\pi\cdot\frac43\pi r^3\rho\Rightarrow E_{\rm in}=\dfrac{4\pi\rho\,r}3$. - $r>R$: enclosed $Q_{\rm tot}=\frac43\pi R^3\rho+4\pi R^2\sigma$, so $E_{\rm out}=\dfrac{Q_{\rm tot}}{r^2}=\dfrac{4\pi\rho R^3}{3r^2}+\dfrac{4\pi\sigma R^2}{r^2}$.

The jump at $R$: $E_{\rm out}(R)-E_{\rm in}(R)=4\pi\sigma$ ✓ (2.68) — the volume charge contributes continuously, the sheet does the jumping. (b) The force per area on the surface layer is $\sigma$ times the field acting on it, which excludes its own (discontinuous) self-field: by the disk/remainder argument (2.158)–(2.159) that is the average of the fields on the two sides:

directed along $+\hat r$ when positive. Checks: $\rho=0$ gives the conductor-style $2\pi\sigma^2$ (2.161); $\sigma=0$ gives zero — the volume charge feels a body force, not a surface one; and the expression is exactly $\sigma E_{\rm remainder}(R)$, the sheet riding in the field of everything else (volume charge $+$ its own smooth continuation). $\blacksquare$

With every other conductor grounded, (2.170) gives

and $W=0$ forces $\vec E\equiv0$ everywhere. Could a nonzero $(V_A,V_B)$ give $\vec E\equiv0$? No: zero field means $\Phi$ constant in each connected component of the vacuum; the grounded conductors (or the boundary at infinity, $\Phi\to0$, which is part of every open system's boundary) pin that constant to $0$, forcing $V_A=V_B=0$. So $W$ is positive definite on $(V_A,V_B)$. A quadratic form $\frac12(c_{11}v_1^2+2c_{12}v_1v_2+c_{22}v_2^2)$ is positive definite iff $c_{11}>0$ and $\det\begin{psmallmatrix}c_{11}&c_{12}\\c_{12}&c_{22}\end{psmallmatrix}>0$: take $(1,0)$ to get $C_{AA}>0$; then minimize over $V_B$ at fixed $V_A$ (complete the square):

i.e. $C_{AB}^2<C_{AA}C_{BB}$. $\blacksquare$ Boundary case worth flagging: the idealized closed two-conductor system (e.g. infinite parallel plates, (2.179)) has $C_{AB}^2=C_{AA}C_{BB}$ — there $V_A=V_B$ produces no field and no grounded reference exists to pin the constant; the strict inequality is restored the moment infinity (or any third grounded surface) belongs to the boundary. The concentric-shell numbers confirm strictness: $C_{ab}^2/C_{aa}C_{bb}=\frac{a(c-b)}{b(c-a)}<1$.

(a) State 1: $V_i=1$, all others $0$ — surface charges give $Q_k=C_{ki}$ by the definition of the coefficients. State 2: $V_j=1$, others $0$ — $Q'_k=C_{kj}$. Reciprocation with no volume charge, $\sum_k Q_kV'_k=\sum_kQ'_kV_k$: the left side keeps only $k=j$ (the only conductor with $V'\neq0$): $Q_j\cdot1=C_{ji}$; the right keeps $k=i$: $Q'_i\cdot1=C_{ij}$. Hence $C_{ji}=C_{ij}$. $\blacksquare$ (b) The (2.166) representation writes each coefficient as

(unit potential on $S_j$ enters through $-\partial_{n'}G_D/4\pi$ as in (2.98), and the induced density on $S_i$ through $\sigma=\partial_n\Phi/4\pi$, (2.92)). The kernel is symmetric, $G_D(\vec x,\vec x')=G_D(\vec x',\vec x)$ (2.107), and the two normal derivatives act on separate slots, so swapping $(i,\vec x)\leftrightarrow(j,\vec x')$ is a relabeling of integration variables: $C_{ij}=C_{ji}$. $\blacksquare$ (c) Put $V_i=1$, all others grounded. Then $W=\frac12\sum_{kl}C_{kl}V_kV_l=\frac12C_{ii}$. But $W=\frac1{8\pi}\int E^2>0$ strictly — the field cannot vanish identically while conductor $i$ sits at unit potential against grounded company (P31's pinning argument). Hence $C_{ii}>0$: a conductor at positive potential above its grounded surroundings carries positive charge. $\blacksquare$

Charges $q_a,q_b,q_c$ produce (isolated-shell superposition, each shell an equipotential):

i.e. $V=P\,q$ with the elastance matrix (u=1/a etc.) $P=\begin{psmallmatrix}u&v&w\\ v&v&w\\ w&w&w\end{psmallmatrix}$, $u=\frac1a,v=\frac1b,w=\frac1c$. Then $C=P^{-1}$. Row-reducing, $\det P=(u-v)(v-w)w$, and the adjugate gives

Translate with $u-v=\frac{b-a}{ab}$, $v-w=\frac{c-b}{bc}$, $u-w=\frac{c-a}{ac}$:

Physical reading: the $(a,b)$ gap is the standard spherical capacitor $ab/(b-a)$; $C_{ac}=0$ because shell $b$, grounded, completely screens $a$ from $c$; $C_{cc}=c+\frac{bc}{c-b}$ is "sphere to infinity" plus the inner $(b,c)$ gap. Verifications. - Schwarz (P31): $\dfrac{C_{ab}^2}{C_{aa}C_{bb}}=\dfrac{a(c-b)}{b(c-a)}<1$ (since $b>a$, $c-b<c-a$); $\dfrac{C_{bc}^2}{C_{bb}C_{cc}}=\dfrac{b-a}{c-a}<1$; $C_{ac}^2=0<C_{aa}C_{cc}$. All strict ✓. - (2.171), $j=a$: $C_{aa}+C_{ba}+C_{ca}=\frac{ab}{b-a}-\frac{ab}{b-a}+0=0$ ✓ (every field line from the innermost shell lands on its neighbors — closed interior). $j=b$: over the common denominator $(b-a)(c-b)$: $-ab(c-b)+b^2(c-a)-bc(b-a)=b[-ac+ab+bc-ab-bc+ac]=0$ ✓. - (2.178): $j=c$ is not closed — the outer surface sees infinity: $C_{ac}+C_{bc}+C_{cc}=\frac{c^2-bc}{c-b}=c=C_{\infty\infty}$, the isolated capacitance of a sphere of radius $c$ ✓ — exactly the book's sum rule relating interior coefficients to the external capacitance. $\blacksquare$

Ground B and put A at $V_A=1$: the field is confined to the cavity between A and B (outside B, the grounded shell and infinity are both at $\Phi=0$ with no enclosed net driving — by uniqueness $\Phi\equiv0$ there). Gauss over a surface inside B's metal gives $Q_B^{\rm inner}=-Q_A$, and B's outer surface carries nothing; hence the totals in this state, $Q_A=C_{AA}$ and $Q_B=C_{BA}$, satisfy

(This is $\sum_iC_{ij}=0$, (2.171), applied to the closed cavity subsystem.) Now invoke the strict Schwarz inequality (P31) on the pair $(A,B)$ — strict because infinity is available as the zero reference outside B:

Since $C_{AA}>0$ (P32c), divide: $C_{BB}>C_{AA}$. $\blacksquare$ Interpretation: $C_{BB}=C_{\rm gap}+C_{\rm out}$ — at $V_B=1,V_A=0$ conductor B charges both its inner surface (against A, the gap capacitance, which equals $C_{AA}$) and its outer surface (against infinity, $C_{\rm out}>0$); the hollow conductor holds strictly more. Spheres make it concrete: $C_{AA}=\frac{ab}{b-a}$, $C_{BB}=\frac{ab}{b-a}+b$, difference exactly $b$.

(a) The state is defined by charges, so invert: $V=C^{-1}q$ with $C^{-1}=\dfrac1{\det C}\begin{psmallmatrix}C_{BB}&-C_{AB}\\-C_{AB}&C_{AA}\end{psmallmatrix}$, $\det C=C_{AA}C_{BB}-C_{AB}^2$. With $q=(Q,-Q)$:

Note $\sum_{ij}C_{ij}>0$ (it is $2W$ for $V_A=V_B=1$, a legitimate nonzero state in open space) and $\det C>0$ (P31): the system capacitance is positive. (b) The energy at prescribed charges:

This is why $C$ as defined is the "capacitor" of circuit theory: charge $\pm Q$, voltage $\Delta V=Q/C$, energy $Q^2/2C$ — with the matrix formula exposing how the self- and induction coefficients combine (used numerically in P37–P39).

Work at fixed charges, where P26 (Exercise 2.10.1) applies. Before: A alone with charge $Q$; $W_{\rm before}=Q^2/2C_A$. After: B introduced, uncharged and insulated; the new equilibrium energy at charges $q=(Q,0)$ is, in terms of the new matrix $C'$ (inverting as in P35),

P26: introducing the uncharged conductor strictly lowers the energy, $W_{\rm after}<W_{\rm before}$:

And by P31 (Exercise 2.12.1) the subtracted term is positive whenever B actually couples ($C'_{AB}\neq0$), so

Two readings of $C'_{AA}$: it is A's charge at $V_A=1$ with B grounded — the grounded neighbor pulls extra charge onto A; while $\det C'/C'_{BB}$ is the effective capacitance with B floating (uncharged), which is also $\ge C_A$ but smaller than the grounded case. Concentric instance: $C_A=a$; $C'_{AA}=ab/(b-a)>a$ always; floating shell $(b,c)$: $\det C'/C'_{BB}=\bigl(\frac1a-\frac1b+\frac1c\bigr)^{-1}$, between $a$ and $ab/(b-a)$. The test verifies the full chain numerically.

(a) For $d\ll R$ the gap field is the uniform slab field of (2.182): charge $\pm Q$ spreads at $\sigma=\pm Q/\pi R^2$, $E_{\rm gap}=4\pi\sigma$, $\Delta V=E\,d=4\pi\sigma d$, so

diverging as the plates close — edge (fringe) corrections are relative $O\bigl(\frac dR\ln\frac Rd\bigr)$ and ignorable at leading order. (b) For $d\gg R$ each disk is, to leading order, an isolated disk plus the monopole potential of its partner (the P39 elastance model):

The $\tfrac12$ is the series combination of each disk's capacitance "to infinity": pulling the plates apart does not send $C\to0$ but to a floor set by the plates' own size — the physical content of the isolated-disk $2R/\pi$ (P12). The two regimes meet at $R^2/4d\sim R/\pi$, i.e. $d\sim\pi R/4$ — comparable to $R$, where neither expansion is trustworthy. $\blacksquare$

Keep the P37(b) model to first order in $R/d$ — disk at potential (own charge)/(own capacitance) plus partner monopole $\mp Q/d$:

so

Structure of the correction: $\frac R\pi\cdot\frac{2R}{\pi d}=C_{\rm sys}^0\,C_{\rm cir}/d$ — the partner's monopole reduces the voltage needed for the same charge (opposite charges attract each other's potential down), so $C$ rises as the plates approach, smoothly heading toward the $R^2/4d$ regime. Higher orders would need the disk's response to a nonuniform field (induced multipoles, $O(d^{-3})$), consistently dropped here. The test checks the expansion error scales as $(R/d)^2$.

(a) To leading order each conductor is a point source to its partner (shape enters only through its isolated capacitance):

i.e. $V=Pq$ with the symmetric elastance matrix above. Then $C=P^{-1}$:

Physical version of the argument: put conductor 1 at $V_1=1$ with 2 grounded. Conductor 1 carries $\approx C^0_{11}$; at distance $d$ it lifts 2's potential by $C^0_{11}/d$; grounded 2 must neutralize that with charge $Q_2\approx-C^0_{22}\cdot C^0_{11}/d=C_{21}$. Symmetric in $1\leftrightarrow2$ ✓. (b) From the same inverse,

using (a) to rewrite $-C^0_{11}C^0_{22}/d^2=C_{12}/d$ — the book's form, exact within the monopole model. Iterative reading: 2's own charge at unit potential is $C^0_{22}$; its partner's induced $C_{12}$ feeds back potential $C_{12}/d$, requiring a compensating geometric series $C^0_{22}\bigl[1-C_{12}/d\cdot(-1)\cdots\bigr]$ that sums to the displayed ratio. (c) Disks: $C^0_{11}=C^0_{22}=C_{\rm cir}=2R/\pi$. System capacitance (P35):

(the middle identity — $C_{\rm sys}=1/(P_{11}+P_{22}-2P_{12})$ — follows from $\Delta V=(P_{11}-P_{12})Q-(P_{12}-P_{22})(-Q)$), recovering P38 exactly. $\blacksquare$

(a) Both $\Psi$ and the true $\Phi$ equal the same constants on every boundary piece, so $\eta\equiv\Psi-\Phi$ vanishes on all of $S$. Expand:

Cross term by Green's first identity (P16) with $\phi=\eta$, $\psi=\Phi$:

($\eta|_S=0$; $\nabla^2\Phi=0$ in the charge-free region). Hence

with equality iff $\vec\nabla\eta\equiv0$, i.e. $\Psi=\Phi$. Dirichlet trials can only overestimate: any deviation from harmonicity costs positive field energy at fixed voltage. $\blacksquare$ (b) Trial $\Psi=(b-\rho)/(b-a)$: matches $\Psi=1$ at $\rho=a$, $0$ at $\rho=b$; $|\vec\nabla\Psi|=1/(b-a)$:

At $b=2a$: $C[\Psi]=\frac{3a}{4a}L=0.750\,L$. The exact solution $\Phi=\ln(b/\rho)/\ln(b/a)$ gives

Indeed $0.750>0.7213$ — the linear trial overshoots by $4.0\%$: it misallocates field energy toward the outer radius where the true field is weaker ($E\propto1/\rho$). The companion principle (§3.10) with charges fixed brackets $C$ from below, sandwiching the true value — the standard variational squeeze for capacitances. $\blacksquare$

(a) Energy per unit length of a set of line charges on equipotential surfaces (the 2-D copy of (2.169)):

directly from the definition of $C_L$ — the "standard form" $Q^2/2C$ with $\lambda$ playing $Q$. $\blacksquare$ (b) For $d\gg a$, each wire's potential is its own surface value plus the partners' line potentials at distance $d$ (P6d: a line $\lambda_j$ contributes $-2\lambda_j\ln(s/K)$ at distance $s$):

Since $\sum_j\lambda_j=-\lambda-\lambda+2\lambda=0$, the arbitrary $K$ cancels — the potentials are well defined without a far cutoff. Evaluate (take $K=1$):

Then

Cross-check via the energy: $w=\frac12\sum\lambda_iV_i =\frac12[2\lambda^2+2\lambda^2+8\lambda^2]\ln(d/a)=6\lambda^2\ln(d/a) =\lambda^2/(2C_L)$ ✓ consistent. $\blacksquare$ (Each wire being $\ll d$ guarantees the equipotential surfaces are circles to $O(a/d)$; corrections would enter as image-line displacements, shifting $N$ at relative order $a/d$ — negligible by hypothesis. Note the combination $2V_3-V_1-V_2$ is exactly the voltage conjugate to $\lambda$ under the charge pattern $(-1,-1,+2)\lambda$: the mode actually excited.)

Reflection in $z=0$ sends a charge $q$ at height $h$ to $-q$ at $-h$; reflection in $z=a$ sends it to $-q$ at $2a-h$. Starting from $+1$ at $z'$ and iterating the two reflections until the set closes generates

(only the $n=0$, $+$ member is the physical charge; all others lie outside the gap). Boundary check: at $z=0$ the $+$ charge labelled $n$ and the $-$ charge labelled $-n$ are mirror partners through the plate, so their terms cancel pairwise and $G_D|_{z=0}=0$ exactly; at $z=a$ the partners are $(+,n)$ and $(-,1-n)$ — again pairwise cancellation. Uniqueness (§2.7) then makes this the Green function. Absolute convergence: group the two $n$-labelled terms. With $f(s)=(\rho^2+s^2)^{-1/2}$, $s_\pm=z\mp z'-2na$, the mean value theorem gives

so $\sum_n|{\rm pair}_n|$ converges by comparison with $\sum1/n^2$. (The ungrouped $\pm1/r$ terms decay only like $1/n$ — the grouping is what makes the convergence absolute.) Numerically the grouped tail is $\sim zz'/(2N^2a^3)$: the test measures $n^2|{\rm pair}_n|\to$ const and the $1/N^2$ approach of $G|_{z=a}\to0$ under symmetric truncation.

(a) Reflection in $x=0$ and in $y=0$ closes after the double reflection (the group is $\mathbb Z_2\times\mathbb Z_2$):

with $\vec x'_{\pm\pm}=(\pm x',\pm y',z')$. On $x=0$ the $++/-+$ and $--/+-$ terms cancel in pairs; on $y=0$ the $++/+-$ and $-+/--$ pairs cancel. Both walls at zero, correct $-4\pi\delta$ source, $G\to0$ at infinity: uniqueness does the rest. (b) The energy of the charge in the presence of the conductor is half the naive charge–image Coulomb sum, $U=\tfrac12 q\,\Phi_{\rm ind}(\vec x')$ — the factor $\tfrac12$ because the induced charge builds up proportionally as the source is brought in (equivalently: images move when the charge moves, so $\int\vec F\cdot d\vec l$ picks up half). The image distances are $2x'$, $2y'$, $2\sqrt{x'^2+y'^2}$, so

The test verifies the factor of $\tfrac12$ the honest way: it integrates the full image force along a radial escape path and reproduces exactly this $W$. (c) With no volume charge, (2.98) leaves the wall term $\Phi=-\frac1{4\pi}\oint\Phi_s\,\partial G_D/\partial n'\,da'$. Because the geometry is a wedge, the answer must be the bounded harmonic function of the polar angle alone that interpolates the two wall values: $\nabla^2\phi=0$, so $\Phi=A+B\phi$ with $\Phi(\phi{=}0)=V$, $\Phi(\phi{=}\pi/2)=-V$ gives

The quadrature route confirms it: reduce the two wall integrals with the 2-D corner Green function of P3 (the $z'$ integration is already absorbed), i.e. $\Phi=-\frac1{4\pi}\sum_{\rm walls}\int_0^\infty d\rho'\,\Phi_{\rm wall}\, (\pm\tfrac1{\rho'})\partial_{\phi'}G$, with $\partial_{\phi'}G|_{\phi'=0}=8\rho^2\rho'^2\sin2\phi/(\rho^4+\rho'^4-2\rho^2\rho'^2\cos2\phi)$ and its $\phi'=\pi/2$ mirror — the code does exactly this integral.

In 2-D each line charge contributes $-2\ln d$ (Eq. 3.11; the $1/L$ scale cancels in the alternating sum). With the same $\mathbb Z_2\times\mathbb Z_2$ image set as P2,

where $d_{\pm\pm}$ is the transverse distance from $\vec x_\perp$ to $(\pm x',\pm y')$. Now use complex notation $w=\rho e^{i\phi}$, $w'=\rho'e^{i\phi'}$: the four image positions are $w',-w',\bar w',-\bar w'$ and

Squaring, $|w^2-w'^2|^2=\rho^4+\rho'^4-2\rho^2\rho'^2\cos2(\phi-\phi')$ and $|w^2-\bar w'^2|^2=\rho^4+\rho'^4-2\rho^2\rho'^2\cos2(\phi+\phi')$, so

On $\phi=0$ ($w$ real) and on $\phi=\pi/2$ ($w$ imaginary, $w^2$ real) the numerator and denominator coincide, so $G_D=0$ — the walls are grounded. (The $w\mapsto w^2$ structure is the conformal unfolding of the quadrant into a half-plane; compare P32/P33 and ~MA-06.)

At swing angle $\theta$ the charge sits a height $h(\theta)=D-L\cos\theta$ above the plane. The conductor interaction energy is half the charge–image Coulomb energy (image $-e$ at depth $h$, separation $2h$; same $\tfrac12$ as P2b):

Expand with $\cos\theta\simeq1-\theta^2/2$: $h\simeq(D-L)+L\theta^2/2$, so

The string constrains the motion to a circle, so the generalized coordinate is $\theta$ with inertia $I=mL^2$, and the quadratic term gives the torque constant $k_\theta=U''(0)=e^2L/4(D-L)^2$. Hence

(Because gravity is absent, the only restoring force is the image attraction: the equilibrium is the string hanging straight down toward the plane, where $h$ is minimal and $U$ lowest.) The code takes the exact $U(\theta)$, differentiates numerically, and recovers both $k_\theta$ and $\omega$.

For $z>0$ the half-space Green function is (3.36), and its inward-normal derivative on the plane is (3.37) with field/source roles exchanged (use symmetry $G_D(\vec x,\vec x')=G_D(\vec x',\vec x)$):

With no volume charge, (2.98) gives

the integrand being nonzero only on the patch. Now recognize the geometry: $z\,da'/r^3=(\hat r\cdot\hat z)\,da'/r^2=d\Omega$ is precisely the projection of the patch element onto the unit sphere around $\vec x$ — the solid-angle element. Hence $\Phi=(V/2\pi)\,\Omega(\vec x)$ for $z>0$. For $z<0$ run the identical argument with the lower half-space Green function (normal $+\hat z$): the sign of $z$ flips twice, leaving $\Phi=(V/2\pi)|\Omega|$ on both sides. Sanity limits: point just above the patch interior, $\Omega\to2\pi$, so $\Phi\to V$ (continuity with the boundary data); far away $\Omega\to({\rm area})\,z/r^3$, a dipole-sheet falloff. The code checks the disk axis value $2\pi(1-z/\sqrt{z^2+R^2})$, the rectangle closed form $4\arctan\!\bigl[w_xw_y/(z\sqrt{w_x^2+w_y^2+z^2})\bigr]$, and the $\Phi\to V$ limit.

With $\mu=\cos\theta$ and the regulator in place, the $k$ integral is elementary because $\mathrm{Re}(\epsilon-iR\mu)>0$:

Then

the imaginary part vanishing by oddness in $\mu$ (as it must — $G$ is real). The remaining integral is an arctangent:

This is the $\epsilon$-regularized route to (3.1); the book's in-text derivation (3.20)–(3.21) instead does the $\mu$ integral first and uses $\int_0^\infty\sin x/x\,dx=\pi/2$ — the regulator makes the interchange of integration order legitimate (dominated convergence), which is the point of the exercise.

(a) Transverse Fourier transform of $\nabla^2G_N=-4\pi\delta$ gives the reduced equation (3.22), $\left(k^2-\partial_z^2\right)g=\delta(z-z')$, now with the Neumann wall condition $\partial_z g|_{z=0}=0$ and boundedness as $z\to\infty$. For $z\ne z'$ the solutions are $e^{\pm kz}$; boundedness picks $C_1e^{-kz}$ above the source, and the wall condition picks the even combination $C_2\cosh kz$ below it. Continuity at $z'$ and the jump $-\partial_zg\,|^{z'^+}_{z'^-}=1$ (integrate the ODE across the source) give $C_1=\cosh(kz')/k\,e^{-kz'}\cdot$… solving the $2\times2$ system,

which is the Dirichlet result (3.30) with the image sign flipped. Fourier inversion term by term is the same $I_1$ integral as (3.32)–(3.34), giving $G_N=1/|\vec x-\vec x'|+1/|\vec x-\vec x''|$, $\vec x''=(x',y',-z')$: the image method result (3.5), recovered from the ODE side. (b) Equation (2.115) instructs: first put the source on the boundary, then evaluate the normal field as the field point approaches it. With $z'=0$ the charge and its image coincide, $G_N=2/\sqrt{\rho^2+z^2}$, and with the outward normal $\hat n=-\hat z$,

by the delta-family identity (3.38) (the book's Ex. 2.2.2b): the expression integrates to exactly $1$ over the plane for every $z>0$ (the code shows this analytically-sharp normalization numerically) while its support shrinks like $z$. Since the plane has infinite area, the $-4\pi/S$ constant of (2.106) is zero here, so (2.115) reads pure surface delta — confirmed.

(a) As in §3.2, transverse translation invariance reduces (3.14) to $\left(k^2-\partial_z^2\right)g=\delta(z-z')$ with $g=0$ at both $z=0$ and $z=a$. Build $g$ from homogeneous solutions vanishing at the near wall: $g=A\sinh(kz)$ for $z<z'$ and $g=B\sinh[k(a-z)]$ for $z>z'$. Continuity at $z=z'$ and the unit jump $-\partial_zg|^{+}_{-}=1$ give $A\sinh(kz')=B\sinh[k(a-z')]$ and $kB\cosh[k(a-z')]+kA\cosh(kz')=1$; solving, $A=\sinh[k(a-z')]/(k\sinh ka)$, $B=\sinh(kz')/(k\sinh ka)$ — i.e. the quoted $z_<,z_>$ form (its $a\to\infty$ limit is (3.30), and the code checks it against P1's image ladder — two independent constructions of the same $G$). (b) The total charge on the bottom plate is $Q|_{z=0}=\oint\sigma\,da=-\frac1{4\pi}\oint\partial_zG|_{z=0}\,da$ (outward normal $-\hat z$). The area integral of $e^{i\vec k\cdot(\vec x-\vec x')_\perp}$ produces $(2\pi)^2\delta^2(\vec k)$, so only the $k\to0$ limit of $g$ survives:

For the bottom plate ($z\to0$, so $z=z_<$): $\partial_z g|_{z=0}=(a-z')/a$, giving $Q|_{z=0}=-(1-z'/a)$; for the top ($z=z_>$): $\partial_z g|_{z=a}=-z'/a$ with normal $+\hat z$, giving $Q|_{z=a}=-z'/a$. They sum to $-1$: every field line from the unit charge ends on a plate, and the shares are the linear interpolation weights of the source height — the Green-reciprocity statement (weight $=$ the linear potential each plate would create at $z'$), which is the Ex. 2.7.2 route the book mentions. (c) Hold the bottom plate at $V$, top at $0$, no interior charge. By (b) plus symmetry of $G_D$, the surface integral of (2.98) evaluates to $\Phi(z)=V\,[-Q|_{z=0}(z'{=}z)]=V(1-z/a)$ — verified by quadrature in the code — i.e. the uniform field $E=V/a$. Then $\sigma=\frac1{4\pi}E=V/4\pi a$ uniformly, and restricting to area $A$ (the book's hint): $Q=\sigma A=AV/4\pi a$, so

the Gaussian parallel-plate capacitance.

The reduced equation is again $\left(k^2-\partial_z^2\right)g=\delta(z-z')$, bounded as $z\to\infty$. The outward normal of the region $z>0$ is $\hat n=-\hat z$, so $\partial G/\partial n=-\partial G/\partial z$ and the Robin condition $\left(g+h\,\partial g/\partial n\right)|_{z=0}=0$ becomes

Make the ansatz

which already solves the ODE with the correct unit jump (the first term is the free 1-D Green function; the second is a source-free homogeneous solution decaying at $\infty$). At the wall,

so $g(0)=h\,g'(0)$ forces $(1+\beta)=hk\,(1-\beta)$, i.e.

Unlike the pure D/N cases, the "image strength" is now $k$-dependent: in real space the image is a point charge plus a continuum tail (an image line charge extending below the plane) — the price of the mixed condition. The code solves the BVP independently on a grid with the Robin row $g_0-h(g_1-g_0)/\Delta z=0$ and confirms the closed form, plus both limits.

(a) The image algebra of §3.3 never used $r'>a$: demanding $\Phi(a\hat n)=0$ for a charge $q$ at $\vec r\,'$ and an image $q'$ at $\vec r\,''\parallel\vec r\,'$ again factors as (3.45), giving

now with $r'<a$ so the image lies outside — outside the physical region, as an image must. So the interior Green function is exactly (3.48). For the grounded shell the exterior is a charge-free region bounded by an equipotential at zero and vanishing at infinity, so $\Phi_{\rm out}\equiv0$ (uniqueness): the ground wire supplies a net $-1$; integrating $\sigma=\frac1{4\pi}\partial G/\partial n$ over the inner surface indeed gives $-1$ (test). (b) Cut the wire and demand neutrality: the inner surface must still carry $-q$ (Gauss on a surface inside the metal), so the outer surface carries $+q$. The outer problem — conductor surface $r=a$, total charge $q$, nothing outside — has the uniformly distributed solution $\Phi_{\rm out}=q/r$, regardless of where $q$ sits inside: the metal screens the interior asymmetry (the inner-surface density is nonuniform, the outer is exactly uniform). Inside, the grounded solution $qG_D$ vanishes at $r=a$, while the true shell potential is $\Phi(a)=q/a$; adding a constant (harmonic, no new sources) fixes it:

Continuity at $r=a$ is then automatic (test). (c) From outside one sees $q/r$ exactly — no information about $\vec r\,'$. Generally: for a neutral hollow conductor of any shape, the interior problem (cavity walls + enclosed charge) and the exterior problem decouple; the outer surface charge arranges itself as the solution of "isolated conductor of this shape carrying total charge $q$", which is unique and independent of the cavity contents. Exterior measurements determine only $q$ and the conductor's shape — the position of the enclosed charge is unobservable. (This is electrostatic shielding run in reverse; compare ~EM-04's Faraday-cage discussion.)

Two mirror operations are in play: inversion in the sphere ($q\mapsto-\frac{a}{r}q$ at $a^2\vec r/r^2$) and reflection in the plane ($q\mapsto-q$ at $\vec r_m$, $z\to-z$). Because $\vec r\,'_m$ has the same radius $r'$ and mirrored polar angle, the two operations commute on this configuration, and the group closes with four members:

On the sphere $r=a$: the 1st+2nd terms cancel (that is the grounded-sphere identity for $\vec r\,'$) and the 3rd+4th cancel (same identity for $\vec r\,'_m$, whose radius is also $r'$). On the plane: the 1st+3rd cancel (mirror pair) and the 2nd+4th cancel (their positions are mirror images with equal strengths $a/r'$). All four boundary pieces at zero, one physical singularity in the region: uniqueness certifies $G_D$. (a) For a source above the plane and outside the boss, all three images lie either below the plane or inside the sphere — outside the physical region ✓. (b) For a source inside the dome ($r'<a$, $z'>0$) the same formula works: the sphere image now sits outside the sphere (P10a), its mirror below the plane, and the plane mirror below the plane — again all images outside the dome region. The code checks both cases on random boundary points. (The region below the plane is field-free in both problems; the images are fictions valid only in the vacuum region, as in ~EM-04.)

Superpose the point-image rule (3.46) over the elements of $\rho$: the element $dq=\rho(\vec x')\,d^3x'$ at radius $r'$ acquires an image

The image element occupies the inverted volume: with $r''=a^2/r'$ and angles fixed, $d^3x''=r''^2dr''\,d\Omega=\bigl(\frac{a^2}{r'}\bigr)^2\frac{a^2}{r'^2}dr'\,d\Omega =(a/r')^6\,d^3x'$. Therefore

and eliminating $r'=a^2/r''$ in favor of the image-point radius gives the first stated form; reading the same equation at the direct point gives the second. On the surface each $dq$–$dq^$ pair separately produces zero potential (that is what (3.46) was built to do), so $\Phi|_{r=a}=0$ for the superposition; a surface of constant potential has no tangential field, i.e. $\vec E\times\hat n|_{r=a}=0$ — the stated boundary condition. (b) For $\rho(\vec x)=q\,\delta^3(\vec x-d\hat z)$, the element rule gives a single image $-\,(a/d)\,q$ at $a^2/d\,\hat z$ — precisely (3.46). Totals: $Q^=\int\rho^*d^3x=-\int\frac{a}{r'}\rho\,d^3x'$ (each element weighted by $a/r'$) — the code verifies this integral identity by two independent quadratures for a Gaussian shell density, and verifies the surface condition with a random point cloud.

Work with the potential coefficients $p_{ij}$ ($V_i=\sum_jp_{ij}Q_j$), which are transparent at large separation: $p_{11}=1/a$, $p_{22}=1/b$ (isolated spheres), and $p_{12}=p_{21}\simeq1/d$ — exactly $1/d$ at leading order because an uncharged sphere in the field of a distant point charge takes the potential of its center (mean-value theorem; this is the reciprocation-theorem result the book flags via its Ex. 2.12.5, and the "prove it first" it demands). (a) Invert the $2\times2$ matrix:

(b) With $(Q_1,Q_2)=(Q,-Q)$:

so

The neglected physics is polarization: each sphere sees the other's field gradient and grows an induced dipole, shifting $p_{12}$ at $O(1/d^3)$ and the self terms at $O(ab^2/d^3)$-type order. The code builds the exact answer by the classical iterated-image ladder (charge $a$ at the center of sphere 1, image $-b\,q/(d-x)$ at $d-b^2/(d-x)$, image back in sphere 1, …), sums the charge series to machine precision, and confirms both asymptotics and their $d^{-2}$-rate of approach.

(a) With line charges, $G=-2\ln d_1+2\ln d_2+{\rm const}$, where $d_1,d_2$ are transverse distances to source and trial image at $\rho''$ on the same ray. On the cylinder ($\rho=a$, angle difference $\Delta$):

So $-2\ln d_1+2\ln d_2=2\ln(a/\rho')$ — constant on the surface: the cylinder is equipotential. This is the cylindrical twin of the sphere image, same inversion point $a^2/\rho'$, but the image strength is exactly $-1$ (not $-a/\rho'$): logarithms trade the sphere's strength rescaling for an additive constant. (b) A potential plus a constant is still a potential; subtract the surface value by adding $2\ln(\rho'/a)$:

and $d_2^2\rho'^2=\rho^2\rho'^2+a^4-2a^2\rho\rho'\cos\Delta$ gives the boxed form. It vanishes identically at $\rho=a$, has the correct $-2\ln$ singularity at the source, and is manifestly symmetric (the numerator and denominator are symmetric polynomials in $\rho\leftrightarrow\rho'$) — consistent with $G_D(\vec x,\vec x')=G_D(\vec x',\vec x)$. (c) Read the same formula for $\rho>a$ with an exterior source $\rho'>a$: it still solves the 2-D Green equation with the right singularity (the image point $a^2/\rho'$ now lies inside the cylinder — outside the physical region, as required), still vanishes on $\rho=a$, and matches the interior form continuously there. Since the Dirichlet data and source structure fix the solution up to the 2-D large-$\rho$ freedom (a constant — logs need not vanish at infinity), the exterior Green function is the identical expression; the code checks boundary zero and harmonicity with an exterior source. (This $G$ feeds P27's Poisson integral.)

Everything is spherically symmetric, so $\Phi=A_i+B_i/r$ region by region. Let $q_a$ be the induced charge on the grounded sphere. Gauss: for $a<r<b$, $E_r=q_a/r^2$; for $r>b$, $E_r=(q_a+Q)/r^2$. Integrate inward from infinity:

with $C$ fixed by continuity at $b$: $C=Q/b$. Grounding pins $\Phi(a)=0$:

which plugged back gives exactly the three-region answer (and $\Phi\equiv0$ for $r\le a$, the field-free grounded interior). Green-function route (the module's check): each shell element $dq$ at radius $b$ has grounded-sphere weight $G_D$ of (3.48), so $\Phi(\vec x)=\oint\sigma_b\,G_D(\vec x,\vec x')\,da'$ with $\sigma_b=Q/4\pi b^2$; the code does this quadrature and lands on the closed form — including the exterior region, where the image content of $G_D$ automatically supplies the $-Qa/b$. Notice $q_a$ is exactly the sum of the per-element images $-(a/b)\,dq$ (P12's rule with $r'=b$ for every element).

(a) Every ring element sits at radius $b$, so each $dq$ has image $-(a/b)\,dq$ at radius $a^2/b$ (the image of a concentric equatorial ring is a concentric equatorial ring of radius $a^2/b$). Summing,

(b) On the axis every point of a ring of radius $R$ is at distance $\sqrt{z^2+R^2}$, so a ring of charge $q$ contributes $q/\sqrt{z^2+R^2}$. Direct ring: $Q_{\rm ring}=2\pi b\lambda$ at radius $b$. Image ring: $-2\pi a\lambda$ at radius $a^2/b$. Therefore

At $z=\pm a$: $\sqrt{a^2+a^4/b^2}=\frac ab\sqrt{a^2+b^2}$, so the second term is $2\pi a\lambda\cdot\frac{b}{a\sqrt{a^2+b^2}}=2\pi b\lambda/\sqrt{a^2+b^2}$ — exactly the first: $\Phi(\pm a)=0$, the axis meeting the grounded surface. Far away, $\Phi\to(2\pi b\lambda-2\pi a\lambda)/|z|$: the monopole is the total (ring + induced) charge, as it must be. The code integrates $\lambda\,b\,d\phi'\,G_D$ around the ring with (3.48) and matches the closed form at $10^{-10}$.

The exterior solution (3.53) is $q\,G_D+Va/r$: i.e. the field of three point charges — the source $q$; the grounded-sphere image $q'=-qa/d$ at $r''=a^2/d$; and a fictitious center charge $Va$ carrying the $Va/r$ term. The force on $q$ is $q$ times the field of the other two at $r=d$ (its own field exerts nothing):

using $d-a^2/d=(d^2-a^2)/d$. (a) The image term is always attractive; a $V$ of the same sign as $q$ repels. Setting $F=0$:

(Near contact $V_0$ diverges — the image attraction is too singular to beat; far away $V_0\to q/d$, the value that just cancels the induced-dipole pull.) (b) Neutral isolated sphere: (3.55) with $Q=0$ replaces $Va$ by $+qa/d$ at the center (the compensating charge that restores neutrality against the image $-qa/d$). Then

attraction always: the sphere polarizes, the near (opposite-sign) induced charge is closer than the far one. Combining over a common denominator gives the $-q^2a^3(2d^2-a^2)/[d^3(d^2-a^2)^2]$ form; integrating $-F$ from $d$ to $\infty$ gives $U=-q^2a^3/[2d^2(d^2-a^2)]$ (the code checks $F=-dU/dd$ numerically), and expanding for $d\gg a$:

the induced-dipole law — the sphere's polarizability is $a^3$, the induced moment $p=a^3E=a^3q/d^2$, and $F=-2pq/d^3$. All four statements are asserted numerically.

Insert $\Phi=X(x)Y(y)$: $X''/X=-Y''/Y=-\alpha^2$ constant. The homogeneous conditions steer the split: $\Phi(0,y)=\Phi(a,y)=0$ wants oscillation in $x$, so $X=\sin\alpha x$ with $\sin\alpha a=0$, i.e. $\alpha_m=m\pi/a$, $m=1,2,\dots$ — the Fourier sine family of ~MA-11; then $Y''=\alpha^2Y$ gives $\sinh/\cosh$, and $\Phi(x,0)=0$ kills the $\cosh$: $Y_m=\sinh(m\pi y/a)$. Superpose and match the last side:

Sine orthogonality ($\int_0^a\sin\frac{m\pi x}{a}\sin\frac{m'\pi x}{a}dx=\frac a2\delta_{mm'}$, Eq. 3.83) projects out

giving the quoted series. (Even $m$ vanish by the symmetry $x\to a-x$ of the data.) This is the same slot expansion as ~EM-04 §5, with $\sinh$ replacing the decaying exponential because the box is finite. The finite-difference cross-check is the uniqueness theorem run numerically: two utterly different constructions, one harmonic function.

The decomposition does the bookkeeping: $\Phi_2$ is harmonic (linear), matches $V$ at $x=0$ and $0$ at $x=a$ for all $y$, and is the $y\to\infty$ limit (far up the slot the problem is one-dimensional between the two vertical walls, so the potential cannot vanish there — it tends to the linear profile; what decays is $\Phi_1$). $\Phi_1$, built from the grounded-wall modes $\sin(n\pi x/a)e^{-n\pi y/a}$, vanishes on $x=0,a$ and at large $y$; its job is the remaining condition at the floor:

Project with sine orthogonality:

using $\int_0^a(1-x/a)\sin(n\pi x/a)\,dx=a/n\pi$ (integrate the two pieces; the $(-1)^n$ terms cancel between them). Every $n$ contributes — the sawtooth $V(1-x/a)$ has no parity to kill alternate terms. So

whose $y=0$ trace is the familiar Fourier sawtooth identity $\frac{2V}{\pi}\sum\frac1n\sin\frac{n\pi x}{a}=V(1-x/a)$ — P20(b) meets it again from the Green-function side.

(a) Expand both $G_D$ and $\delta(x-x')$ in the wall-adapted sine basis (Eq. 3.89): $\delta(x-x')=\frac2a\sum_n\sin\frac{n\pi x}{a}\sin\frac{n\pi x'}{a}$. Writing $G_D=\sum_n\sin\frac{n\pi x}{a}\sin\frac{n\pi x'}{a}\,h_n(y,y')$ and matching term by term,

with $h_n(0)=0$ and decay as $y\to\infty$. Below the source take $A\sinh(k_ny)$, above take $Be^{-k_ny}$ ($k_n=n\pi/a$); continuity and the jump $-h_n'\,|^+_-=8\pi/a$ give $A=\frac{8\pi}{ak_n}e^{-k_ny'}$, $B=\frac{8\pi}{ak_n}\sinh(k_ny')$, i.e. $h_n=\frac{8}{n}\sinh(k_ny_<)e^{-k_ny_>}$ — the quoted $G_D$. The code's flux integral around the source verifies the $-4\pi$ normalization the ODE jump encodes. (b) Only the charged wall $x'=0$ contributes to (2.98): $\partial G/\partial n'=-\partial G/\partial x'|_{x'=0} =-\frac{8\pi}{a}\sum_n\sin\frac{n\pi x}{a}\sinh\frac{n\pi y_<}{a}e^{-n\pi y_>/a}$. The $y'$ integral splits at $y$ and is elementary:

(the $e^{-ky}\cosh ky+e^{-ky}\sinh ky=1$ collapse). Hence

The "$1$" part is the sawtooth series summing to $\Phi_2=V(1-x/a)$ — the Fourier series the exercise advertises — and the exponential part is exactly $\Phi_1$ with $A_n=-2V/n\pi$: P19 recovered wholesale. (Numerically the sawtooth converges only like $1/N$, which is why the test compares with a $6\times10^4$-term partial sum.)

Expand $G_N=\sum_{n\ge0}\cos\frac{n\pi x}{a}\cos\frac{n\pi x'}{a}h_n(y,y')$. The $n\ge1$ modes obey $h_n''-k_n^2h_n=-\frac{8\pi}{a}\delta(y-y')$ as before, but with the Neumann floor condition $h_n'(0)=0$: even solution $\cosh(k_ny)$ below the source, decay above, so $h_n=\frac8n\cosh(k_ny_<)e^{-k_ny_>}$. The $n=0$ mode is where Neumann problems earn their keep: with $\delta_0=1/2$,

Integrating: $h_0'$ is $0$ below the source (floor condition) and jumps to $-4\pi/a$ above it, so $h_0=-\frac{4\pi}{a}y_>+{\rm const}$ — the secular term. It cannot decay: Gauss's theorem (2.102) demands $\oint\partial_nG_N\,da=-4\pi$, and with rigid walls the only exit is the open end. Indeed the flux there is $a\cdot\partial_y(-4\pi y/a)=-4\pi$ ✓ (the oscillatory modes integrate to zero across the section — the code measures exactly $-4\pi$ at $y=3a$). Equivalently: for this unbounded surface the $-4\pi/S$ of (2.106) is zero on the walls, and the "charge outflow" is carried by the secular mode. The additive constant is the usual Neumann gauge freedom ((2.103) defines $\Phi$ only up to $\langle\Phi\rangle_S$).

The sine projection gives $g_n''-k_n^2g_n=-\frac{8\pi}{a}\delta(y-y')$ with decay in both directions ($k_n=n\pi/a$): the symmetric decaying solution is

(check the jump: $-\partial_yg_n$ jumps by $\frac{8\pi}{a}$ across $y'$ ✓). To sum, write $\sin\sin'=\tfrac12[\cos\theta_--\cos\theta_+]$ with $\theta_\pm=\pi(x\pm x')/a$, and use $\sum_{n\ge1}\frac{t^n}{n}\cos n\theta=-\tfrac12\ln(1-2t\cos\theta+t^2)$ with $t=e^{-\pi|y-y'|/a}$:

dividing through by $2t$. As $x\to x',y\to y'$ the denominator vanishes quadratically, reproducing the $-2\ln$ line-charge singularity; on $x=0$ or $x=a$, $\cos\theta_+=\cos\theta_-$ and $G_D=0$. The same $G_D$ is the 2-D image ladder (P1's picture with line charges): $-2\sum_n[\ln d_n^+-\ln d_n^-]$ with images at $2na\pm x'$ — symmetric truncations converge $\sim1/N$, so the code Richardson-extrapolates before comparing all three representations.

The machinery of P18 carries over unchanged until the last projection, which now meets the data $-Ex$:

With $\int_0^ax\sin\frac{m\pi x}{a}dx=\frac{a^2}{m\pi}(-1)^{m+1}$ (integrate by parts; the boundary term at $x=a$ carries the sign),

giving the stated series — all $m$ present and alternating, the signature of data that is odd about the box midline ($-Ex$ has no $x\to a-x$ symmetry, but its sawtooth extension alternates). Sanity: at $y=b$ the series is the standard Fourier expansion $x=\sum_m\frac{2a(-1)^{m+1}}{m\pi}\sin\frac{m\pi x}{a}$ times $-E$ ✓; and term by term each summand is harmonic, so the sum is the solution (uniqueness against the FD solver in the test).

The bridge is §3.6's image-lattice reading of the box Green function: for a unit charge at the center of a grounded cube of side $a$, the images are unit charges of sign $(-1)^{i+j+k}$ at the points $({\rm center})+a\,(i,j,k)$ — precisely the NaCl arrangement with lattice spacing $a$. Therefore

so the Madelung energy is the regularized Green function at coincidence — remove the self term and take the limit. (Interpreting $M$ through $G_D$ also dissolves the conditional-convergence worry the book raises about (3.108): the grounded box fixes the summation order physically.) Numerically (unit cube): put the two arguments at ${\rm center}\pm(\epsilon/2)\hat z$, sum (3.98) — at the center only odd $n,m$ survive since $\sin^2(n\pi/2)$ selects them — with the exp-stable reduced factor, and form

($\epsilon$-even by the symmetry of the split). Two Richardson steps on $\epsilon=(0.2,\,0.1,\,0.05)$ annihilate $c_2$ and $c_4$:

yielding $-1.7475635$ — five good decimals, comfortably beyond the requested three. Cross-check: the Evjen summation (expanding cubes with $1/2$-weighted faces, the standard cure for the conditional convergence) gives $-1.7475646$ at $N=14$. The interaction energy per reference ion is then $W=M/a$, as the book notes below (3.108).

The full-azimuth interior solution (3.118) keeps only the terms regular at $\rho=0$: $\Phi=a_0+\sum_m\rho^m(A_m\cos m\phi+B_m\sin m\phi)$ (the $\ln\rho$ and $\rho^{-m}$ members would blow up at the axis). Match at $\rho=b$ against the Fourier expansion of the data: for $\delta(\phi)$ on $(-\pi,\pi)$,

(all cosine coefficients equal $\frac1\pi\int\delta\cos m\phi\,d\phi=\frac1\pi$; no sines, $\delta$ being even). Termwise identification gives $a_0=V_0/2\pi$, $A_mb^m=V_0/\pi$, $B_m=0$ — the quoted series. Physical readings: $\Phi(0)=V_0/2\pi$ is the surface average (mean-value theorem, the $m=0$ term); each higher multipole of the pinprick data decays inward as $(\rho/b)^m$ — the cylinder is a low-pass filter in harmonic content, which is why relaxation methods smooth from the boundary in (~EM-04 §6). P31 sums this series to the 2-D Poisson kernel, closing the loop with the Green function of P27.

This is §3.8's corner solution (3.119) promoted to a full boundary-value problem. The angular range is restricted to $(0,\beta)$, so single-valuedness no longer quantizes $\nu$; instead the side conditions do. Keep the members of (3.114)–(3.115) regular at $\rho=0$ (drop $\rho^{-\nu}$ and $\ln\rho$). The $\nu=0$ piece $A_0+B_0\phi$ must satisfy $\Phi=V$ at both $\phi=0$ and $\phi=\beta$: $A_0=V$, $B_0=0$. The $\nu\ne0$ pieces must vanish at both sides (their job is the deviation from $V$): $\sin(\nu\phi)$ with $\sin(\nu\beta)=0$, i.e.

— generally non-integer powers of $\rho$; nothing is wrong with that away from the axis' full neighborhood. Superpose and match the end data: $\Phi(b,\phi)-V=\sum_ma_mb^{m\pi/\beta}\sin(m\pi\phi/\beta)$ is an ordinary Fourier sine series on $(0,\beta)$ (the Sturm–Liouville family of ~MA-11); orthogonality delivers the quoted $a_m$. Endpoint mismatches ($\Phi(b,0^+)\neq V$ etc.) are fine — sine series converge to the data in the open interval and to the mean at jumps. Near the tip the $m=1$ term dominates (Eq. 3.120): $\Phi-V\approx a_1\rho^{\pi/\beta}\sin(\pi\phi/\beta)$ and $|\vec E|=\frac{\pi|a_1|}{\beta}\rho^{\pi/\beta-1}$ (3.123): field concentration at a reentrant tip ($\beta>\pi$), shielding in a slot ($\beta<\pi$) — the lightning-rod exponent, which the test extracts from the series by a log-log fit. (If the two sides carry different constants $V_1,V_2$, add the extra harmonic wedge term $(V_2-V_1)\phi/\beta$ — P2(c) is exactly that case with $\Phi(b,\phi)$ absent.)

With no interior charge, (2.98) keeps only the surface term $\Phi=-\frac1{4\pi}\oint\Phi\,\partial G_D/\partial n'\,dl'$ (2-D: a line integral, the $z'$ length already absorbed in the line-charge normalization). Differentiate P14's closed form radially at $\rho'=b$; writing $D=\rho^2+b^2-2b\rho\cos\Delta$, $\Delta=\phi-\phi'$:

The outward normal of the interior region is $+\hat\rho'$, so $\partial G/\partial n'=\partial G/\partial\rho'|_{\rho'=b}$ and, with $dl'=b\,d\phi'$,

Properties worth logging (all tested): setting $\Phi(b,\cdot)\equiv1$ must return $1$ (an equipotential cylinder has a constant interior), so the kernel has unit angular integral; at $\rho=0$ it is uniform — the mean-value theorem again; and as $\rho\to b^-$ it becomes a delta family in $\phi'-\phi$, reproducing the boundary data pointwise where it is continuous. This kernel is the 2-D twin of the sphere's (3.50).

Split the data as the mean plus an odd square wave: $\Phi(b,\phi')=\frac{V_1+V_2}2+\frac{V_1-V_2}2\,{\rm sq}(\phi')$, with ${\rm sq}=+1$ on the right half, $-1$ on the left. The mean passes through the unit-normalized kernel untouched. For the square-wave part, the needed antiderivative is standard:

(differentiate to verify; $t=\tan\frac{\phi'-\phi}2$ substitution). Evaluate between the jump points $\phi'=\pm\pi/2$ for the two halves and combine the four arctangents with the addition law $\arctan u-\arctan v=\arctan\frac{u-v}{1+uv}$; after the trigonometric dust settles (the $\tan\frac{\pm\pi/2-\phi}2$ pairs combine to $2(\rho/b)\cos\phi/(1-\rho^2/b^2)$) one lands exactly on the boxed (3.131). The module takes the complementary route: it verifies the identity by high-accuracy quadrature of the Poisson integral against the closed form — and P29 gets (3.131) a third way, by summing the Fourier series, so the three derivations (series, Green function, complex sum) triangulate the result.

Term by term, for $|Z|<1$,

Take the imaginary part with $Z=te^{i\psi}$:

so the argument is $\arctan\bigl[2t\sin\psi/(1-t^2)\bigr]$ (the real part $1-t^2>0$ keeps the principal branch honest for $t<1$). Now insert $t=\rho/b$, $\psi=\phi-\pi/2$, $\sin\psi=-\cos\phi$ into (3.130):

which is (3.131). On the symmetry axis ($\phi=0$, toward the $V_1$ half) the double-angle identity $\arctan\frac{2t}{1-t^2}=2\arctan t$ ($t<1$) collapses it further to

a clean arctangent ramp from the mean at the center to $V_1$ at the wall — the form the code checks the series against point by point.

(a) Outside, regularity at infinity replaces regularity at the axis: from (3.118) keep the constant and the $\rho^{-n}$ terms (the $\ln\rho$ term is absent because the cylinder carries zero net line charge — the potential must stay bounded). The boundary matching at $\rho=b$ is the same Fourier computation as the interior one — identical coefficients — so every $(\rho/b)^{2n+1}$ simply becomes $(b/\rho)^{2n+1}$:

(b) The P29 lemma applies verbatim with $t=b/\rho<1$, giving the boxed arctangent. The structural answer to the last question: interior and exterior are exchanged by the inversion $\rho\mapsto b^2/\rho$ — in 2-D, inversion maps harmonic functions to harmonic functions with no radial prefactor (unlike the 3-D Kelvin transform's $a/r$ weight), because $w\mapsto b^2/\bar w$… more plainly, $b^2/z$ is analytic (P33) and harmonicity is conformally invariant. Both series are the same function of the variable $t=\min(\rho,b^2/\rho)/b$; the boundary data (and the $\rho$-independent average $\tfrac{V_1+V_2}2$ at the two extremes $\rho\to0,\infty$) match automatically. P33 rederives exactly this by the conformal map $w=b^2/z$.

With $t=\rho/b$, sum the geometric cosine series through $W=te^{i\phi}$:

Then

so $\Phi=\frac{V_0}{2\pi}\frac{1-t^2}{1-2t\cos\phi+t^2}$, and multiplying numerator and denominator by $b^2$ gives the boxed form. Consistency with P27 is immediate: putting $\Phi(b,\phi')=V_0\,\delta(\phi')$ under Poisson's integral just evaluates the kernel at $\phi'=0$. So the three objects — the separation-of-variables series (P25), the Green-function kernel (P27), and this closed form — are one function; the delta-data problem is the cylinder's own Green identity read backwards (the kernel is the boundary "response function"). This is also the generating identity behind P29: integrating the present formula over half the circle reproduces the split-cylinder arctangent.

(a) Write $w=e^{\pi z/b}$, $z=x+iy$: then $|w|=e^{\pi x/b}$ and $\arg w=\pi y/b$. The box $0<x<a$, $0<y<b$ therefore maps to $1<|w|<e^{\pi a/b}\equiv r_0$, $0<\arg w<\pi$: the upper half-annulus. The exponential is entire with $f'=\frac\pi be^{\pi z/b}\neq0$, so the map is conformal on the closed box, and the conformal-mapping theorem (§3.11: pulled back through an analytic map, constant-Dirichlet data stays constant-Dirichlet) transfers the boundary values piece by piece: $x=0$ (grounded) $\to$ the inner arc $|w|=1$; $x=a$ (grounded) $\to$ the outer arc $|w|=r_0$; $y=0$ (grounded) $\to$ $\arg w=0$, the segment $(1,r_0)$; and the charged side $y=b$ ($\Phi=V$) $\to$ $\arg w=\pi$, i.e. $v=0$ with $-r_0<u<-1$ — exactly the advertised layout. (b) Solving in the $w$-plane is not necessary — invert the map. A point $(r,\theta)$ in the half-annulus pulls back to

and harmonicity survives the pullback, so $\Phi_{\rm ann}(r,\theta)= \Phi_{\rm box}(x,y)$ with P18's series. Substituting (and using $a=\frac b\pi\ln r_0$):

which turns the box series exactly into the quoted half-annulus series. Note the effective coordinates: $\ln r$ and $\theta$ are the real and imaginary parts of $\log w$ — the map's inverse — so the "curvilinear rectangle" $(\ln r,\theta)$ is the box again; conformal mapping never did anything but relabel the harmonics (~MA-06). The code verifies the map numerically (Cauchy–Riemann residuals at interior points, boundary tracing of all four sides) and the potential identity on random interior points.

$f(z)=b^2/z$ is analytic for $z\neq0$ with $f'=-b^2/z^2\neq0$, hence conformal on the exterior $\rho>b$ (the puncture at $\infty$ maps to the regular point $w=0$). Writing $z=\rho e^{i\phi}$:

so the exterior maps onto the punctured interior disk, and the circle $\rho=b$ maps onto itself with $\phi\to-\phi$. Crucially the boundary data survive: reflection $\phi\to-\phi$ maps the right half-cylinder to itself and the left to itself ($\cos\phi$ even), so the $V_1/V_2$ split is preserved. By the conformal-mapping theorem, $H(z)\equiv\Phi_{\rm in}(w(z))$ is harmonic in $\rho>b$ and takes the correct boundary values; it is also bounded at infinity ($w\to0$ gives the average $\tfrac{V_1+V_2}2$ — the right physical limit for a neutral boundary). Uniqueness (bounded exterior Dirichlet problem) then forces $\Phi_{\rm out}=H$:

and $\cos(-\phi)=\cos\phi$ gives the stated form — precisely P30's closed form, now obtained in three lines. The inversion trick is the 2-D image-in-a-circle statement one level up: P14's image point $a^2/\rho'$ is $w=a^2/z$ acting on the source.

(a) Factor the generating function and expand each exponential:

The coefficient of $u^m$ collects all $(j,i)$ with $j-i=m$, i.e. $j=m+n$, $i=n$:

the sum starting at $n=\max(0,-m)$ — which for $m\ge0$ is (4.8), and for $m<0$ reproduces $(-1)^mJ_{|m|}$ after shifting the index. (b) Expand $e^{it\cos\phi}=\sum_k\frac{(it)^k}{k!}\cos^k\phi$ inside (4.13) and use $\cos^k\phi=2^{-k}(e^{i\phi}+e^{-i\phi})^k$. The $\phi$ average kills every term except those where the binomial supplies $e^{+in\phi}$, i.e. $k=n+2j$ with binomial coefficient $\binom{n+2j}{j}$:

(a) Multiply the expansion by its complex conjugate:

Averaging over $\phi\in[0,2\pi)$ keeps only $m=m'$: $1=\sum_mJ_m^2$. (b) Square instead of conjugating: $e^{2it\cos\phi}$ has the same expansion with $t\to2t$, and its $\phi$ average is the $m=0$ term, $J_0(2t)$. On the other side the average of $\bigl(\sum_mi^me^{im\phi}J_m\bigr)^2$ keeps $m'=-m$:

using $J_{-m}=(-1)^mJ_m$.

From the series (4.8), $\frac{d}{dt}\bigl[t^{m+1}J_{m+1}(t)\bigr] =\frac{d}{dt}\sum_n\frac{(-1)^n t^{2m+2n+2}}{2^{m+1+2n}n!(m+n+1)!} =t^{m+1}\sum_n\frac{(-1)^n(t/2)^{m+2n}}{n!(m+n)!}=t^{m+1}J_m(t)$. Substituting $t=k\rho$ and integrating,

For $m>-1$ the lower limit vanishes ($t^{m+1}J_{m+1}\sim t^{2m+2}$). For $m=-1$ it is $J_0(0)=1$, giving $-1/k=-a^{m+1}\delta_{m,-1}/k$ — the quoted Kronecker term. (Numerically: $J_{-1}=-J_1$ and $\int_0^aJ_1\,d\rho=(1-J_0(ka))/k$, consistent.)

In (4.13) substitute $\phi=\pi/2-\psi$ (a shift plus reflection of the periodic integrand, so the domain stays one full period): $\cos\phi=\sin\psi$ and $e^{-in\phi}=e^{-in\pi/2}e^{in\psi}=i^{-n}e^{in\psi}$, so

Now let $\psi\to-\psi$: the integral becomes $\int e^{-i(t\sin\psi-n\psi)}\frac{d\psi}{2\pi}$, whose value must equal its own conjugate (swap back), i.e. it is real; and multiplying by $(-1)^n=J_{-n}/J_n$ bookkeeping is avoided by noting the result equals $\int e^{i(t\sin\psi-n\psi)}d\psi/2\pi$ directly — Eq. (4.15). For (4.16): split $e^{i(\cdot)}=\cos(\cdot)+i\sin(\cdot)$ on $[-\pi,\pi]$. $\sin(t\sin\phi-n\phi)$ is odd under $\phi\to-\phi$ and integrates to zero; $\cos(t\sin\phi-n\phi)$ is even, so the half-range integral doubles: $J_n=\frac1\pi\int_0^\pi\cos(t\sin\phi-n\phi)\,d\phi$.

(a) Directly: $(\vec x-\vec x')_\perp$ has some azimuth $\phi_D$, so $\vec k_\perp\cdot(\vec x-\vec x')_\perp=kD\cos(\alpha-\phi_D)$ and (4.13) with $n=0$ gives $\int_0^{2\pi}\frac{d\alpha}{2\pi}e^{ikD\cos(\alpha-\phi_D)}=J_0(kD)$; the extra $1/2\pi$ tags along. Expanded: write $e^{i\vec k_\perp\cdot\vec x_\perp}e^{-i\vec k_\perp\cdot\vec x'_\perp}$ and expand each factor with (4.14), $e^{ik\rho\cos(\phi-\alpha)}=\sum_mi^me^{im(\phi-\alpha)}J_m(k\rho)$:

Equating the two evaluations proves the theorem. (b) Multiply the theorem by $k/2\pi$ and integrate. With the Gaussian regulator $e^{-p^2k^2}$, Weber's second exponential integral gives in closed form

a normalized 2-D Gaussian of width $p\sqrt2$ centered at $\vec x_\perp=\vec x'_\perp$: unit mass $\int2\pi D\,dD\,(\cdot)=1$ for every $p$, peak $\propto1/p^2$. As $p\to0$ this is exactly $\delta^2\bigl((\vec x-\vec x')_\perp\bigr) =\frac1\rho\delta(\rho-\rho')\delta(\phi-\phi')$ — the completeness relation (4.39) resolved into azimuthal components.

Let $u(\rho)=J_m(\bar k\rho)$, which satisfies Bessel's equation $(\rho u')'+(\bar k^2\rho-m^2/\rho)u=0$. Multiply by $2\rho u'$ and regroup into a total derivative:

Integrate from $0$ to $a$. At $\rho=0$ the bracket vanishes ($u\sim\rho^m$ kills $-m^2u^2$ and $\rho^2u'^2$ separately, including $m=0$). At $\rho=a$, $u'(a)=\bar kJ_m'(y)=0$ by the choice of $\bar k$, leaving $(y^2-m^2)J_m(y)^2$. Hence

Expand the transverse delta with the $m$-resolved completeness (4.39) and the azimuthal exponentials: $4\pi\delta^3=2\!\int_0^\infty\!k\,dk\sum_me^{im(\phi-\phi')}J_m(k\rho)J_m(k\rho')\,\delta(z-z')$ (Eq. 4.108). Seeking $G$ with the same transverse structure and a reduced $g(k;z,z')$ turns $\nabla^2G=-4\pi\delta$ into the 1-D problem

Homogeneous solutions $e^{\pm kz}$: take $\psi_1=\sinh kz$ (wall condition), $\psi_2=e^{-kz}$ (decay). The Wronskian recipe (4.130) with unit jump $-\partial_zg|^+_-=1$ gives $C=1/k$ ($W_z[\sinh kz,e^{-kz}]=-k$), so $g=\sinh(kz_<)e^{-kz_>}/k$, which expands to $\bigl(e^{-k|z-z'|}-e^{-k(z+z')}\bigr)/2k$ — visibly the free part minus its mirror image. The $m$ sum collapses to $J_0(kD)$ by P5's addition theorem, giving the second (Hankel) form; carrying out that integral term by term (free + image, each a Lipschitz integral $\int J_0(kD)e^{-k\Delta}dk=1/\sqrt{D^2+\Delta^2}$) reproduces the image Green function — the numeric check does exactly this comparison.

The wall now cuts the azimuth: Dirichlet conditions at $\phi=0$ and $\phi=\pi$ select $\sin(m\phi)$, integer $m\ge1$, with completeness $\delta(\phi-\phi')=\frac2\pi\sum_m\sin m\phi\sin m\phi'$ on $(0,\pi)$. The radial delta stays $\int k\,dk\,J_mJ_m'$ (valid per $m$), and $z$ is now the unbounded direction, so its reduced equation $(k^2-\partial_z^2)g=\delta(z-z')$ with decay both ways has the free solution $g=e^{-k|z-z'|}/2k$ (P7 without the wall term). Assembling with $\nabla^2G=-4\pi\delta$:

Physically the same object as the plane-image Green function — reflection $y'\to-y'$ is $\phi'\to-\phi'$, and $\sin m\phi\sin m\phi'=\tfrac12[\cos m(\phi-\phi')-\cos m(\phi+\phi')]$ is precisely (source) $-$ (image) of the full-space azimuthal kernel. The test verifies the numerical identity.

No volume charge, so the Neumann representation (2.103) leaves only the surface term $\Phi=\frac1{4\pi}\oint G_N\,\partial_{n'}\Phi\,da'+\langle\Phi\rangle$. Build $G_N$ for the gap by the reduced method (P7 with Neumann walls): $\psi_1=\cosh kz$, $\psi_2=\cosh k(d-z)$, Wronskian $=k\sinh(kd)$, so $g_N=\cosh(kz_<)\cosh(k(d-z_>))/(k\sinh kd)$, and $G_N=2\int k\,dk\,J_0(kD)\,g_N+\dots$ (azimuthally symmetric part suffices — the data is axisymmetric). On the top plate the outward normal is $+\hat z$ and $\partial_{n'}\Phi=\partial_z\Phi|_{z'=d}=-E_z^0$ on the disk. The $\phi'$ integral gives $2\pi J_0(k\rho)J_0(k\rho')$ (addition theorem) and the radial integral is P3 with $m=0$: $\int_0^a\rho'J_0(k\rho')d\rho'=aJ_1(ka)/k$. With $g_N(z,z'{=}d)=\cosh(kz)/(k\sinh kd)$,

(b) Swapping the bottom BC changes only the reduced function: $\psi_1=\sinh kz$ (Dirichlet at 0), $\psi_2=\cosh k(d-z)$ (Neumann at $d$), $W=-k\cosh(kd)$, so $g(z,d)=\sinh(kz)/(k\cosh kd)$ — the quoted mixed form. Consistency: $-\partial_z\Phi|_{z=d}=aE_z^0\int dk\,J_0(k\rho)J_1(ka)=E_z^0\,\theta(a-\rho)$ (the classic discontinuous Weber–Schafheitlin integral) — the imposed plate data, which the test recovers numerically by regularization.

The transverse (bounded) directions get the discrete complete set: $e^{im\phi}/\sqrt{2\pi}$ in azimuth and the Fourier–Bessel functions $\mathcal J_{1m}(k_{mn}\rho)$, which obey Dirichlet at $\rho=a$ ($J_m(x_{mn})=0$) with completeness (4.71) $\sum_n\mathcal J_{1m}(k_{mn}\rho)\mathcal J_{1m}(k_{mn}\rho')=\delta(\rho-\rho')/\rho$. The remaining $z$ problem is P7's half-line equation $(k_{mn}^2-\partial_z^2)g=\delta(z-z')$, $g(0)=0$, $g(\infty)=0$, so $g=\sinh(k z_<)e^{-kz_>}/k$ with $k=k_{mn}$. Assembling ($4\pi\times\frac1{2\pi}=2$):

Every factor is manifestly zero on its wall. The check exploits uniqueness the other way round: the same region is (infinite cylinder) minus (mirror through $z=0$), so $G_D$ must equal $G_{\rm cyl}(\Delta z)-G_{\rm cyl}(z+z')$ built from the completely independent $\rho$-reduced representation (4.122)/P16 — the two expansions agree numerically, certifying both.

Fix $m$ and write the $n$-th term of (4.71) with weight $w_n=2/[aJ_{m+1}(x_{mn})]^2$. At a zero of $J_m$ the asymptotic amplitude form $J_{m+1}(x)\simeq\sqrt{2/\pi x}\,|\!\cos(\cdot)|=\sqrt{2/\pi x_{mn}}$ (the cosine is $\pm1$ at the partner phase), so

i.e. $w_n=\pi k_n/a$ with $k_n=x_{mn}/a$. By (4.98) successive zeros are spaced $\Delta x=\pi$, so $\Delta k=\pi/a$ and

which is (4.39). The numeric check smears both sides with the Gaussian $e^{-p^2k^2}$: the lattice sum then equals Weber's closed form $\frac1{2p^2}e^{-(\rho^2+\rho'^2)/4p^2}I_\nu(\rho\rho'/2p^2)$ to machine precision once $a$ exceeds the Gaussian support, the small-$a$ deficit dying like the image term $e^{-(2a-\rho-\rho')^2/4p^2}$.

Grounded caps make $z$ the oscillatory direction: $Z_n=\sin(n\pi z/L)$, forcing the separation constant sign that turns the radial equation into the modified Bessel equation (4.92); regularity at $\rho=0$ selects $I_\nu$, and axisymmetric data ($V$ independent of $\phi$) kills all but $\nu=0$:

The wall condition $\Phi(a,z)=V$ is a sine expansion of a constant: $A_nI_0(n\pi a/L)=\frac2L\int_0^LV\sin\frac{n\pi z}L\,dz=\frac{2V}{n\pi}\bigl(1-(-1)^n\bigr)$, i.e. $4V/n\pi$ for odd $n$, zero for even. Hence the quoted series. (The code evaluates the $I_0$ ratio with scaled ive to avoid overflow at large $n\pi a/L$.) Uniqueness is exercised numerically: an independent finite-difference solution of the same boundary-value problem agrees pointwise.

(a) Put both points of $W=\frac12\int\!\!\int\sigma\sigma'/|\vec x-\vec x'|$ on the plane $z=z'=0$ and insert (4.120): $e^{-k(z_>-z_<)}\to1$, and the $\phi,\phi'$ integrals of $e^{im(\phi-\phi')}$ keep only $m=0$, each giving $2\pi$:

so $C^{-1}=2W/Q^2$ is the quoted functional — every trial density gives a lower bound on $C$ (Thompson, §3.10). (b) The two radial integrals are P3 closed forms: $\int_0^a\rho J_0=aJ_1(t)/k$ and, integrating $\rho^3J_0$ by parts with the same identity, $\int_0^a\rho(\rho/a)^2J_0=\frac ak\bigl(J_1(t)-\tfrac{2J_2(t)}{t}\bigr)$, $t=ka$. With $Q=\pi a^2A(1+\beta/2)$, $C(\beta)$ is a ratio of quadratics in $\beta$; numerical maximization gives $C=0.62126\,a$ at $\beta=7.0$ — matching the book's $0.6213$ and safely below the exact $2a/\pi$. The uniform member $\beta=0$ lands on $3\pi a/16$, the same number the Ch. 3 notes-check produced by real-space quadrature — two independent representations of one variational value.

The half-space Neumann function adds the same-sign image: reduced $g_N=\bigl(e^{-k|z-z'|}+e^{-k(z+z')}\bigr)/2k$, so on the wall $g_N(z,0)=e^{-kz}/k$. With no volume charge, (2.103) leaves $\Phi=\frac1{4\pi}\oint G_N\,\partial_{n'}\Phi\,da'$; the outward normal of the region is $-\hat z$, so on the patch $\partial_{n'}\Phi=-\partial_{z'}\Phi=E_z^0$. The azimuthal integral gives $2\pi J_0(k\rho)J_0(k\rho')$ and the radial one $aJ_1(ka)/k$ (P3):

On the axis ($J_0=1$) the integral is elementary: $\int_0^\infty J_1(ka)e^{-kz}dk/k=(\sqrt{a^2+z^2}-z)/a$. Gauss's law sanity check: the patch injects flux $\pi a^2E_z^0$ into the half-space, so far away $\Phi\simeq(\pi a^2E_z^0/2\pi)/r=E_z^0a^2/2r$ — the $1/2$ because the flux fills a half-space; the closed form's large-$z$ expansion agrees.

(a) With $u=J_\nu$, $v=J_{-\nu}$ (same equation, same $\nu^2$),

(b) Evaluate $C$ where the functions are simple: $J_{\pm\nu}(t)\to\sqrt{2/\pi t}\cos(t\mp\nu\pi/2-\pi/4)$. Then

where $\theta_\pm=t\mp\nu\pi/2-\pi/4$ and the $O(t^{-3/2})$ derivative corrections drop out of the constant. For the Neumann function use its definition $N_\nu=(J_\nu\cos\nu\pi-J_{-\nu})/\sin\nu\pi$:

finite as $\nu\to m$ — the integer-order limit quoted.

Substituting the ansatz into $\nabla^2G=-4\pi\delta$ and matching the delta expansion mode by mode gives the radial problem (4.123),

whose homogeneous solutions are $I_m(k\rho)$, $K_m(k\rho)$. The jump is $dg/d\rho|^+_-=-1/\rho'$, so in the recipe (4.130) $D=-1/\rho$ while $W_\rho[\psi_1,\psi_2]=k\,W_t[I_m,K_m]\cdot(\text{combo})=-(\text{combo})/\rho$: the constant is $C=1$ whenever $\psi_{1,2}$ are unit-coefficient in the $I$, $K$ that carry the Wronskian. (a) Exterior $(a<\rho<\infty)$: $\psi_1$ must vanish at $a$ — $\psi_1=I_m(k\rho)-K_m(k\rho)I_m(ka)/K_m(ka)$; $\psi_2=K_m(k\rho)$ decays. (b) Interior: $\psi_1=I_m(k\rho)$ is the regular one; $\psi_2=K_m(k\rho)-I_m(k\rho)K_m(ka)/I_m(ka)$ vanishes at $a$. In both, the admixed term contributes nothing to the Wronskian, so $C=1$ and $g=\psi_1(\rho_<)\psi_2(\rho_>)$ — the two quoted forms. Dropping the wall ($a\to\infty$ or $a\to0$) leaves $g=I_m(k\rho_<)K_m(k\rho_>)$, Eq. (4.125) with $A=1$: assembled through the $\cos k\Delta z$ integral it must equal $1/|\vec x-\vec x'|$, which the test verifies — that is the Wronskian derivation of the cylindrical free-space Green function.

Both $z$ (grounded caps) and $\phi$ (periodic) are oscillatory; $\rho$ carries the delta:

Each $(m,n)$ mode obeys the modified radial equation (4.123) with $k=k_n$ — modified because the $z$ eigenvalue entered with the sign that makes $e^{\pm ikz}\to\sin$. Two-wall Dirichlet conditions pick the quoted $\psi_1$, $\psi_2$ (each a $I,K$ combination vanishing at its wall). Their Wronskian: bilinearity and $W[I,K]=-1/\rho$ give

so $C=D/W=(-1/\rho)/W$ gives the quoted denominator. Assembling with the mode weights, $4\pi\cdot\frac1{2\pi}\cdot\frac2L=\frac4L$. (The code evaluates everything in ive/kve scaled form and falls back to the exact $m\gg kb$ power law — the 2-D annulus modes of P23 — where the scaled products underflow.)

(a) Apply the cylindrical Laplacian mode by mode. The $\rho,\phi$ part of each term satisfies Bessel's equation: $\bigl[\frac1\rho\partial_\rho(\rho\partial_\rho)-\frac{m^2}{\rho^2}\bigr]J_m(k\rho)e^{im\Delta\phi} =-k^2J_me^{im\Delta\phi}$. The $z$ factor satisfies $\partial_z^2e^{-k|z-z'|}=k^2e^{-k|z-z'|}-2k\,\delta(z-z')$ — the $+k^2$ cancels the $-k^2$, leaving only the kink term:

where the last equality is exactly the expansion (4.108) of the three-dimensional delta. (b) Integrate (4.120) over the disk: the $\phi'$ integral kills all $m\ne0$ (factor $2\pi\delta_{m0}$), and the radial integral is P3's $m=0$ case, $\int_0^a\rho'J_0(k\rho')d\rho'=aJ_1(ka)/k$:

the quoted form — the potential of a uniformly charged disk (unit surface density), and the workhorse behind P9, P13, P14.

Use $\sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]$ and the elementary complex sum $\sum_{n\ge1}\frac{x^n}n\cos n\theta =-\tfrac12\ln(1-2x\cos\theta+x^2)$ (real part of $-\ln(1-xe^{i\theta})$, $0\le x<1$). With $x=u^{q}$ and $\theta_\mp=q(\phi\mp\phi')$:

the quoted form. Both flats are grounded on inspection: at $\phi=0$ or $\phi=\beta$, $\cos\theta_+=\cos\theta_-$ and the ratio is 1. For $\beta=\pi/2$ ($q=2$) multiply numerator and denominator by $\rho_>^4$:

exactly the four-image quadrant Green function of Ex. 3.1.3 (~MACRO_EM-03 P3) — the $w\mapsto w^{q}$ conformal unfolding in Green-function form.

The angular delta on $(0,\beta)$ expands as (4.132), and each $n$ mode obeys the $k=0$ radial equation (4.134), $\frac1\rho\frac d{d\rho}(\rho\frac{dg_n}{d\rho})-\frac{\gamma^2}{\rho^2}g_n=-\frac{\delta(\rho-\rho')}\rho$, with power-law homogeneous solutions $\rho^{\pm\gamma}$ (4.136). Build the two-wall solutions

Their Wronskian: $W_\rho[\rho^{\gamma},\rho^{-\gamma}]=-2\gamma/\rho$, so by bilinearity $W_\rho[\psi_1,\psi_2]=\frac{2\gamma}\rho\bigl[(b/a)^{\gamma}-(a/b)^{\gamma}\bigr]$. With $D=-1/\rho$ the recipe gives $C=-1/\bigl(2\gamma[(b/a)^\gamma-(a/b)^\gamma]\bigr)$, and expanding $C\,\psi_1(\rho_<)\psi_2(\rho_>)$ produces the four-power form quoted (dividing through by $(b/a)^\gamma$ to make every base $\le1$ — the numerically stable grouping the code uses). As $a\to0$ only the first and third powers survive, $g_n\to\frac1{2\gamma}u^\gamma[1-(\rho_>/b)^{2\gamma}]$ — Eq. (4.141) with outer wall $b$, whose $b\to\infty$ limit is P19's open wedge.

Full azimuth: modes $e^{im\Delta\phi}$, radial equation (4.134) with $\gamma\to m$ and, for $m=0$, solutions $\{1,\ln\rho\}$. Regularity at the axis and $g(a)=0$ give

each with the $-1/\rho'$ jump ($W[\ln\rho,1]=-1/\rho$; $W[\rho^m,\rho^{-m}]=-2m/\rho$). Assembling $G=2g_0+4\sum_{m\ge1}\cos m\Delta\phi\,g_m$ (the $4\pi\delta$ normalization with $\delta(\phi)=\frac1{2\pi}\sum e^{im\Delta\phi}$) gives the quoted series. Extra: apply $\sum\frac{x^m}m\cos m\theta=-\frac12\ln(1-2x\cos\theta+x^2)$ twice, with $x_1=\rho_</\rho_>$ (free + ) and $x_2=\rho_<\rho_>/a^2$ (image $-$):

Reading the numerator as $a^2\times$(distance to the image at $a^2/\rho'$ weighted by $\rho'/a$) recovers the cylinder-image construction (~MACRO_EM-03 P14).

The correct start (the hint) is

i.e. Dirichlet sine modes on the restricted azimuth and a cosine transform in the free $z$ direction. The matching ansatz for $G$ turns $\nabla^2G=-4\pi\delta$ into the radial problem (4.123) with $m^2\to\nu^2$, $\nu=n\pi/\beta$ — Bessel's modified equation of fractional order, since nothing in its derivation required integer order (the book's footnote 7). Regularity at the axis and the wall zero at $\rho=a$ give exactly P16(b)'s $g^{\rm in}_\nu=I_\nu(k\rho_<)\bigl[K_\nu(k\rho_>)-I_\nu(k\rho_>)K_\nu(ka)/I_\nu(ka)\bigr]$, $C=1$ as before. Collecting weights, $4\pi\cdot\frac2\beta\cdot\frac1\pi=\frac8\beta$. For $\beta=\pi$ the region is the half-cylinder $y>0$: uniqueness says $G$ must equal $G_{\rm cyl}(\phi-\phi')-G_{\rm cyl}(\phi+\phi')$ built from the full interior cylinder kernel — the test confirms the fractional-order sum collapses onto that integer-order image difference.

Identical machinery to P20 with the angle unrestricted (integer $m$, cosine assembly as in P21). For $m\ge1$ the two-wall power solutions are P20's with $(a,b)\to(b,a)$ and $\gamma\to m$. The new case is $m=0$, where the homogeneous solutions are $1$ and $\ln\rho$: take $\psi_1=\ln(\rho/b)$ (zero at the inner wall) and $\psi_2=\ln(\rho/a)$ (zero at the outer). Then $W_\rho[\psi_1,\psi_2]=\ln(\rho/b)\frac1\rho-\ln(\rho/a)\frac1\rho =\frac{\ln(a/b)}\rho$, and $D=-1/\rho$ gives $C=-1/\ln(a/b)$, i.e. the quoted $g_0$ — the 2-D analog of the $\sinh$ ladder, linear in $\ln\rho$ on each side of the source. Consistency limits: $b\to0$ sends $g_0\to-\ln(\rho_>/a)$ and $g_m\to$ P21's modes (the approach is $\propto1/\ln(a/b)$, which the test removes by extrapolation); $\ln$-mode charge bookkeeping: $-\partial g_0/\partial\rho$ jumps by $1/\rho'$, the induced charge splitting between the walls in the ratio $\ln(\rho'/b):\ln(a/\rho')$ — the 2-D version of P45's plate split.

(a) Expand $(x^2-1)^\ell=\sum_r\binom{\ell}{r}(-1)^rx^{2\ell-2r}$ and differentiate $\ell$ times term by term: $\frac{d^\ell}{dx^\ell}x^{2\ell-2r}=\frac{(2\ell-2r)!}{(\ell-2r)!}x^{\ell-2r}$ (zero once $2\ell-2r<\ell$, truncating at $r=[\ell/2]$). Dividing by $2^\ell\ell!$ gives the first form. For the second, split $(2\ell-2r)!=(2\ell-2r-1)!!\,(2\ell-2r)!!$ and $(2\ell-2r)!!=2^{\ell-r}(\ell-r)!$; then $\frac{(2\ell-2r)!}{2^\ell\,r!\,(\ell-r)!}=\frac{(2\ell-2r-1)!!}{2^r\,r!} =\frac{(2\ell-2r-1)!!}{(2r)!!}$, using $(2r)!!=2^rr!$. (b) At $x=0$ only the constant term $\ell-2r=0$ survives: $\ell$ even, $r=\ell/2$, giving $(-1)^{\ell/2}(\ell-1)!!/\ell!!$ (with $(2r)!!=\ell!!$ and $(\ell-2r)!=1$); odd $\ell$ has no constant term. (c) The $m$-th derivative shifts the surviving term to $\ell-2r=m$ (so $\ell-m$ even, $r=(\ell-m)/2$), with factor $m!$ from $\frac{d^m}{dx^m}x^m$:

after writing $(2\ell-2r-1)!!=(\ell+m-1)!!$ and $(2r)!!=(\ell-m)!!$ at $r=(\ell-m)/2$.

Use the Schwinger/Rodrigues form (4.167): $Y_{\ell m}\propto e^{im\phi}(\sin\theta)^{-m} \bigl(\tfrac{d}{d\cos\theta}\bigr)^{\ell-m}(\cos^2\theta-1)^\ell$. Under the inversion: $\sin\theta$ is unchanged; $\cos\theta\to-\cos\theta$ flips each derivative, giving $(-1)^{\ell-m}$ from the $\ell-m$ derivatives of the even function $(\cos^2\theta-1)^\ell$; and $e^{im(\phi\pm\pi)}=(-1)^me^{im\phi}$ (either sign of the shift, since $e^{\pm im\pi}=(-1)^m$ — which is why the two branches give one answer). Total phase $(-1)^{\ell-m}(-1)^m=(-1)^\ell$. Equivalently: $Y_{\ell m}$ is a degree-$\ell$ homogeneous harmonic polynomial in $\hat r$ divided by $r^\ell$, and any degree-$\ell$ polynomial picks up $(-1)^\ell$ under $\vec r\to-\vec r$.

(a) Take $n>m$ and integrate by parts $n$ times:

because every boundary term contains an underdifferentiated $(x^2-1)^{n-j}$, which vanishes at $\pm1$, and the surviving integrand needs the $n$-th derivative of the degree-$m<n$ polynomial $P_m$, which is zero. (b) Let $N_n=\int P_n^2$. Multiply (4.181) by $P_{n+1}$ and integrate — by (a) only the $P_{n+1}^2$ term survives on the right: $(2n+1)\int xP_nP_{n+1}=(n+1)N_{n+1}$. Now write (4.181) again with $n\to n+1$, multiply by $P_n$ and integrate: $(2n+3)\int xP_{n+1}P_n=(n+1)N_n$. The same cross-integral appears in both, so

Induction: $N_0=\int_{-1}^1dx=2=\frac2{2\cdot0+1}$; if $N_n=\frac2{2n+1}$ then $N_{n+1}=\frac{2n+1}{2n+3}\cdot\frac2{2n+1}=\frac2{2(n+1)+1}$. $\square$

Differentiate the closed form:

(a) Insert the series into (i): $\sum_n\bigl[xP_n'-P_{n-1}'\bigr]t^n=\sum_n nP_nt^n$, so $xP_n'-P_{n-1}'=nP_n$; relabeling $n\to n+1$ gives (4.180). (b) Multiply the $t$-derivative identity through by the root: $(1-2xt+t^2)\,\partial_tG=(x-t)\,G$. Substituting the series and collecting $t^n$:

which is (4.181) — the three-term recursion behind every stable Legendre evaluation in the module's code.

(a) Write $D=d/dx$ and $u_n=(x^2-1)^n$, so $P_n=D^nu_n/(2^nn!)$. Since $Du_{n+1}=2(n+1)x\,u_n$,

by Leibniz. Similarly $Du_n=2nx\,u_{n-1}$ gives

Also $u_n=(x^2-1)u_{n-1}$, so $D^{n+1}u_n=D^{n+1}\bigl[(x^2-1)u_{n-1}\bigr] =(x^2-1)D^{n+1}u_{n-1}+2(n+1)xD^nu_{n-1}+n(n+1)D^{n-1}u_{n-1}$. Eliminating $D^{n+1}u_{n-1}$ between the two expressions and substituting into (i) collapses, after dividing by $2^nn!$, to $P'_{n+1}=(2n+1)P_n+P'_{n-1}$. (The module verifies this identity exactly on polynomial coefficients — the honest replacement for the last page of algebra.) (b) $(x^2-1)^n=(x-1)^n(x+1)^n$. In the Leibniz expansion of $D^n$ at $x=1$, any term leaving a positive power of $(x-1)$ vanishes; only the term with all $n$ derivatives on $(x-1)^n$ survives: $D^n u_n|_{x=1}=n!\,(x+1)^n|_{x=1}=n!\,2^n$, so $P_n(1)=1$.

Apply the addition theorem (4.223) to each factor:

The $d\Omega'$ integral is the orthonormality relation $\int Y_{\ell m}Y^*_{\ell m'}d\Omega'=\delta_{mm'}$, collapsing the double sum:

the last step being the addition theorem read backwards. (This "reproducing" property is the projector identity: $\frac{2\ell+1}{4\pi}P_\ell(\hat u\cdot\hat v)$ is the kernel of the projection onto the $\ell$-th harmonic subspace, and a projector composed with itself is itself.)

(a) Both source points of $W=\frac12\int\int\sigma\sigma'/|\vec x-\vec x'|$ sit on the shell $r=r'=a$, where (4.225) gives $|\vec x-\vec x'|^{-1}=\sum_{\ell m}\frac{4\pi}{(2\ell+1)a}Y_{\ell m}(\hat n)Y^*_{\ell m}(\hat n')$. Axisymmetry keeps only $m=0$; with $Y_{\ell0}=\sqrt{\tfrac{2\ell+1}{4\pi}}P_\ell$ and $da=a^2d\Omega$,

(the $x$ integral stops at 0 because $\sigma$ lives on the upper hemisphere only). Then $C^{-1}=2W/Q^2$ is the quoted sum — every trial is a lower bound on $C$ (§3.10). (b) With $\sigma=A(1+\beta x)$: $\int_0^1P_\ell\sigma=A(c_\ell+\beta d_\ell)$ where $c_\ell=\int_0^1P_\ell dx$ (P33's numbers) and $d_\ell=\int_0^1xP_\ell dx$, computed stably from $xP_\ell=\frac{(\ell+1)P_{\ell+1}+\ell P_{\ell-1}}{2\ell+1}$; and $Q=2\pi a^2A(1+\beta/2)$. Maximizing the ratio numerically (the $\ell$ sum taken to $6000$, converging like $1/\ell^2$ per term) gives $C=0.80515\,a$ at $\beta^=-0.684$ — the book's $0.8052$, and $1.6\%$ under Loh's exact $0.8183$. Since $x=\cos\theta$ decreases toward the rim, $\beta^<0$ means $\sigma$ is largest at the rim: the linear family is doing its best to imitate the true edge singularity.

(a) The Schwinger construction (4.168) expands, for the null vector $\vec a=(\tfrac{\xi_+^2-\xi_-^2}{2i},\,\tfrac{\xi_+^2+\xi_-^2}{2},\,i\xi_+\xi_-)$-type parametrization, the $\ell$-th power of the scalar $\vec a\cdot\vec r$ as a generating polynomial in $\xi_+/\xi_-$ whose $Y_{\ell m}$ coefficients carry fixed numerical weights. But $\vec a\cdot\vec r$ is a scalar: computing it in the primed frame (where $\vec a$ has parameters $\xi'_\pm$ and $\vec r$ has angles $\theta',\phi'$) must give the same number, whence (4.289) — one linear relation between the $2\ell+1$ unprimed harmonics and the $2\ell+1$ primed ones, with coefficients polynomial in the ratio $t'=\xi'_+/\xi'_-$. (b) Each choice of $\vec a$ gives such a relation with a different $t'$; stacking $2\ell+1$ of them, the left sides are known and the coefficient matrix on the primed side is $V_{jm}=(t'_j)^{\ell+m}$ — a Vandermonde matrix, invertible whenever the $t'_j$ are distinct, since $\det V=\prod_{i<j}(t'_j-t'_i)\ne0$. Inverting expresses each $Y_{\ell m}(\theta',\phi')$ as a fixed combination of the $Y_{\ell m'}(\theta,\phi)$ — Eq. (4.288). No $\ell'\ne\ell$ enters: rotations cannot mix degrees. (c) Take the primed frame with $\hat z'\parallel\vec r$. Then $\theta'$ of the point $\vec r\,''$... more directly: $P_\ell(\cos\gamma)$, with $\gamma$ the angle of $\hat n$ from $\hat z'$, equals $\sqrt{4\pi/(2\ell+1)}\,Y_{\ell0}(\theta',\phi')$ evaluated in the primed frame; by (4.288) that is a linear combination of the unprimed $Y_{\ell m}(\theta,\phi)$ — which is (4.219), the input the chapter needed for completeness (4.220) and, after computing the coefficients, the addition theorem (4.223). The numeric check closes the loop by verifying (4.223) itself on random direction pairs.

Insert the Coulomb expansion (4.225), $\frac1{|\vec r-\vec r'|}=\sum_{\ell m}\frac{4\pi}{2\ell+1}\frac{r_<^\ell}{r_>^{\ell+1}}Y_{\ell m}(\hat r)Y^_{\ell m}(\hat r')$. (a) $\int d\Omega'\,Y^_{\ell m}(\hat r')=\sqrt{4\pi}\,\delta_{\ell0}\delta_{m0}$ (orthogonality against $Y_{00}=1/\sqrt{4\pi}$): only the monopole survives,

(This is the shell theorem in kernel form: a uniform shell looks like a point charge outside, a constant potential inside.) (b) $\cos\theta'=\sqrt{4\pi/3}\,Y_{10}(\hat r')$, so the integral projects out $\ell=1$, $m=0$:

Physically: the potential of the shell density $\sigma\propto\cos\theta'$ — uniform field inside ($\propto r\cos\theta$), pure dipole outside ($\propto\cos\theta/r^2$), continuous at $r=r'$.

(a) Interior/exterior harmonic expansions matched to the surface data $V\,\mathrm{sgn}(x)$, $x=\cos\theta$: $\Phi_{\rm in}=\sum A_\ell(r/a)^\ell P_\ell$, $\Phi_{\rm out}=\sum A_\ell(a/r)^{\ell+1}P_\ell$, with

Integrating (4.182) from $x$ to $1$ and using $P_\ell(1)=1$ (Ex. 4.11.2b) gives the workhorse identity

The surface charge adds the interior and exterior normal derivatives: $\sigma=\frac1{4\pi}\bigl(-\partial_r\Phi_{\rm out}+\partial_r\Phi_{\rm in}\bigr)_{r=a} =\sum_\ell\frac{2\ell+1}{4\pi a}A_\ell P_\ell$. The upper-cap charge is then

i.e. $C=Q/V$ as quoted with $\ell=2n+1$. (b) Insert $P_{2n}(0)=(-1)^n\frac{(2n-1)!!}{(2n)!!}$ (P24b):

whose bracket is $4n+3$ — the quoted form. Stirling gives $t_n\to4/\pi n$, so $C_N\simeq\frac{2a}\pi\ln N$: log-divergent, confirmed numerically ($\frac2\pi\ln2$ per doubling).

Expand in Legendre modes (azimuthal symmetry): $G_N=\sum_\ell(2\ell+1)g_\ell(r,r')P_\ell(\cos\theta)$ with the radial equation (4.230). For the exterior region the bounding "surface" is the sphere plus the sphere at infinity, so $S\to\infty$ and the Neumann boundary condition (2.106), $\partial_nG_N=-4\pi/S$, becomes homogeneous: $\partial_rg_\ell|_{r=a}=0$ for every $\ell$. From $r^\ell$, $r^{-\ell-1}$ build

with jump $-1/r'^2$ in $g'_\ell$ fixing $C=1/(2\ell+1)$; multiplying by $(2\ell+1)P_\ell$ gives the book's series (the $\ell=0$ member being $1/r_>$: a Neumann sphere cannot hide the monopole). Closed form: split $\frac{\ell}{\ell+1}=1-\frac1{\ell+1}$. The "$1$" part sums by the generating function to the Kelvin image $\frac{a}{r'}\frac1{|\vec r-\vec r_a|}$; the $\frac1{\ell+1}$ part uses $\sum_\ell\frac{t^{\ell+1}}{\ell+1}P_\ell(x)=\int_0^t\frac{ds}{\sqrt{1-2sx+s^2}} =\ln\frac{t-x+\sqrt{1-2tx+t^2}}{1-x}$, a uniform line-charge image of density $-1/a$ stretched from the center to the inverse point — the standard trick to repair the flux a lone image would push through the Neumann wall. The test also confirms the $\vec r\leftrightarrow\vec r'$ symmetry that the $t=a^2/rr'$ variable makes manifest.

(a) For $\ell\ge1$ the $-4\pi/S$ flux can be (and, for the expansion to close, must be) carried entirely by the $\ell=0$ mode, so the higher modes obey homogeneous Neumann walls: from $r^\ell,r^{-\ell-1}$,

(each normalized so $\psi'=r^{\ell-1}-a^{2\ell+1}/r^{\ell+2}$-type vanishes at its wall). Radial equation $-(r^2g')'+\ell(\ell+1)g=\delta(r-r')$ demands the jump $g'|^+_-=-1/r'^2$; $W_r[\psi_1,\psi_2]=\frac{(2\ell+1)\bigl(b^{2\ell+1}-a^{2\ell+1}\bigr)}{\ell(\ell+1)\,r^2}$, so $C=\ell(\ell+1)/[(2\ell+1)(b^{2\ell+1}-a^{2\ell+1})]$ — the quoted $g_\ell$ (the module evaluates it in ratio form to avoid overflow at large $\ell$). $b\to\infty$: $\psi_2\to\frac{b^{2\ell+1}}{(\ell+1)r^{\ell+1}}$ and $C\psi_1\psi_2\to$ P34's mode, term by term. The $\ell=0$ mode: Gauss's law forces total flux $-4\pi$ through the two walls, and (2.106) distributes it as $-4\pi/S$ per unit area: $\partial_rg_0|_a=+4\pi/S$, $\partial_rg_0|_b=-4\pi/S$ with $S=4\pi(a^2+b^2)$ — one checks $-a^2(+\frac1{a^2+b^2})\cdot4\pi+b^2(-\frac1{a^2+b^2})\cdot4\pi=-4\pi$. Solving $-(r^2g_0')'=\delta(r-r')$ with those wall slopes gives $g_0=\frac1{r_>}-\frac{a^2}{a^2+b^2}\bigl(\frac1r+\frac1{r'}\bigr)+{\rm const}$: determined up to the constant — the §2.8 zero mode of the interior Neumann problem (only potential differences are physical; equivalently $\langle\Phi\rangle_S$ in (2.103) absorbs it). It is not unique, and no boundary condition can make it so. (b) The (2.115) refinement says the normal derivative of $G_N$ acquires a surface delta when the source point lies on the wall. Its integrated signature: a source on a Neumann wall coincides with its same-sign image, so the Green function doubles, $G_N\to2/|\vec x-\vec x'|$, as $\vec x\to\vec x'$ with $r'=a$. The test drives the 500-mode sum to the wall and watches $G_N\,|\vec x-\vec x'|\to2$.

(a) Outside, harmonicity + decay force $\Phi=\sum_{\ell m}c_{\ell m}r^{-\ell-1}Y_{\ell m}$; matching at $r=a$ and projecting with $\oint Y^*_{\ell m}$ (orthonormality) gives $c_{\ell m}=a^{\ell+1}B_{\ell m}$ — the quoted form. Inside, regularity at the origin swaps the radial factor for $(r/a)^\ell$ with the same $B_{\ell m}$. (Equivalently: feed the sphere's Dirichlet Green function into the surface term of (2.98); the modes of $-\frac1{4\pi}\partial_{n'}G_D$ are exactly these radial ratios.) (b) Resum. Insert $B_{\ell m}$, use the addition theorem to collapse the $m$ sum, and evaluate the $\ell$ sum with the generating-function derivative identity: for $t=r/a<1$,

Then $\Phi_{\rm in}=\frac1{4\pi}\oint d\Omega'V\sum_\ell(2\ell+1)t^\ell P_\ell =\frac{a(a^2-r^2)}{4\pi}\oint\frac{d\Omega'V}{(a^2+r^2-2ar\cos\gamma)^{3/2}}$, restoring dimensions by $a^3$ from $t$-powers. Outside, $t=a/r$ gives the same kernel with $(r^2-a^2)$ — the overall $\mp$ sign quoted. As $r\to a^\pm$ the kernel is a normalized, concentrating family on the sphere (unit mass by the $\ell=0$ term), so $\Phi\to V$: the boundary data is recovered, which is the numeric check.

Write $\Phi=V+u$: then $u$ is harmonic with $u=0$ on the base and $u=-V$ on the dome. Harmonics regular at the origin that vanish on the base $\theta=\pi/2$ are exactly the odd-$\ell$ family $r^\ell P_\ell(\cos\theta)$ (odd $P_\ell(0)=0$), so $u=-V\sum_{\rm odd}c_\ell(r/a)^\ell P_\ell$. The dome condition needs $\sum_{\rm odd}c_\ell P_\ell(x)=1$ on $0<x<1$ — i.e. the odd-$\ell$ Legendre expansion of $\mathrm{sgn}(x)$ restricted to the upper half:

by P33's integral identity. Hence the quoted series. Sanity limits: at the center ($r\to0$) only the "1" survives, $\Phi\to V$ — correct, the center sits on the base plate; on the axis the series is alternating with $(r/a)^\ell$ decay. (The coefficients are evaluated with the stable $P_{2n}(0)$ recursion — the double-factorial form overflows past $\ell\sim300$, an implementation lesson the code encodes.)

Conductors are equipotentials, so only the $\ell=0$ sector of the §4.12 Green function is engaged (all $Y_{\ell m}$ surface moments of constant data vanish for $\ell\ge1$). Between the spheres the general monopole solution is $\Phi=\alpha/r+\beta$; the data $\Phi(a)=V_a$, $\Phi(b)=V_b$ give

Gauss's law reads the charges off: the inner sphere carries $Q_a=\alpha=\frac{ab}{b-a}(V_a-V_b)$, so $C_{aa}=\frac{ab}{b-a}$, $C_{ab}=-\frac{ab}{b-a}$. The outer conductor's inner face carries $-\alpha$ (the field between must terminate), while its outer face sees the exterior solution $\Phi=bV_b/r$ and carries $bV_b$:

giving $C_{ba}=C_{ab}$ (reciprocity, automatic) and the quoted $C_{bb}$. Limits: $b\to\infty$: $C_{aa}\to a$ (isolated sphere), $C_{bb}\to b$; $b-a\to0$: the parallel-plate divergence. The matrix is symmetric with $C_{aa}C_{bb}-C_{ab}^2>0$ — the §2.12 energy positivity, which the test asserts.

Inside, $\Phi=\sum_\ell A_\ell(r/a)^\ell P_\ell(\cos\theta)$; at the origin only the $\ell=1$ term has a nonzero gradient ($\nabla[rP_1\cos]=\hat z$; higher $\ell$ vanish like $r^{\ell-1}$, and $\ell=0$ is constant). Project the boundary data:

Then $\vec E(O)=-\nabla\Phi|_0=-\frac{A_1}{a}\hat z =-\frac{3V}{2a}\sin^2\theta_0\,\hat z$. Sanity: for $\theta_0\to0$ or $\pi$ the sphere is (almost) all one potential and the field dies like the cap area; the maximum is the symmetric split. The test differentiates the 800-term series numerically at the origin and matches.

(a) Rotational symmetry: put $\hat z$ along $\hat r'$, so $G=G(\gamma)$ with the $+$ charge at $\gamma=0$ and the $-$ at $\gamma=\pi$. Away from both, $\frac1{\sin\gamma}\frac d{d\gamma}\bigl(\sin\gamma\,\frac{dG}{d\gamma}\bigr)=0$ integrates to $\sin\gamma\,G'=c_1$. Try $G=Q_0(\cos\gamma)$: $Q_0'(x)=1/(1-x^2)$, so $\frac{dG}{d\gamma}=-\sin\gamma\cdot\frac1{\sin^2\gamma}=-\frac1{\sin\gamma}$ and $\sin\gamma\,G'=-1$: harmonic in the punctured sphere. Near $\gamma=0$, $1-x\simeq\gamma^2/2$ gives $G\simeq\ln(2/\gamma)$: exactly the 2-D point charge $-\ln(\text{distance})$ demanded by the $-2\pi$ delta (flux bookkeeping: $\oint(\partial G/\partial\gamma)\,\sin\gamma\,d\phi=-2\pi$ ✓); near $\gamma=\pi$, $G\simeq-\ln(2/(\pi-\gamma))\cdot$… i.e. $+\ln$ divergence of the opposite sign — the $-1$ charge. Uniqueness on the compact surface holds up to a constant; $Q_0$ is the zero-average representative ($\int_{-1}^1Q_0\,dx=0$ by oddness). (b) Expand $G=\sum_\ell g_\ell\frac{2\ell+1}{4\pi}\cdot4\pi\,(\dots)$ — cleanly: use $\delta^2(\hat n-\hat n')=\sum_{\ell m}Y_{\ell m}(\hat n)Y^*_{\ell m}(\hat n')$ and $Y_{\ell m}(-\hat n')=(-1)^\ell Y_{\ell m}(\hat n')$ (P25): the source difference keeps odd $\ell$ only, with the $m$ sums collapsing by the addition theorem to $\frac{2\ell+1}{4\pi}P_\ell(\cos\gamma)\times2$. Since $\nabla^2_{\theta\phi}P_\ell=-\ell(\ell+1)P_\ell$,

The Cesàro-averaged numerical sum matches $Q_0(\cos\gamma)$ with coefficient exactly 1 — the printed extra factor 2 in the book's (b) would double the charge and contradict its own part (a).

(a) The exterior monopole reads the total charge: $\Phi_{\rm out}=\sum A_\ell(a/r)^{\ell+1}P_\ell$ has $Q_{\rm tot}=aA_0=a\,\langle V\rangle$, the average of the surface potential. Zero net charge ($+Q-Q$) therefore forces

the cap area fractions being $(1\mp x_0)/2$. (b) Project the two-level data: $A_\ell=\frac{2\ell+1}2\bigl[V_1I_+ + V_2I_-\bigr]$ with $I_+=\int_{x_0}^1P_\ell$, $I_-=\int_{-1}^{x_0}P_\ell$. Then $\sigma=\sum\frac{2\ell+1}{4\pi a}A_\ell P_\ell$ and the top-piece charge is $Q=2\pi a^2\int_{x_0}^1\sigma\,dx=\frac a2\sum(2\ell+1)A_\ell I_+$. Using (a) to write $V_1,V_2$ in terms of $V_1-V_2$ ($V_1=\frac{1+x_0}2(V_1-V_2)$, $V_2=-\frac{1-x_0}2(V_1-V_2)$) gives

the printed form ($\ell=0$ drops: $A_0=\langle V\rangle=0$). Integrals to sums: integrate (4.182) as in P33: $I_+=[P_{\ell-1}(x_0)-P_{\ell+1}(x_0)]/(2\ell+1)$, and $I_-=-I_+$ because $\int_{-1}^1P_\ell=0$ for $\ell\ge1$ — which also collapses the bracket to $2I_+$ and the sum to $\frac a4\sum(2\ell+1)^2I_+^2$. (c) At $x_0=0$, $(2\ell+1)I_+=P_{\ell-1}(0)-P_{\ell+1}(0)$ (odd $\ell$), so $C=\frac a4\sum_{\rm odd}[P_{\ell-1}(0)-P_{\ell+1}(0)]^2$ — exactly half of Ex. 4.12.1's $\frac a2\sum[\cdot]^2$: there $C=Q/V$ with $V_{1,2}=\pm V$, here $C=Q/2V$. Same physics, factor-two of definition.

Give the halves independent potentials $V_1$, $V_2$. Since $\int_{-1}^0P_\ell=(-1)^\ell\int_0^1P_\ell$ (parity of $P_\ell$), the projection is

The upper-half charge (P41's bookkeeping at $x_0=0$):

and reading off $Q_1=C_{11}V_1+C_{12}V_2$ gives both quoted series; $C_{22}=C_{11}$ and $C_{21}=C_{12}$ by the up-down symmetry. Consistency: odd-$\ell$ terms cancel in $C_{11}+C_{12}$, and even $\ell\ge2$ have $I_\ell=0$ (their $P_{\ell\mp1}(0)$ vanish), leaving only $\ell=0$: $C_{11}+C_{12}=\frac a4\cdot1\cdot1\cdot2=\frac a2$ exactly — set $V_1=V_2=V$: the full sphere holds $Q_{\rm tot}=2Q_1=(C_{11}+C_{12})2V=aV$ ✓. Setting $V_2=-V_1$ instead isolates $C_{11}-C_{12}=\frac a2\sum_{\rm odd}(2\ell+1)^2I_\ell^2$ — Ex. 4.12.1's series, so the gap capacitance $\frac12(C_{11}-C_{12})$ diverges logarithmically as it must.

Image route: $G_D=\frac1{|\vec r-\vec r'|}-\frac1{|\vec r-\vec r''|}$ with $\vec r''$ the mirror of $\vec r'$ ($\theta'\to\pi-\theta'$, same $\phi'$, same $r'$). Expand both by the Coulomb expansion (4.225); the image term carries $Y^_{\ell m}(\pi-\theta',\phi')=(-1)^{\ell+m}Y^_{\ell m}(\theta',\phi')$ (the $\theta$-half of P25's parity: $P_\ell^{|m|}(-x)=(-1)^{\ell+|m|}P_\ell^{|m|}(x)$, no azimuth flip here). Subtracting kills $\ell+m$ even and doubles $\ell+m$ odd:

the quoted form. On the plane $\theta=\pi/2$: $Y_{\ell m}(\pi/2,\phi)\propto P_\ell^{|m|}(0)$, which vanishes exactly when $\ell+m$ is odd — every surviving term is individually zero on the conductor. (Equivalently, one can derive it without images: expand $G_D$ in the half-space harmonic basis — the $\ell+m$-odd $Y_{\ell m}$, complete for functions vanishing at $\theta=\pi/2$ — and solve the radial jump problem; the doubling $8\pi$ appears because the half-basis resolves only the odd part of the delta.)

Completeness of $\{\psi_n\}$ means every admissible $f$ expands as $f(\vec x)=\sum_nc_n\psi_n(\vec x)$; orthonormality ($\int\psi^_n\psi_m=\delta_{nm}$) fixes the coefficients: $c_n=\int d^3x'\,\psi^_n(\vec x\,')f(\vec x\,')$. Substituting the coefficients back and exchanging sum and integral,

An object that reproduces every test function under the integral is the delta distribution — Eq. (4.267). (The equality is distributional: the sum does not converge pointwise, but its action on smooth functions does, which is precisely what the numeric check exercises with a truncated sine basis.) Conversely, if (4.267) holds, expanding any $f$ succeeds — completeness and the delta relation are equivalent.

The Laplacian eigenfunctions of the slab that vanish on the plates are $\psi_{\vec k,n}=\frac{e^{i\vec k\cdot\vec x_\perp}}{2\pi}\sqrt{\frac2a}\sin\frac{n\pi z}a$ with eigenvalue $-(k^2+k_n^2)$, $k_n=n\pi/a$; they are orthonormal and complete. The eigenfunction recipe (4.268-style) inverts $-\nabla^2G=4\pi\delta$ mode by mode:

The angular part of $d^2k$ is P5's addition-theorem average, $\int_0^{2\pi}d\alpha\,e^{ikD\cos\alpha}=2\pi J_0(kD)$, leaving the quoted single $k$ integral. It is the same object as Ex. 3.1.1's image ladder and §3.2's reduced-$g$ $k$-integral — three representations, and the numeric check pins the eigenfunction one against the ladder. (Practical note: at $z=z'$ the raw $k$ integrand decays only like $J_0(kD)/2k$ — conditionally convergent; doing the $n$ sum first (P48) and pulling out $e^{-k|z-z'|}/2k\to1/|\vec x-\vec x'|$ analytically leaves an exponentially convergent remainder. The code does exactly this.)

The reduced equation (P16) is $\bigl[\mathcal L_\rho-k^2\bigr]g_m=-\delta(\rho-\rho')/\rho$ with $\mathcal L_\rho=\frac1\rho\partial_\rho(\rho\partial_\rho)-\frac{m^2}{\rho^2}$ and $g_m(a)=0$. The $\mathcal J_{1m}(k_{mn}\rho)$ are exactly the Dirichlet eigenfunctions of $\mathcal L_\rho$: $\mathcal L_\rho\mathcal J_{1m}=-k_{mn}^2\mathcal J_{1m}$, orthonormal with weight $\rho$ (4.69) and complete (4.71). Expand $g_m=\sum_nc_n\mathcal J_{1m}(k_{mn}\rho)$ and the delta by (4.71); matching coefficients,

the quoted sum. Its closed resummation is P16(b)'s $I_m(k\rho_<)[K_m(k\rho_>)-I_m(k\rho_>)K_m(ka)/I_m(ka)]$ — the identity the test verifies numerically; it is the radial analog of P48's sine sum rule.

Eigenfunctions of $d^2/dx^2$ with Dirichlet ends: $\psi_n=\sqrt{2/L}\sin k_nx$, eigenvalues $-k_n^2$, complete and orthonormal. The (2.124) normalization has no $4\pi$: $G_D=\sum_n\psi_n(x)\psi_n(x')/k_n^2$, the quoted series. Closed form: solve directly — $G$ is linear on each side of $x'$ (no source), vanishes at both ends, is continuous at $x'$, and its slope drops by 1 there:

Physical reading (§2.9): a string pulled transversally at $x'$ — the triangle profile; or the potential of a unit charge between grounded points in 1-D. The test verifies series $=$ triangle, the exact zeros, and the unit kink.

The box eigenfunctions are the products quoted (Dirichlet Fourier–Bessel radially, exponentials azimuthally, sines axially), with $-\nabla^2\psi=\lambda\psi$. The 4.267/4.268 recipe gives $G_D=4\pi\sum\psi(\vec x)\psi^(\vec x')/\lambda$. Now fix $(m,n)$ and do the $p$ sum: it is exactly $\frac2L\sum_p\sin(k_pz)\sin(k_pz')/(k_{mn}^2+k_p^2)$ — and comparing with the reduced-Green-function* route to the same $G_D$ (Eq. (4.79), built in §4.4 from $g=\sinh(kz_<)\sinh(k(L-z_>))/(k\sinh kL)$, which solves $(k^2-\partial_z^2)g=\delta$ with grounded ends) forces, mode by mode,

— the quoted rule (multiply by $L/2$). Direct proof without (4.79): expand $g(k;z,z')$ itself in the sine basis; its coefficients are $\int_0^Lg\sin k_pz\,dz=\sin(k_pz')/(k^2+k_p^2)$ by two integrations by parts (the boundary terms vanish, the kink supplies the numerator). The rule turns P45's $n$ sum into the exponentially convergent (4.79) form — the identity the code's G_plates_bessel/G_finite_cyl_79 exploit, and the triple-sum test confirms end to end.

(a) Separate the Helmholtz eigenproblem $\nabla^2\psi=-k^2\psi$ in spherical coordinates: the angular operator is $-\ell(\ell+1)$ on $Y_{\ell m}$ (the §4.13 separation), leaving the radial equation quoted — Bessel's equation in disguise: substituting $j_\ell(t)=\sqrt{\pi/2t}\,J_{\ell+1/2}(t)$ maps it to the half-odd-order Bessel equation. Of the two solutions $j_\ell$ (regular, $\sim t^\ell$) and $n_\ell$ (singular, $\sim t^{-\ell-1}$), the origin inside the domain keeps only $j_\ell$. (b) The Dirichlet wall demands $\psi(a,\theta,\phi)=0$ for all angles, i.e. $j_\ell(ka)=0$: the spectrum is $k_{\ell n}=z_{\ell n}/a$, discrete and $\ell$-dependent (degenerate in $m$: $2\ell+1$ each). These are the modes that would assemble the sphere's interior $G_D$ as $4\pi\sum\psi\psi^*/k^2$ — and, in Ch. 10, the resonant frequencies of the spherical cavity. The $\ell=0$ tower $k=n\pi/a$ is the 1-D box in radial disguise ($rj_0(kr)=\sin(kr)/k$ obeys P47's problem on $[0,a]$).

Let the moments about the old origin be $q=\int\rho$, $\vec p=\int\vec x'\rho$. About a new origin at $\vec R$ the position vectors are $\vec x'-\vec R$, so

If $q\neq0$ the choice $\vec R=\vec p/q$ is well defined and gives $\vec p\,'=0$. (Conversely, when $q=0$, $\vec p\,'=\vec p$ for every $\vec R$: the dipole moment of a neutral distribution is intrinsic — Eq. (5.39) footnote's general pattern: the lowest nonvanishing multipole is origin-independent.) $\blacksquare$

In the new frame the source point is $\vec y=\vec x'-\vec R$. Monopole: $q'=\int\rho=q$. Dipole: $\vec p\,'=\int(\vec x'-\vec R)\rho =\vec p-q\vec R$ (P1). Quadrupole: insert $\vec y$ into the definition,

Expand: $3x_i'x_j'-3R_ix_j'-3R_jx_i'+3R_iR_j$ and $(\vec x'-\vec R)^2=r'^2-2\vec x'\cdot\vec R+R^2$. Integrating term by term,

which is the boxed form. Consistency checks: (i) trace: $\sum_iQ'_{ii} =0-6\vec R\cdot\vec p+6\vec R\cdot\vec p+q(3R^2-3R^2)=0$ — still traceless; (ii) if $q=0$ and $\vec p=0$, then $Q'_{ij}=Q_{ij}$: the quadrupole is origin-independent exactly when all lower moments vanish (the general rule stated after (5.39)).

(a) Monopole: $q=K\int x\,dV=0$ (odd in $x$). Dipole: $p_x=K\int x^2dV=K\big(\int_{-L/2}^{L/2}x^2dx\big)L^2=K\frac{L^3}{12}L^2 =\frac{KL^5}{12}$; $p_y=K\int xy\,dV=0$ (odd in $y$), likewise $p_z=0$. Quadrupole: every component is an integral of $Kx\times(\text{quadratic})$: - $Q_{11}=K\int x(3x^2-r^2)dV$ — integrand odd in $x$ → 0; same for $Q_{22},Q_{33}$; - $Q_{12}=3K\int x^2y\,dV=0$ (odd in $y$), $Q_{13}=3K\int x^2z\,dV=0$, $Q_{23}=3K\int xyz\,dV=0$ (odd in each).

So the quadrupole vanishes identically — not by symmetry of the shape alone but because $\rho$ is odd under $x\to-x$ while every quadrupole integrand is even-degree in the coordinates times $x$, hence odd under inversion of one axis. (b) With $q=0$, $Q=0$, the expansion (5.12) starts and — through $\ell=2$ — ends with the dipole term, $\Phi=\vec p\cdot\vec x/r^3$, so $\vec E=[3(\vec p\cdot\hat r)\hat r-\vec p]/r^3$ (5.15). Because $\rho$ is odd under $\vec x\to-\vec x$... more precisely under $x\to-x$, all even-$\ell$ moments vanish, so the next correction is the $\ell=3$ (octopole) term of P4, smaller by $O(L^2/r^2)$.

The $\ell=3$ Taylor term of $1/|\vec x-\vec x'|$ is $-\frac1{3!}(\vec x'\cdot\vec\nabla)^3\frac1r$ (5.4–5.5). Two gradients gave (5.8); the third gradient of $(3x_ix_j-\delta_{ij}r^2)/r^5$ gives

symmetric and traceless on every pair (contract $j=k$: $-15x_ir^2/r^7+3(x_i+x_i+3x_i)/r^5=0$ — harmonicity of $1/r$). With the $(-1)^\ell$ of (5.4) the $\ell=3$ contribution to $\Phi$ is

Regroup source and field factors as in (5.10): the $\delta$-terms contract two source indices into $r'^2$ and leave one $x$-factor, e.g. $\sum_{jk}x_j'x_k'\delta_{jk}\,x_i x_i'\cdot 3/r^5 =3r'^2(\vec x\cdot\vec x')/r^5=\sum_{ijk}\frac{x_ix_jx_k}{r^7}\, r'^2x_i'\delta_{jk}\,$-pieces after symmetrization, so

(Because the field tensor is traceless, only the traceless part of $x_i'x_j'x_k'$ survives the contraction — so $R_{ijk}$ may be, and here is, defined traceless with no loss.) Counting: a symmetric rank-3 tensor in 3-D has $\binom{3+2}{3}=10$ independent entries. Tracelessness $\sum_iR_{iik}=0$ is one condition for each $k$ (and by symmetry every other trace coincides with these): 3 conditions, leaving $10-3=7=2\cdot3+1$ — exactly the number of $Y_{3m}$'s, as the spherical expansion (5.20) predicts.

(a) $q=1+1-2=0$; $\vec p=(+1)(a\hat k)+(+1)(-a\hat k)=0$. Quadrupole ($Q_{ij}=\sum_nq_n(3x_ix_j-\delta_{ij}r^2)_n$): the two $\pm a$ charges each give $3z^2-r^2=2a^2$ on the 33 component and $-a^2$ on 11 and 22; the central charge sits at $r=0$:

Then

(b) Exactly, with $\cos\theta=\hat r\cdot\hat k$ and $r>a$ the Legendre generating function gives

whose leading term is (a); odd $\ell$ cancels between $\pm a$ and the monopole cancels against $-2/r$. The next correction is down by $(a/r)^2$ — the prolate ($Q_{33}>0$) case of Fig. 5.3.

(a) A dipole is the limit of charges $\pm q$ at $d\hat k\pm\frac{\vec\delta}2$ with $\vec p=q\vec\delta$. Each charge images as usual: $q\to-q$ reflected through $z=0$. The image pair sits at $-d\hat k\mp\frac{\vec\delta^{\,}}2$ where $\vec\delta^{\,}=(\delta_x,\delta_y,-\delta_z)$ is the reflection of $\vec\delta$, with the charges swapped in sign. The image dipole is therefore $\vec p\,''=-q\vec\delta^{\,}=(-p_x,-p_y,+p_z)$: the transverse components flip, the vertical one survives. On $z=0$ the two dipole potentials cancel pointwise (each charge cancels its own mirror), so $\Phi=0$ there and uniqueness makes this the* $z>0$ solution. (b) With $\vec p=p_z\hat k$, $\Phi=p_z(z-d)/R_1^3+p_z(z+d)/R_2^3$, $R_{1,2}^2=\rho^2+(z\mp d)^2$. On the plane $R_1=R_2=R=(\rho^2+d^2)^{1/2}$ and

The conductor's surface charge is $\sigma=\frac1{4\pi}E_n=-\frac1{4\pi} \partial_z\Phi|_{0^+}$, giving the boxed $\sigma$. For $p_z>0$ the induced charge is positive in the near zone directly beneath the dipole and switches sign on the circle $\rho=\sqrt2\,d$. The total, $\int_0^\infty\sigma\,2\pi\rho\,d\rho\propto\int_{d^2}^\infty \frac{(3d^2-u)}{u^{5/2}}du=0$: a neutral source induces zero net charge.

(a) The quadrupole energy in an external field is (from (5.25), written with derivatives at the quadrupole's location) $W_Q=\frac16\sum_{ij}Q_{ij}\,\partial_i\partial_j\Phi$. With the symmetric $Q_{ij}$ and $\nabla^2\Phi=0$ at the (source-free) quadrupole position,

The force is $\vec F=-\vec\nabla W=-\frac14Q_{33}\vec\nabla\Phi_{zz} =\frac14Q_{33}\partial_z^2(-\vec\nabla\Phi)=\frac14Q_{33}\, \partial^2\vec E/\partial z^2$ (derivatives commute). (b) The dipole's field is $\vec E=[3(\vec p\cdot\hat s)\hat s-\vec p]/s^3$ with $\vec s=\vec x-\vec x_1$. Rather than differentiate blindly, use the interaction energy with $\vec x$ the separation from dipole to quadrupole (magnitude $x_1$ along $-\hat 1$): from (5.28)-type algebra, or directly from (a): $W(\vec x_0)=\frac14Q_{33}\,\partial_z^2\Phi_{\rm dip}(\vec x_0)$. On the 1-axis ($\vec s=-x_1\hat 1$), $\Phi_{\rm dip}=\vec p\cdot\vec s/s^3$ gives $\partial_z^2\Phi_{\rm dip}|_{\vec x_0=0}=3p_xx_1/x_1^5\cdot\ldots$ — carrying it out (or by the code's algebra): $W(x_1)=\frac34\,Q_{33}\,p_x/x_1^4$. Moving the quadrupole toward the dipole decreases the separation, so

Only $p_x$ (the component along the line of centers) contributes at this order; for $Q_{33}p_x>0$ the quadrupole is pulled away from the dipole.

Use P7(a) with $\Phi_2$ = potential of quadrupole 2. From P5's algebra, a symmetric quadrupole's potential is

with $\vec y$ measured from quadrupole 2. Then $W=\frac14Q^{(1)}_{33}\,\partial^2\Phi_2/\partial y_3^2$ evaluated at quadrupole 1's position. Perpendicular case $\vec y=(\mp r,0,0)$: with $f=(2y_3^2-y_1^2-y_2^2)/y^5$,

Collinear case $\vec y=(0,0,r)$: on the axis $f=2/y_3^3$, so $\partial_{y_3}^2f=24/r^5$ and $W=\frac14Q_{33}^{(1)}\frac{Q^{(2)}_{33}}4\frac{24}{r^5} =\frac32Q^{(1)}_{33}Q^{(2)}_{33}/r^5$. Sign verdict: the microscopic check (two $(+1,-2,+1)$ triplets) gives $W=+9a^4/r^5\cdot(1+O(a^2/r^2))$ — positive — matching $+9/16\,(4a^2)^2/r^5$; the printed minus sign cannot be reproduced by any orientation consistent with Fig. 5.23. Physically: two prolate charge "dumbbells" side by side present like charges broadside and repel; end-to-end they also repel (positive tips adjacent), consistent with both signs found. Both configurations are unstable extrema in orientation — tilting one axis lowers $W$ (the T-shaped $45°$ arrangements are where quadrupoles attract).

The induced surface charge of the grounded sphere is exactly equivalent, for all exterior fields, to the image $q'=-qR/D$ at $\vec x''=(R^2/D)\hat z$ (taking the charge on $+\hat z$). Take $\rho_<$ = image charge (interior), $\rho_>$ = the real charge at $D$: the configuration is separable in the book's sense (all of $\rho_<$ lies at radius $R^2/D<$ all of $\rho_>$ at $D$). The moments (5.33)–(5.34) of on-axis point charges carry only $m=0$:

(using $\sqrt{4\pi/(2\ell+1)}\,Y_{\ell0}(0)=P_\ell(1)=1$). Then (5.37):

Note this $W$ is the charge–surface interaction energy (the full image-Coulomb value), not the $\frac12$-weighted work of assembly — exactly what the product formula (5.37) computes. Each $\ell$ term is the interaction of the charge with the sphere's induced $2^\ell$-pole; the $\ell=0$ term $qq'/D=-q^2R/D^2$ is the "grounded monopole" attraction that survives at $D\gg R$.

(a) Start from $\vec F=\int\rho\vec E^{(0)}$ (5.41) and Taylor expansion (5.42). Dipole term: $\int\rho\,(\vec x\cdot\vec\nabla')\vec E^{(0)}$ is already $(\vec p\cdot\vec\nabla)\vec E^{(0)}$ — the identity (5.44), $\vec\nabla(\vec x\cdot\vec E^{(0)})=(\vec x\cdot\vec\nabla)\vec E^{(0)}$, needs $\vec\nabla\times\vec E^{(0)}=0$ and converts it to the gradient form of (5.52); running it backwards is where curl-freeness enters. Quadrupole term: the raw Taylor coefficient is $\frac12\sum_{ij}x_ix_j\partial_i\partial_j\vec E$; to replace $x_ix_j\to\frac13(3x_ix_j-r^2\delta_{ij})$ (which builds $Q_{ij}$) one must add $-\frac16r^2\delta_{ij}\partial_i\partial_j\vec E =-\frac16r^2\nabla^2\vec E$, and $\nabla^2\vec E^{(0)} =\vec\nabla(\vec\nabla\cdot\vec E^{(0)})-\vec\nabla\times(\vec\nabla\times\vec E^{(0)})=0$ requires both $\vec\nabla\cdot\vec E^{(0)}=0$ (no external charge inside the distribution) and curl-freeness (statics). With those assumptions the term is $\frac16\sum Q_{ij}\partial_i\partial_j\vec E^{(0)}$. So (5.52) and (5.53) agree in electrostatics whenever the external sources don't overlap the distribution — the book's remark after (5.53). (b) $\vec N=\int\rho\,\vec x\times\vec E^{(0)}(\vec x)\,d^3x$. Expand $E^{(0)}_k(\vec x)=E^{(0)}_k(0)+x_m\partial_mE^{(0)}_k+\dots$:

with $\overline{x_jx_m}\equiv\int\rho x_jx_m=\frac13\big(Q_{jm}+\delta_{jm}\int\rho r^2\big)$. The $\delta_{jm}$ piece contributes $\epsilon_{ijk}\partial_jE_k^{(0)}\propto(\vec\nabla\times\vec E^{(0)})_i=0$, leaving $N_i=(\vec p\times\vec E^{(0)})_i+\frac13\epsilon_{ijk}Q_{jm}\partial_mE^{(0)}_k$. Since $\partial_mE_k=\partial_kE_m$ (curl-free), swapping them gives the second quoted form — showing one is sufficient. (The $r^2$ moment — the "scalar radius" — never exerts torque or force in a source-free external field: only the traceless moments are mechanically visible.)

Summing P10(a)'s quadrupole force over atoms and smoothing,

Integrate by parts in $x_i$: $F_k=\frac16\oint da\,\hat n_i\,q_{ij}\,\partial_jE_k -\frac16\int(\partial_iq_{ij})\,\partial_jE_k$; integrate the volume term by parts again in $x_j$:

Reading off: the volume piece defines $\rho^Q_{\rm eff}$; the middle surface piece is an ordinary surface charge $\sigma^Q_{\rm eff}=-\frac16\hat n_j\partial_iq_{ij}$; the first surface piece has the structure $\oint(\vec{\mathcal P}\cdot\vec\nabla)E_k$ with $(\mathcal P^Q_{\rm eff})_j=\frac16\hat n_iq_{ij}$ — the surface analog of the bulk dipole-force integrand $(\vec P\cdot\vec\nabla)\vec E$ in (5.55), i.e. a surface polarization (dipole layer), not a surface charge. Note the bookkeeping is exactly one derivative deeper than the dipole case: there $-\vec\nabla\cdot\vec P$ plus $\sigma=\vec P\cdot\hat n$; here two derivatives distribute into (volume) + (surface charge) + (surface dipole layer). The book's bracketed note — that $\vec{\mathcal P}^Q_{\rm eff}$ can be regrouped as a volume polarization $\vec P^Q_{\rm eff}$ with $(P^Q)_j=-\frac16\partial_iq_{ij}$ plus the same surface term — is the single-integration-by-parts reading of the same identity.

(a) Inside the material, $\vec\nabla\cdot\vec E=4\pi(\rho_f+\rho_b)$ (5.61) while $\vec\nabla\cdot\vec D=4\pi\rho_f$ (5.65) with $\vec D=\epsilon\vec E$ and constant $\epsilon$ gives $\vec\nabla\cdot\vec E=4\pi\rho_f/\epsilon$ (5.70). Subtracting,

Each embedded free charge is screened by a co-located bound charge of opposite sign and fraction $(\epsilon-1)/\epsilon$. (b) The dielectric is neutral: its total bound charge (volume + surface) vanishes, $\int\rho_b\,d^3x+\oint\sigma_b\,da=0$. With (a),

The screening charge "pushed off" every interior free charge reappears on the surface, restoring the full $Q_{\rm free}$ as seen by a distant Gaussian surface — the mechanism behind P25's far-field monopole.

With $y=\rho\sin\phi$, the driving potential is $-E_0\rho\sin\phi$: pure $m=1$, $\sin\phi$ symmetry. The §3.7 separable solutions with this symmetry are $\rho\sin\phi$ and $\rho^{-1}\sin\phi$. Regularity at the axis and the far-field condition force

Match at $\rho=a$: continuity of $\Phi$ (tangential $E$): $Aa=-E_0a+B/a$; continuity of $D_\rho$: $\epsilon A=-E_0-B/a^2$. Solving,

the quoted forms. Physics: the interior field is uniform and reduced, $\vec E_{\rm in}=\frac2{\epsilon+1}E_0\hat y$ — the 2-D depolarization factor is $\frac12$ (vs $\frac13$ for the sphere, P24: $\frac3{\epsilon+2}$); the exterior disturbance is a 2-D (line-)dipole $\propto\sin\phi/\rho$ with moment per unit length $\frac{\epsilon-1}{\epsilon+1}\frac{a^2E_0}2\cdot2$. Limits: $\epsilon\to1$ kills the disturbance; $\epsilon\to\infty$ gives $\Phi_{\rm in}\to0$, $B\to E_0a^2$ — the conducting cylinder. This transverse polarizability is the engine of P37's rod force.

(a) Superpose: $d\vec p=\vec P(\vec x')d^3x'$, so

and $\vec P\cdot\vec\nabla'f=\vec\nabla'\cdot(f\vec P)-f\,\vec\nabla'\cdot\vec P$ with the divergence theorem gives the two stated terms: an effective surface charge $\sigma=\vec P\cdot\hat n$ and volume charge $-\vec\nabla'\cdot\vec P$ — precisely (5.59) and (5.58), now derived from the potential rather than the force. (b) Uniform $\vec P$ has $\vec\nabla\cdot\vec P=0$: only surfaces contribute. The bubble wall carries $\sigma_b=\vec P\cdot\hat n'$ with $\hat n'=-\hat r$ (outward from the material means into the bubble): $\sigma_b=-P_0\cos\theta$. A $\cos\theta$ surface charge on a sphere produces a uniform interior field: the interior potential must be $\propto r\cos\theta=z$ (the only regular $\ell=1$ harmonic), and matching the standard boundary conditions gives $\Phi_{\rm in}=-\frac{4\pi}3P_0z$, i.e. $\vec E_P=+\frac{4\pi}3P_0\hat z$ — anti-parallel to the wall dipoles' interior field... directed along $+\hat z$ because the negative bound charge sits on top ($\sigma_b<0$ at $\theta=0$). The slab's remote outer face also carries bound charge, but its (position-independent) sheet field is part of the macroscopic field $\vec E$ in the Clausius–Mossotti bookkeeping of P15(c); the exercise's $\vec E_P$ is the local cavity (Lorentz) term, and that is exactly the $\frac{4\pi}3\vec P$ evaluated here. (The exact interior integral is done by the code as a brute-force surface quadrature — it is uniform to the quoted accuracy at every sampled interior point.)

(a) By P14(a) only the hole's walls contribute (uniform $\vec P$). On the four side faces $\vec P\cdot\hat n'=0$; the top face ($z'=+L/2$, material's outward normal $\hat n'=-\hat z$) carries $\sigma_b=-P_0$ and the bottom $+P_0$. The field at the center is $\vec E(0)=-\oint\sigma_b\,\hat x'\,da'/r'^3$ (pointing from the walls to the origin); the two charged faces contribute equally by the symmetry $\vec x'\to-\vec x'$, so

(b) The transverse components integrate to zero on the square; the $z$-component is $2P_0\int_{\rm top}(z'/r'^3)da'=2P_0\,\Omega_{\rm face}$, where $\Omega_{\rm face}=\int\cos\theta'\,da'/r'^2$ is the solid angle the face subtends at the center. Six congruent faces tile the full sphere: $\Omega_{\rm face}=4\pi/6=2\pi/3$. Hence $\vec E_P(0)=2P_0\cdot\frac{2\pi}3\hat z=\frac{4\pi}3P_0\hat z$ — identical to the spherical cavity of P14(b): the $\frac{4\pi}3$ Lorentz factor is shared by any cavity with cubic-or-better symmetry. (c) Each molecule polarizes in the total local field: $\vec P=N\alpha\vec E_T=N\alpha(\vec E+\tfrac{4\pi}3\vec P)$. Insert $\vec P=\chi\vec E$:

and with $\chi=(\epsilon-1)/4\pi$ (5.67) this is the familiar $\alpha=\frac3{4\pi N}\frac{\epsilon-1}{\epsilon+2}$. The correction factor $(1+\frac{4\pi}3\chi)^{-1}$ is exactly the cavity field of (a)–(b): dense media polarize more easily than the naive $\chi=N\alpha$ because each atom feels its neighbors' polarization.

Construction. A single plane interface reflects with coefficient $\beta$ and transmits with $\tau=1+\beta$ (Eqs. 5.106–5.107). Compose the two perpendicular interfaces as for the conductor corner (§3.1): each reflection multiplies by $\beta$, so the vacuum-side images at $(-x',y')$, $(x',-y')$ carry $\beta$ and the doubly reflected $(-x',-y')$ carries $\beta^2$; crossing an interface multiplies by $\tau$. Verification. On $x=0,y>0$: with $R_{++}=R_{-+}$ and $R_{+-}=R_{--}$ there, $G_{Q1}|_0=(1+\beta)(1/R_{++}+\beta/R_{+-})=\tau(\dots)=G_{Q2}|_0$ — continuous; the normal derivatives give $(1-\beta)=\epsilon\tau\iff\beta=(1-\epsilon)/(1+\epsilon)$ — satisfied. The same works on $y=0,x>0$, and on the internal seams provided the media ratio across them is again $\epsilon$ — i.e. third quadrant $\epsilon^2$. For uniform $\epsilon$ the internal-seam $D$ condition fails by $-2\epsilon\tau\beta\,x'/R^3$-type terms: pure $O(\beta)$. No-go for the uniform corner: the dielectric region is connected, so its potential is one harmonic function whose only allowed image singularity is the source point itself, $G_d=t/R_{++}$; matching term-by-term on the two interfaces then forces $b+c=0,\;b-c=0$ from $x=0$ but $b=t-1\neq0$ from $y=0$ — a contradiction unless $\epsilon=1$. Hence the exercise's expected $\beta$-composition answer is exact only in the factorizable sense above, or to first order in $\beta$ (weak dielectric) for the drawn geometry — the 2-D dielectric wedge is in fact a classic problem with no closed image solution. Both readings are quoted in the answer; the code pins down each statement numerically.

Fourier-transforming in the transverse plane turns $\nabla^2G_D=-4\pi\delta$ into the 1-D problem $(k^2-\partial_z^2)g=\delta(z-z')$ per mode (§3.2 / Eq. 5.84). The piecewise ansatz respecting the boundary conditions is

with four conditions: continuity at $z'$; slope jump $g'(z'^-)-g'(z'^+)=1$; continuity at $d$; and $D_n$ continuity $g'(d^-)=\epsilon g'(d^+)$ (vacuum below, $\epsilon$ above). Eliminating $A,B,C,D$ (the code does it by linear algebra; by hand one propagates the wall solution $\sinh(kz)$ upward and the decaying solution downward, gluing at $z'$) yields exactly the quoted $f(z_>)$, $K(z')$ — the $e^{k(2d-z)}$ writing of the outer branch absorbs $e^{2kd}$ into the normalization so that $c\equiv\frac{1+\epsilon}{1-\epsilon}$ appears symmetrically. Notes: (i) $c<-1$ for $\epsilon>1$, so the denominator $1+ce^{2kd}$ never vanishes; (ii) the structure is a two-mirror ladder: expanding $[1+ce^{2kd}]^{-1}$ in powers of $c^{-1}e^{-2kd}$ regenerates the infinite image series between the conductor ($-1$ reflections) and the dielectric surface ($\beta$ reflections); (iii) both P40's force integral and P42's $\epsilon\to\infty$ plate problem are direct corollaries of this $g$.

The swapped problem solves $\vec\nabla\cdot[\epsilon_s(z)\vec\nabla G]=-4\pi\delta$ with $\epsilon_s=\epsilon$ for $z<d$ and $1$ for $z>d$. Divide the whole equation by the constant $\epsilon$:

But $\epsilon_s/\epsilon$ equals $1$ below $d$ and $1/\epsilon$ above — the original P17 medium pattern with dielectric constant $1/\epsilon$, and a source of strength $1/\epsilon$. By linearity and uniqueness, $G^{\rm swapped}(\epsilon)=\frac1\epsilon G^{\rm P17}(1/\epsilon)$. Only the ratio of permittivities across an interface (and the value at the source) matters. Sanity limits: $\epsilon\to\infty$ makes the swapped $G\to0$ (charge frozen inside a conductor-like slab); $\epsilon\to1$ recovers P17's $\epsilon\to1$ conductor-only kernel; and the $\beta$ coefficient maps as $\beta(1/\epsilon)=-\beta(\epsilon)$ — reflections off the interface flip sign, as they must when approaching from the dense side.

Piecewise-linear ansatz (1-D harmonic functions are lines): $G=\alpha x$ for $x<d$ (wall at 0), $G=\gamma+\beta x$ for $d<x<x'$, $G=B(L-x)$ for $x>x'$ (wall at $L$). Conditions: (i) continuity at $d$; (ii) $D$ continuity at $d$: $\epsilon\,\alpha=\beta$ (the 1-D $D=-\epsilon\,dG/dx$); (iii) continuity at $x'$; (iv) unit source in vacuum: $G'(x'^-)-G'(x'^+)=1$, i.e. $\beta+B=1$. From (i)–(ii): $G=\alpha d+\epsilon\alpha(x-d)$ in the middle. From (iii): $\alpha[d+\epsilon(x'-d)]=B(L-x')$. Substituting $B=1-\epsilon\alpha$ and solving,

(The Answer block's second expression for $B$ simplifies to this; the code uses these forms.) Limits: $\epsilon=1$: $\alpha=(L-x')/L$, $B=x'/L$ — exactly (2.136); $d\to0$: same. Interpretation: the slab acts like a series capacitor — the "electrical thickness" of the dielectric layer is $d/\epsilon$ when referred to the vacuum side (rewrite $\Delta/\epsilon=d/\epsilon+(L-d)$, the denominator of P43), so the charge's image weight shifts exactly as the reduced distance dictates.

Ansatz: in the charge's region use the source (screened by $\epsilon_2$) plus one image at the mirror point; in the far region a single effective charge at the true source position:

Both are harmonic in their half-spaces (images live outside), and $\nabla\cdot(\epsilon\nabla\Phi)$ has exactly the $-4\pi q\delta$ source at $\vec x'$. On $z=0$, $R_1=R_2\equiv R$, so: continuity: $\frac1{\epsilon_2}(q+q')=\frac{q''}{\epsilon_1}$; normal $D$: $\partial_z(1/R_1)|_0=+z'/R^3$ and $\partial_z(1/R_2)|_0=-z'/R^3$, so $D_z$ continuity ($\epsilon_2\partial_z\Phi_2=\epsilon_1\partial_z\Phi_1$) gives $q-q'=q''$. Solving the pair,

Limits: $\epsilon_1=\epsilon_2=\epsilon$: $q'=0$, $\Phi=q/(\epsilon R_1)$ everywhere ✓; $\epsilon_1\to\infty$: $q'=-q$ — the grounded conductor; $\epsilon_2=1,\epsilon_1=\epsilon$: the book's (5.106)–(5.107). Sign physics: for $\epsilon_1>\epsilon_2$ the image $q'$ is negative — the charge is attracted toward the optically denser medium (bound charge of opposite sign accumulates at the interface, cf. (5.111) and P36).

Each region is source-free (except the delta at $z'$), so $g''=k^2g$ with solutions $e^{\pm kz}$; boundedness kills the growing exponential in the two outer regions (leaving $A$ and $F$), while both exponentials survive in the bounded regions ($B,C$ and $D,E$): six constants. The matching conditions are the standard four interface equations (continuity + $D_n$ at $z=0$ and $z=-d$) plus the two source conditions at $z'$ (continuity + jump $g'(z'^-)-g'(z'^+)=1$). Six linear equations, six unknowns; solvability: the associated homogeneous problem (no source) admits only $g\equiv0$ (a nontrivial bounded solution would violate uniqueness of the Dirichlet problem at infinity), so the determinant is nonzero and the inhomogeneous system has exactly one solution — "enough information." Physically the solved kernel (code) shows the two-interface ladder: the charge sees an image train with alternating reflection coefficients $\beta$ at $z=0$ and $-\beta$-weighted internal reflections between $0$ and $-d$, summing to a geometric series in $\beta^2e^{-2kd}$; at $kd\gg1$ single reflection dominates → half-space; at $\epsilon\to1$, $\beta\to0$ → free space.

In 2-D the Green equation is $\vec\nabla\cdot(\epsilon\vec\nabla G)=-4\pi\delta^2$, free solution $-2\ln(\rho/K)$ (Eq. 3.11; $K$ is the line-charge length scale — physics is $K$-independent). Expand $G$ and $\delta(\phi-\phi')=\frac1{2\pi}[1+2\sum_m\cos m(\phi-\phi')]$ in azimuthal modes; with the book's prefactor $4$, the mode problems are

$m=0$: inside the dielectric $g_0=$ const (regular); $D$ continuity forces the middle region's log coefficient to vanish too, so $g_0$ is constant out to $\rho'$ and $-\tfrac12\ln(\rho/K)$ beyond — i.e. $g_0=-\tfrac12\ln(\rho_>/K)$, independent of $\epsilon$: a neutral dielectric leaves no monopole image (total bound charge zero, P12). $m\ge1$: $g_m=A\rho^m$ ($\rho<a$), $B\rho^m+C\rho^{-m}$ ($a<\rho<\rho'$), $D\rho^{-m}$ ($\rho>\rho'$). Interface at $a$: continuity $Aa^m=Ba^m+Ca^{-m}$ and $D_\rho$: $\epsilon mAa^{m-1}=mBa^{m-1}-mCa^{-m-1}$ give $C=\beta a^{2m}B$, $A=(1+\beta)B$ with $\beta=(1-\epsilon)/(1+\epsilon)$; the jump at $\rho'$ fixes $B=\rho'^{-m}/2m$, $D=B[\rho'^{2m}+\beta a^{2m}]$. Assembling gives exactly the quoted branches: outside, the induced part is the source's $m$-th image weight $\beta(a^2/\rho\rho')^m$ — the dielectric analog of the line-charge-in-cylinder image of §3.3, with strength $\beta$ instead of $-1$; inside, the transmitted mode $(1+\beta)/2m=(1/m(1+\epsilon))$. At $\epsilon=1$ the sum telescopes to $-2\ln\rho_>+2\sum_m\frac1m(\rho_</\rho_>)^m\cos m\Delta\phi =-2\ln|\vec x_\perp-\vec x'_\perp|$ ✓. (P44 sums the coincidence limit into the force on the line charge.)

Insert the expansions into $\vec\nabla\cdot(\epsilon\vec\nabla G)=-4\pi\delta$: each $(m,k)$ mode obeys the modified Bessel equation

whose homogeneous solutions are $I_m(k\rho)$ (regular at 0, grows) and $K_m(k\rho)$ (decays, singular at 0) — "Bessel functions of imaginary argument". Regularity on the axis keeps only $I_m$ inside the dielectric; decay at $\rho\to\infty$ keeps only $K_m$ outside the source; both survive in the annulus: four constants. The four conditions above are two interface conditions at the dielectric surface (tangential $E$ ↔ continuity of $g$; normal $D$ ↔ the $\epsilon$-weighted slope) and the two source conditions (the jump normalization follows by integrating the mode equation across $\rho'$). Uniqueness: the homogeneous system would need a bounded solution of the modified Bessel problem vanishing in all three regions' boundary senses; none exists (the Wronskian $I_mK_m'-I_m'K_m=-1/x$ never vanishes), so the $4\times4$ determinant is nonzero: the coefficients are determined — "enough information," as the exercise asks. The code's $\epsilon=1$ cross-check $G\to1/|\vec x-\vec x'|$ validates the normalization ($g_m^{\rm free} =I_m(k\rho_<)K_m(k\rho_>)$ by the Wronskian, exactly the classic identity).

Put $q=E_0R^2$ at $z=-R$ ($R\to\infty$): near the origin its field is $E_0\hat z$ with corrections $O(r/R)$, and its potential is $q/R+E_0z+\dots$ — after dropping the constant, the applied $-E_0z$ (sign per the geometry: the code handles the bookkeeping; here quote the standard orientation). In the exterior Green function (5.141) each induced term scales as $q\,a^{2\ell+1}/(rR)^{\ell+1}=E_0R^{1-\ell}(\cdots)$: only $\ell=1$ survives $R\to\infty$,

the field of the induced dipole $\vec p=\frac{\epsilon-1}{\epsilon+2}a^3E_0\hat z$ (Eq. 5.144). Likewise the interior kernel (5.140) keeps only $\ell=0$ (constant, dropped) and $\ell=1$: $q\cdot\frac{3}{\epsilon+2}\frac{r}{R^2}P_1\to\frac3{\epsilon+2}E_0r\cos\theta$ with the sign arranging to $\Phi_{\rm in}=-\frac{3}{\epsilon+2}E_0z$ — a uniform interior field $\vec E_{\rm in}=\frac3{\epsilon+2}E_0\hat z$ (5.146), reduced by the spherical depolarization factor $\frac13$: $E_{\rm in}=E_0/[1+\frac{\epsilon-1}3]$. Boundary sanity: $\Phi$ continuous at $r=a$ ($-E_0a+p/a^2$ vs $-3E_0a/(\epsilon+2)$ — equal by the $p$ value) and $\epsilon\partial_r\Phi_{\rm in}=\partial_r\Phi_{\rm out}$ ✓ (code). This is the sphere twin of P13's cylinder ($\frac2{\epsilon+1}$ vs $\frac3{\epsilon+2}$).

Radial ansatz per $\ell$ (source in the dielectric): $g_\ell=Ar^\ell$ ($r<r'$), $Br^\ell+Cr^{-\ell-1}$ ($r'<r<a$), $Dr^{-\ell-1}$ ($r>a$). Source conditions at $r'$ (in medium $\epsilon$): continuity and $\epsilon r'^2[g'(-)-g'(+)]=1\cdot r'^0\Rightarrow g'(-)-g'(+)=1/(\epsilon r'^2)$, giving directly $C=r'^\ell/[\epsilon(2\ell+1)]$ — the Coulomb-in-medium direct part. Interface at $a$: continuity and $\epsilon g'(a^-)=g'(a^+)$ give

and $A=B+C\,r'^{-2\ell-1}$. The $D$ coefficient is exactly the $(r\leftrightarrow r')$ transpose of (5.140) — the hinted symmetry; $B$ is the interior "reaction" kernel (the engine of P48's force). Far field: $\ell=0$: $D_0=1$ → $\Phi\to1/r$: at large $r$ the full unit charge is visible (the interior screening cloud $\rho_b=(1-\epsilon)/\epsilon$ at the charge is exactly cancelled by the surface bound charge, P12). $\ell=1$: $(2\cdot1+1)g_1=3r'/[(\epsilon+2)]\cdot r^{-2}P_1$ → $\vec p_{\rm eff}=3\vec x'/(\epsilon+2)$: the displacement of the charge off center is what the outside world sees, screened by the same $3/(\epsilon+2)$ as P24 (reciprocity: source inside ↔ observer inside). Total surface charge: $\sigma_b=\frac1{4\pi}[E_r(a^+)-E_r(a^-)]$; only $\ell=0$ survives the angular integral: $E_r(a^+)=1/a^2$, $E_r(a^-)=1/(\epsilon a^2)$, so $\oint\sigma_b\,da=1-\frac1\epsilon=\frac{\epsilon-1}\epsilon$ ✓ — P12(b) from the explicit solution, independent of where the charge sits inside.

The bubble problem solves $\vec\nabla\cdot[\epsilon_b(r)\vec\nabla G]=-4\pi\delta$ with $\epsilon_b=1$ ($r<a$), $\epsilon$ ($r>a$). Divide by $\epsilon$ (P18's trick): the medium pattern becomes $1/\epsilon$ inside, $1$ outside — a "dielectric sphere" of constant $1/\epsilon$ in vacuum with source strength $1/\epsilon$. Uniqueness gives the substitution rule. Explicitly, with $\epsilon'=1/\epsilon$ in (5.141): $(\epsilon'-1)\ell/[\ell(1+\epsilon')+1] =(1-\epsilon)\ell/[\ell(1+\epsilon)+\epsilon]$, whose sign flip converts the sphere's attractive image series into the bubble's repulsive one (P46): a cavity in a dielectric behaves like a region of deficit polarizability — an "anti-sphere" with effective $\ell=1$ polarizability $-\frac{\epsilon-1}{1+2\epsilon}a^3$ (per unit medium-screened source). The $\ell$-dependence of the denominators means there is again no single image point (only $\epsilon\to\infty$ collapses the sphere series to one; the bubble has no conducting limit).

Write the dipole as the limit of charges $\pm q$ at $\vec x'\pm\vec\delta/2$ and superpose the one-charge Green function. (a) Far away, each source charge contributes its own effective system: the bare charge (monopoles cancel between $\pm q$, leaving $\vec p_0$) and its sphere-induced $\ell=1$ image. The induced dipole responds linearly to the field at the sphere's center, so in the limit $\vec p_{\rm ind}=\alpha\vec E_{\rm dip}(0)$ with $\alpha=\frac{\epsilon-1}{\epsilon+2}a^3$ (5.143–5.144). The two dipoles do not sit at the same point ($\vec p_0$ at $\vec x'$, $\vec p_{\rm ind}$ at the origin), but the displacement only affects the $\ell=2$ (quadrupole) term: to leading order $\Phi\simeq(\vec p_0+\vec p_{\rm ind})\cdot\vec x/r^3$. (b) For $r'\gg a$ the dipole's field is uniform across the sphere, $\vec E_{\rm dip}(0)+O(a/r')$, so P24's screening applies verbatim: $\vec E_{\rm in}=\frac3{\epsilon+2}\vec E_{\rm dip}(0)$. From the kernel: the $\ell=0$ term of (5.140) contributes only a constant ($p_0$-projected $-\vec p_0\cdot\hat n'/r'^2$), and the $\ell=1$ term differentiates to exactly the uniform interior gradient — the code confirms by finite-differencing the series. Everything here is the polarizability algebra reused in P49's force.

Regularity at the origin keeps only $r^\ell$ in the dielectric; between $a$ and $b$ both radial solutions appear, split at the source radius into two two-parameter branches. Counting: $1+2+2=5$ constants. The five stated conditions are all of them: the two dielectric-interface conditions (tangential $E$, normal $D$), the two source conditions (the jump follows by integrating $-(r^2g')'+\ell(\ell+1)g=\delta(r-r')$ across $r'$), and the grounded conductor $g_\ell(b)=0$ ($\Phi=0$ on the wall for every mode). The homogeneous problem has no nontrivial solution — a source-free potential regular at 0, matching at $a$, vanishing at $b$ must vanish identically (uniqueness inside a grounded shell) — so the determinant is nonzero and the system pins all five coefficients: "enough information." The $\epsilon=1$ limit collapses $A=B$, $C=0$ and reproduces the image-in-sphere kernel; $\epsilon\to\infty$ freezes the core ($g(a)\to0$), giving the conductor-in-conductor annulus kernel.

With $\rho=\sigma\,\delta(z'-d)$ and the interior kernel (5.140),

(azimuthal symmetry kills $m\neq0$). Parametrize the plane by $r'\in[d,\infty)$ with $\cos\theta'=d/r'$, $da'=2\pi r'dr'$:

For $\ell\ge2$, $P_\ell(t)t^{\ell-2}$ has even parity, so $\int_0^1=\frac12\int_{-1}^1P_\ell(t)\,t^{\ell-2}\,dt=0$ by Legendre orthogonality ($t^{\ell-2}$ has degree $<\ell$): every higher multipole of an infinite uniform plane vanishes. $\ell=1$: $\int_0^1P_1t^{-1}dt=1$, so $I_1=2\pi$; $\ell=0$ diverges — the familiar infinite constant of an infinite plane's potential (gauge; drop it). Hence $\Phi_{\rm in}=\text{const}+\frac{3}{\epsilon+2}2\pi\sigma\,r\cos\theta$ and the interior field is uniform: magnitude $\frac3{\epsilon+2}\times2\pi\sigma$, directed away from the plane (for $\sigma>0$) — the plane's field $|E|=2\pi\sigma$ screened by P24's factor. The Green-function route thus derives rather than assumes the equivalence to the uniform-field problem.

Superpose the interior kernel (5.140) over the ring. For a source at $(b,\theta'=\pi/2,\phi')$ the addition theorem reduces the azimuthal average of $P_\ell(\cos\gamma)$ to $P_\ell(\cos\theta)P_\ell(\cos\theta')=P_\ell(\cos\theta)P_\ell(0)$ ($m\neq0$ terms integrate away around the uniform ring). Hence

Ex. 4.9.1(b): $P_\ell(0)=0$ for odd $\ell$ and $P_{2n}(0)=(-1)^n\frac{(2n-1)!!}{(2n)!!}$; keeping $\ell=2n$, $2\ell+1\to4n+1$ and $\ell(1+\varepsilon)+1\to2n(1+\varepsilon)+1$ gives the quoted series verbatim. Structure worth noting: at $n=0$ the term is $Q/b$ — the ring's constant potential leaks in unscreened (constants feel no dielectric); the first nontrivial term is $n=1$ with the $\frac5{2(1+\varepsilon)\cdot?}$-type screened $P_2$ — the sphere reduces each even harmonic by $\frac{4n+1}{2n(1+\varepsilon)+1}$ relative to vacuum's $1$, the $\ell$-dependent generalization of $3/(\epsilon+2)$.

Let the shells carry potentials $V_1$ (upper), $V_2$ (lower); the surface value is the step function $V(\theta)$ with Legendre projection

using $\int_{-1}^0P_\ell=(-1)^\ell I_\ell$. The potential is $\sum V_\ell(r/a)^\ell P_\ell$ inside (dielectric) and $\sum V_\ell(a/r)^{\ell+1}P_\ell$ outside — continuity at $r=a$ is built in (the shell fixes $\Phi$ there). The free charge on the shell is the $D$-jump:

the exterior contributes $(\ell+1)$, the dielectric interior $\varepsilon\ell$. Integrate over the upper hemisphere ($\int_{\rm up}P_\ell\,da=2\pi a^2I_\ell$):

Then $C_{11}=\partial Q_1/\partial V_1$ and $C_{12}=\partial Q_1/\partial V_2$ give

($+$ for $j{=}1$, $(-1)^\ell$ for $j{=}2$). Since $(\ell+1)+\varepsilon\ell=(2\ell+1)+(\varepsilon-1)\ell$, this is the vacuum $(2\ell+1)^2$ sum plus the stated $(\varepsilon-1)\ell(2\ell+1)I_\ell^2$ shift. Only odd $\ell$ contribute ($I_{2n}=0$ for $n\ge1$), so the $C_{12}$ shift is negative: the dielectric strengthens the inductive coupling. The symmetric-system capacitance $C=\frac12(C_{11}-C_{12})$ then grows by $\frac a4(\varepsilon-1)\sum_{\rm odd}\ell(2\ell+1)I_\ell^2$ — dielectric filling raises $C$, as it must.

(a) The mode count: regularity at the axis (1 constant in the core), two 2-parameter annuli split at the source — five per $m$, matching the five conditions listed (two interface, two source, one wall). For $m=0$ the radial solutions are $\{1,\ln\rho\}$; $D$-continuity at $b$ with a regular core forces the middle log coefficient to vanish (as in P22), and the wall then fixes the outer branch to $-\frac12\ln(\rho/a)$-type — the $m=0$ mode is again $\epsilon$-blind. Uniqueness of the grounded-cavity Dirichlet problem makes the homogeneous determinant nonzero: "enough information." (b) The dielectric Green function is symmetric (5.78), so no new solve is needed: for $\rho'<b$ evaluate the (a) coefficients at the swapped arguments; concretely the $\rho<b$ branch $A(\rho')\rho^m$ of (a) becomes, read as a function of its second argument, the $b<\rho<a$ field of an interior source. (c) Divide the PDE by $\epsilon$ exactly as in P18/P26: media $(\epsilon\ \text{core},1\ \text{gap})\to(1\ \text{core},1/\epsilon\ \text{gap})$ with source weight $1/\epsilon$ — the swapped problem with $\epsilon\to1/\epsilon$, overall factor $1/\epsilon$. (The wall condition $g(a)=0$ is $\epsilon$-independent and survives both maps.)

The $\ell=2$ induced term of the exterior Green function (5.141) is

(coefficient $(\epsilon-1)\ell/[\ell(1+\epsilon)+1]$ at $\ell=2$; note $2(1+\epsilon)+1=3+2\epsilon$). Compare with the quadrupole potential $\frac12\sum Q_{ij}x_ix_j/r^5$: for an axisymmetric tensor about $\hat x'$, $\frac12\sum Q_{ij}x_ix_j/r^5=\frac{Q_{33}}{2}P_2(\cos\gamma)/r^3$ (P5's algebra with the 3-axis along $\hat x'$), so

Meanwhile the gradient of the charge's field at the origin is

an axisymmetric traceless tensor whose 33-component (along $\hat x'$) is $2/r'^3$. Hence $Q_{ij}=\frac{2a^5(\epsilon-1)}{3+2\epsilon}\big(-\partial E'_j/\partial x'_i\big)$ reproduces $Q_{33}$ above and, being proportional to the traceless symmetric $3\hat n\hat n-\mathbb 1$, all other components — the tensor identity holds, with $a^5$ (a "quadrupole polarizability" $\propto a^5$, just as the dipole's is $\propto a^3$; each extra $\ell$ costs $a^2$). Since $Q_{ij}\propto\partial E'_j/\partial x'_i$, a spatially uniform applied field (zero gradient) induces no quadrupole — the book's parenthetical remark. The code's bound-charge quadrature ($\int(3z^2-r^2)\sigma_b\,da$ over the exact series solution) confirms coefficient and scaling; $a^3$ would be off by $a^2$ and dimensionally inconsistent ($Q\sim$ charge·length², and $\partial E'/\partial x'\sim$ charge/length³ leaves length⁵ to supply).

(a) With no free charge $\vec\nabla\cdot\vec D=0$ (5.65), and statics gives $\vec E=-\vec\nabla\Phi$:

the surface term dying because $\Phi\sim1/r$ ($1/r^2$ for neutral bodies) and $D\sim1/r^3$ while the sphere area grows only as $r^2$. (b) Summing the pairwise energies over dipole pairs and passing to the continuum, the interaction energy of the polarized matter is $W^{\rm int}=-\frac12\int\vec P\cdot\vec E$ — the $\frac12$ correcting the double count of pairs, with $\vec E$ the total (self-consistent) field. Now eliminate $\vec P$ via $4\pi\vec P=\vec D-\vec E$ and apply (a):

(The book's comment: the continuum limit quietly adds self-energy pieces — the $i=j$ terms excluded from the sum reappear as the local-field part of the integral; that is precisely why $W^{\rm int}$ ends up expressible as the manifestly positive total field energy.) Worked instance (code): uniform $\vec P_0$ sphere: $\vec E_{\rm in}=-\frac{4\pi}3\vec P_0$, $\vec D_{\rm in}=+\frac{8\pi}3\vec P_0$ — anti-parallel $E$, $D$! — and the exterior is the pure dipole $p=\frac{4\pi}3a^3P_0$: $\int_{\rm in}\vec E\cdot\vec D=-\frac{32\pi^2}9P_0^2V$ exactly cancels $\int_{\rm out}E^2=+\frac{8\pi}3p^2/a^3$; both $W^{\rm int}$ forms give $\frac{2\pi}3P_0^2V$.

(a) The force on the surface layer is (charge)×(average field), the §2.11/§5.9 average-field rule (5.185): the layer cannot push on itself, so the acting field is $\frac12(\vec E_1+\vec E_2)$. The total surface charge (free+bound) is $4\pi\sigma_{\rm tot}=E_{2n}-E_{1n}$ (tangential components are continuous and contribute no normal force):

which is also the jump of the Maxwell normal stress $T_{nn}=E_n^2/8\pi$ (plus continuous tangential pieces that cancel in the difference). (b) Only bound charge: $D_n$ continuous, $E_{2n}=\epsilon E_{1n}$, and $4\pi\sigma_b=E_{2n}-E_{1n}=(\epsilon-1)E_{1n}$:

Extra: now $E_{2n}-E_{1n}=4\pi(\sigma_f+\sigma_b)$ and $E_{2n}-\epsilon E_{1n}=4\pi\sigma_f$; solving, $E_{1n}=4\pi\sigma_b/(\epsilon-1)$, $E_{2n}=4\pi[\sigma_f+\sigma_b\epsilon/(\epsilon-1)]$, so

identical to the printed form (note $(1-\varepsilon)^2=(\varepsilon-1)^2$). Factored, $2\pi(\sigma_f+\sigma_b)[\sigma_f+\sigma_b(\epsilon+1)/(\epsilon-1)]$: total charge times average field, as it must be.

(a) The vacuum-side solution is charge $+$ image $q'=-\frac{\epsilon-1}{\epsilon+1}$ at $-z'$ (P20 with $\epsilon_1=\epsilon,\epsilon_2=1$). The force on the charge is $qq'/(2z')^2=-\frac{\epsilon-1}{\epsilon+1}\frac1{4z'^2}$ (toward the surface). By Newton's third law the dielectric (the physical carrier of the bound charge that the image summarizes) feels the equal and opposite pull $+\frac{\epsilon-1}{\epsilon+1}\frac1{4z'^2}$ toward the charge. (b) The coincidence limit of the induced part: $G_D-G_D^0=q'/R_2\to q'/(2z')$ at the charge, so

matching (a) (the $\frac12$ of $\Delta W$ against the full image force is exactly compensated by the $z'$-dependence of the image distance — the same bookkeeping as the conductor's $W=-q^2/4z'$). (c) From (5.111)/P20, $\sigma_b(\rho)=-\frac1{2\pi}\frac{\epsilon-1}{\epsilon+1} \frac{z'}{(\rho^2+z'^2)^{3/2}}$. P35(b):

Same number, and now with a location: half the force accumulates inside $\rho\lesssim0.64\,z'$ — the stress picture shows where the pull acts, which the energy methods cannot (the book's §5.9 moral).

The rod sees the charge's field $E_0=1/z^2$ transverse to its axis. P13: a cylinder in a transverse field carries the uniform interior field $E_{\rm in}=\frac{2E_0}{\epsilon+1}$, so its polarization is $\vec P=\frac{\epsilon-1}{4\pi}\vec E_{\rm in} =\frac{\epsilon-1}{2\pi(\epsilon+1)}\vec E_0$ and the total induced moment

An induced dipole in a nonuniform field has energy $W=-\frac12\alpha E_0^2(z)=-\frac{\alpha_\perp}{2z^4}$ (the $\frac12$ because the moment builds with the field), hence

negative = toward the charge: attraction for $\epsilon>1$ (high-$\epsilon$ matter is always drawn toward stronger field — the $\nabla E^2$ force). The $z^{-5}$ law is generic for induced dipoles ($F\propto\partial_zE_0^2$).

For a needle along the field, the interface condition is continuity of tangential $E$ along the (dominant) lateral surface: the interior field equals the applied one, $E_{\rm in}\simeq E_0$, up to end corrections of order $(a/L)^2\ln$-small — equivalently, the prolate depolarization factor $n_z\to0$ as the aspect ratio diverges ($E_{\rm in}=E_0/[1+n_z(\epsilon-1)]$). Then

and the induced-dipole energy/force machinery of P37 gives $W=-\alpha_\parallel/2z^4$, $F_z=-2\alpha_\parallel/z^5=-\frac{(\epsilon-1)a^2L}{2z^5}$: attraction. Comparing polarizabilities: $\alpha_\parallel/\alpha_\perp=(\epsilon+1)/2>1$ — no depolarization along the needle versus the factor $\frac2{\epsilon+1}$ broadside. (A free rod therefore also feels a torque aligning it with the local field before it is pulled in — needle first.)

The two gap regions are capacitors in parallel: $C(x)=\frac{W}{4\pi D}[\epsilon x+(L-x)]$, $dC/dx=\frac{(\epsilon-1)W}{4\pi D}$. (a) Fixed $V$ (battery attached): by (5.179) the force is $+\partial W_{\rm field}/\partial x|_V$ with $W_{\rm field}=\frac12CV^2$ (the battery supplies $VdQ=V^2dC$, twice the field-energy gain, so the net system energy falls): $F_x=\frac{V^2}2\frac{dC}{dx}=\frac{(\epsilon-1)WV^2}{8\pi D}>0$: sucked in, uniformly (edge effects neglected). (b) Fixed $Q=\frac{LWV}{4\pi D}$ (the vacuum-capacitor charge at voltage $V$): $W_{\rm field}=Q^2/2C(x)$ and $F_x=-\frac{\partial}{\partial x}\frac{Q^2}{2C} =\frac{Q^2}{2C^2}\frac{dC}{dx} =\frac{(\epsilon-1)W}{8\pi D}\Big[\frac{4\pi DQ}{W\,[L+(\epsilon-1)x]}\Big]^2 \cdot\frac{W^2}{(4\pi D)^2}\cdot\ldots =\frac{(\epsilon-1)WV^2L^2}{8\pi D\,[L+(\epsilon-1)x]^2},$$ i.e. the same formula with $V\to V_{\rm inst}=Q/C(x)=VL/[L+(\epsilon-1)x]$. Both are positive: dielectrics are always drawn into stronger fields, whatever is held fixed. At the same instantaneous state ($x$, same plate charge) the forces are identical — the sign flip between $-\partial_x W|_Q$ and $+\partial_x W|_V$ is exactly the battery's ledger (5.174)–(5.179), not a physical difference; that is the soapbox point.

The dielectric's position enters only through $d$, so at fixed charge the force on it is $F_z=-\partial\Delta W/\partial d$ with

($g^0=\sinh(kz')e^{-kz'}/k$ is the conductor-only kernel; the transverse $k$-integral of $4\pi\int\frac{d^2k}{(2\pi)^2}$ at coincidence is $2\int_0^\infty k\,dk$). From P17's closed form at $z=z'$, with $c=\frac{1+\epsilon}{1-\epsilon}$:

(the $c\,e^{2kd}e^{-kz'}$ pieces cancel between numerator and the $-e^{-kz'}$ term, leaving $e^{kz'}-e^{-kz'}$). Only the denominator carries $d$, so per mode

Multiplying by $k\,dk$ and integrating gives the boxed $F_z$. Sign: $c<-1$ for $\epsilon>1$ makes the integrand negative: $F_z<0$, the slab is pulled down toward the charge — consistent with P36's half-space limit ($d\to z'^+$ dominated by $k\sim1/(d-z')$) and with the $\epsilon\to\infty$ conductor limit of P42.

(a) With $\beta=\frac{1-\epsilon}{1+\epsilon}$ small, $c=1/\beta$ and $[1+ce^{2kd}]^{-2}\simeq\beta^2e^{-4kd}(1-2\beta e^{-2kd}+\dots)$, so to first order

(using $\sinh^2=\frac14(e^{2kz'}+e^{-2kz'}-2)$), and $\int_0^\infty k\,e^{-2ks}dk=1/4s^2$ gives the quoted bracket. (b) To first order in $\beta$ the dielectric surface reflects each vacuum source once: sources are the charge $+1$ at $z'$ and its conductor image $-1$ at $-z'$; their dielectric images are $\beta$ at $2d-z'$ and $-\beta$ at $2d+z'$. The force on the dielectric is minus the total force its response exerts... i.e. $-\sum$(forces on the two vacuum sources from the two $\beta$-images):

(each pair separated by twice its distance to the plane $z=d$), so

For $\epsilon>1$ ($\beta<0$) the bracket is positive (the $(d-z')^{-2}$ attraction to the near charge wins), so $F<0$: net pull toward the charge. The $-2/d^2$ piece is the repulsion bookkeeping between the dielectric's response and the conductor image — a genuinely three-body term that a naive "charge attracts dielectric" estimate misses.

(a) At $\epsilon\to\infty$ P17's kernel becomes the two-plate Dirichlet kernel $g=\sinh(kz_<)\sinh(k(d-z_>))/[k\sinh(kd)]$ ($g(d)=0$). The plate charge is $\sigma=\frac1{4\pi}\vec E\cdot\hat n$ with $\hat n=-\hat z$ (into the gap): $\sigma=-\frac1{4\pi}\partial_zG|_{z=d^-}\cdot(-1)^{\dots}$ — carefully: $\sigma=\frac1{4\pi}E_n=-\frac1{4\pi}E_z(d^-)\cdot\ldots$ with $E_z=-\partial_zG$: $\sigma=+\frac1{4\pi}\partial_zG|_{d^-}$. Using $G=2\int_0^\infty k\,dk\,J_0(k\rho)\,g$ (the $m$-sum done by Ex. 4.2.1) and $\partial_zg|_{z=d}=-\sinh(kz')/\sinh(kd)$:

(b) The plate is pulled with pressure $2\pi\sigma^2$ (§2.11) toward the gap:

Insert (a) and use the Hankel closure $\int_0^\infty\rho J_0(k\rho)J_0(k'\rho)d\rho=\delta(k-k')/k$:

Limit check: P40 with $c\to-1$: $4c\,e^{2kd}/[1+ce^{2kd}]^2\to-4e^{2kd}/(e^{2kd}-1)^2=-1/\sinh^2(kd)$ — exactly (b)'s integrand ✓. And integrating (a): $\int\sigma\,da=-z'/d$, the standard result that a charge between grounded plates deposits fractions $z'/d$ and $1-z'/d$ on the far and near plates (cf. the §3.2/Ex. 3.2.3 induced-charge computation).

Series capacitors: vacuum gap $(x-d)$ plus dielectric layer of "reduced thickness" $d/\epsilon$ (P19's lesson):

(a) Fixed $\Delta V$: $F=+\partial_x(\tfrac12C'\Delta V^2)|_V =\frac{\Delta V^2}2\,\frac{dC'}{dx}$ and $dC'/dx=-C'^2\cdot4\pi$:

attractive and growing as the gap closes. (b) Disconnecting at separation $L$ freezes $\sigma=\frac{\Delta V}{4\pi[\,d/\epsilon+(L-d)\,]}$. At fixed charge the plate feels the field of everything else, $F=-2\pi\sigma^2$ per area (equivalently $W/A=2\pi\sigma^2[\,d/\epsilon+(x-d)\,]$ and $F=-\partial_xW$): constant in $x$,

At $x=L$ (the disconnect instant) the two answers coincide, as they must — the force is a property of the instantaneous state; thereafter the fixed-$V$ force outgrows the fixed-$Q$ one as the plates approach because the battery keeps feeding charge.

At coincidence the direct parts cancel and only the induced (image-series) parts of P22's $g_m$ survive:

($m=0$ contributes nothing — P22's $\epsilon$-blind monopole mode; the $\sum x^m/m=-\ln(1-x)$ resummation is exact for $\rho'>a$). Then

$\beta<0$ for $\epsilon>1$: attraction, always. Limits: far field $F\to2\beta a^2/\rho'^3$ — the 2-D induced-dipole law ($W\sim-\alpha_{2D}E_0^2$ with $E_0=2/\rho'$, $\alpha_{2D}=\frac{\epsilon-1}{\epsilon+1}\frac{a^2}2$ per unit length, consistent with P13); near contact the image series piles up into the half-space result: $F\simeq-\frac{\epsilon-1}{\epsilon+1}/(\rho'-a)$ per... precisely $\beta/(\rho'-a)$ — the 2-D twin of P36 with the line-charge normalization. The $\epsilon\to\infty$ limit reproduces the conducting cylinder's image force.

(a) With $\vec E=-\vec\nabla\Phi$,

the surface term vanishing for localized fields (and on the confinement boundary $\vec D\cdot\hat n=0$ — the homogeneous Neumann condition of the flux tube — so $\delta\vec D\cdot\hat n=0$ there too). Fixing $\vec\nabla\cdot\delta\vec D=0$ is fixing $\delta\rho=0$: among all fields with the given free charge, the physical one extremizes this particular $W$ — which is why the $\alpha E^2$ addition to $\frac12\vec E\cdot\vec D$ is the right energy functional here (for the nonlinear $\epsilon(E^2)$, $\frac1{8\pi}\int\vec E\cdot\vec D$ alone is not a perfect integral of $\vec E\cdot\delta\vec D$, cf. (5.162)–(5.163)). (b) Write $u=E^2$ and the energy density $4\pi w=\frac12\epsilon(u)u+\alpha u=\alpha[u\ln(u/K^2)+u]\equiv\alpha f(u)$, with $f'=\ln(u/K^2)+2$, $f''=1/u$. Under $\vec E\to\vec E+\delta\vec E$, $\delta u=2\vec E\cdot\delta\vec E+|\delta\vec E|^2$, so collecting the second-order pieces,

In the confinement region the model has $E^2>K^2$, so $\ln(E^2/K^2)>0$ and both terms are positive for any $\delta\vec E\neq0$: $\delta W^{(2)}>0$, a strict local minimum. (This convexity is what makes the flux-tube solution stable and underwrites the linear-confinement bound (5.195)–(5.197): $W>\frac K{8\pi}\int|\vec D|\ge\frac12K|Q|(R-2r)$ — energy growing linearly with quark separation.)

Repeat (5.198)–(5.202) with P26's kernel, reference medium $G^0=1/(\epsilon|\vec x-\vec x'|)$:

whose leading $\ell=1$ term is the boxed $\Delta W$ (denominator $1+2\epsilon$). Equivalently by the P26 substitution rule applied to (5.202): $\Delta W_{\rm sphere}=-\frac{\epsilon-1}{\epsilon+2}\frac{a^3}{2r_0^4} \to\frac1\epsilon\Big[-\frac{\frac1\epsilon-1}{\frac1\epsilon+2}\Big]\frac{a^3}{2r_0^4} =+\frac{\epsilon-1}{\epsilon(1+2\epsilon)}\frac{a^3}{2r_0^4}$ — "what one has to do to (5.202)". Since $\Delta W>0$ increases as the charge approaches ($r_0\downarrow$), the force pushes the charge away: repulsion for $\epsilon>1$. Physics: the bubble is a hole in a polarizable medium — a region of polarizability deficit. Field energy in a dielectric is lowered by concentrating field in high-$\epsilon$ material; pushing the low-$\epsilon$ hole away from the strong-field region (equivalently the charge away from the hole) lowers the energy. This is the electrostatic version of why bubbles in a dielectric liquid migrate to weak-field regions (and its converse is P47's attraction of a dielectric sphere in vacuum) — same $1/r_0^5$ law, opposite sign, $\epsilon$-dependent weight.

For $r'\gg a$ the charge's field is uniform across the sphere, $\vec E_0\simeq\hat r'/r'^2$ in magnitude $E_0=1/r'^2$. Inside, (5.146): $\vec E_{\rm in}=\frac3{\epsilon+2}\vec E_0$, so the (uniform) polarization is

Then

which is precisely the $\ell=1$ (leading) term of the exact interaction energy (5.201) — the Green-function route and the polarization-energy route are term-for-term the same physics ($-\frac12\alpha E_0^2$ with $\alpha=\frac{\epsilon-1}{\epsilon+2}a^3$, the induced-dipole energy with the build-up factor $\frac12$). The force follows: $F_r=-\partial_{r'}\Delta W=-\frac{\epsilon-1}{\epsilon+2}\frac{2a^3}{r'^5}$, attractive for $\epsilon>1$, the $1/r'^5$ induced-dipole law (contrast the grounded conductor's $1/r'^3$: there an $\ell=0$ monopole image exists; the neutral dielectric starts at $\ell=1$ — the book's remark under (5.202)).

The self-interaction energy of the charge with the droplet's response is $\Delta W(r_0)=\frac12G_{\rm ind}(\vec r_0,\vec r_0)$, where $G_{\rm ind}$ is the regular (induced) part of P25's inside–inside kernel — the reference $G^0=1/(\varepsilon|\vec x-\vec x'|)$ (charge in unbounded fluid) subtracts the Coulomb-in-medium self-energy. From P25,

so, at coincidence ($P_\ell(1)=1$),

The $\ell=0$ term is an $r_0$-independent constant (the droplet's overall self-energy shift); differentiating term by term,

the quoted series. Every term is negative for $\varepsilon>1$: the charge is driven to the center. Physically, looking out from inside the dense medium, the interface reflection coefficient has the sign of $(\varepsilon-1)/(\varepsilon+1)>0$ — the image charge is of the same sign (P20 with the media swapped), so the wall repels the charge; the most distant wall point pulls least, and the stable point is the center (restoring force $\propto r_0$ for small $r_0$: $F_r\simeq-\frac{2(\varepsilon-1)}{\varepsilon(\varepsilon+2)}\frac{r_0}{a^3}$, a harmonic trap). This is the exact converse of the charge-outside-bubble repulsion of P46 — both say: the system pushes free charge into the high-$\varepsilon$ region and keeps it away from interfaces with lower $\varepsilon$.

The sphere responds as an induced dipole $\vec p_{\rm ind}=\alpha\vec E_p(0)$, $\alpha=\frac{\varepsilon-1}{\varepsilon+2}a^3$ (5.144), in the dipole's field at the center. For the dipole at $\vec x_d$ ($|\vec x_d|=d$),

and the induced-response interaction energy (the $\frac12$ of P47) is

Force on the dipole: $\vec F=-\vec\nabla_{x_d}W$. Differentiate:

and at $\vec x_d=d\hat z$ ($\vec p\cdot\vec x_d=p_3d$):

which is exactly the boxed form (note the double negative for $i=1,2$). Reading it: the radial component is always attractive ($3p_3^2+p^2>0$) with the strongest pull when $\vec p$ is radial ($|E_p(0)|^2=4p^2/d^6$) and weakest broadside ($p^2/d^6$) — factor 4, the induced-dipole anisotropy; the transverse force $\propto p_3p_{1,2}$ torques the trajectory unless $\vec p$ is purely radial or purely transverse (those are the symmetry axes where $\hat x_d$ is a principal direction of $\vec p\otimes\vec p$). The $d^{-7}$ law completes the module's force ladder: charge–sphere $r^{-5}$ (P47), dipole–sphere $d^{-7}$ — each source derivative costs $d^{-1}$ twice in $W=-\frac\alpha2E^2$.

Only the immersed portion feels a force, so from (6.10)

Since $\vec B$ is one constant vector throughout the region, it factors out of the integral:

because $\int d\vec x$ along any path is just the endpoint displacement — here from the entry point (current entering the field region) to the exit point. All the wiggles of the immersed arc cancel telescopically. With $\vec B$ perpendicular to the circuit plane and $\vec L$ in the plane, $\vec F$ lies in the plane, perpendicular to the chord, with magnitude $ILB/c$. (Had the whole circuit been immersed, $\vec L=0$ — a closed loop in a uniform field feels no net force, only a torque.)

The current density is $J=I/\pi a^2$. Ampère's law (6.36) on a coaxial circle: $2\pi\rho B_\phi=\frac{4\pi}{c}I_{\rm enc}$ with $I_{\rm enc}=I\rho^2/a^2$ inside, $I$ outside, so

The force density is $\vec f=\frac1c\vec J\times\vec B =\frac1c J\hat z\times B_\phi\hat e_\phi=-\frac{JB_\phi}{c}\hat\rho$ — radially inward. A shell $(\rho,\rho+d\rho)$ of unit length feels $dF_\rho=-\frac{JB_\phi}{c}\,2\pi\rho\,d\rho$, and with $A(\rho)=2\pi\rho$ per unit length the book's relation gives

Each shell's inward squeeze is borne by the material beneath it, so integrating in from the free surface ($P(a)=0$):

and $P\equiv0$ for $\rho>a$ (no current, no body force). The pressure is compressive (inward): parallel current filaments attract (§6.1) — this is the classic $z$-pinch profile.

The moving surface charge is the superposition of two surface currents: rotation gives $K_\phi=\sigma v_\phi=\sigma\omega R$ (a solenoid sheet), translation gives $K_z=\sigma v_0$ (an axial sheet). Solenoid part. By symmetry $\vec B=B_z(\rho)\hat z$; $B_z(\infty)=0$ and a rectangular Ampère loop straddling the sheet with legs parallel to $z$ picks up only the sheet current: $B_z^{\rm in}-B_z^{\rm out}=\frac{4\pi}{c}K_\phi$, while a loop entirely inside (or outside) shows $B_z$ is constant in each region. Hence $B_z^{\rm out}=0$ and $B_z^{\rm in}=\frac{4\pi}{c}\sigma\omega R$. Wire part. By symmetry $\vec B=B_\phi(\rho)\hat e_\phi$; a coaxial circular Ampère loop encloses $I_{\rm enc}=K_z\cdot2\pi R=2\pi R\sigma v_0$ when $\rho>R$ and nothing when $\rho<R$:

Superpose: uniform axial field inside, azimuthal wire field outside. (Each part also satisfies the jump condition (6.43) across the sheet.)

Think of the sheet as a stack of parallel wires along $z$ at $(0,y')$, each carrying $dI=K_0\,dy'$. The wire at $y'$ produces $d\vec B=\frac{2\,dI}{c\,d}\hat e_\phi$ with $d=\sqrt{x^2+(y-y')^2}$. By the $y'\to2y-y'$ mirror symmetry the components along $\hat x$ cancel pairwise and the $\hat y$ components add:

So $\vec B=\frac{2\pi K_0}{c}\mathrm{sign}(x)\,\hat y$: uniform on each side (the sheet has no length scale), tangential, and odd in $x$. Consistency with (6.43), $\hat n=\hat x$: $\hat x\times[\vec B(0^+)-\vec B(0^-)]=\hat x\times\frac{4\pi K_0}{c}\hat y =\frac{4\pi K_0}{c}\hat z=\frac{4\pi}{c}\vec K$ ✓. (This is the magnetostatic twin of the charged sheet's $\pm2\pi\sigma$ field, cf. §6.4's $\pm\frac{2\pi}{c}K$ disk argument (6.48)–(6.49).)

Let $\hat c$ be an arbitrary constant vector and apply ordinary Stokes' theorem to the field $\vec A\times\hat c$:

Left side: the cyclic identity $d\vec\ell\cdot(\vec A\times\hat c)=\hat c\cdot(d\vec\ell\times\vec A)$ gives $\hat c\cdot\oint d\vec\ell\times\vec A$. Right side: for constant $\hat c$, $\vec\nabla\times(\vec A\times\hat c)=(\hat c\cdot\vec\nabla)\vec A-\hat c\,(\vec\nabla\cdot\vec A)$, so

Since $\hat c$ is arbitrary, the vectors multiplying it are equal:

where the last equality is the $\varepsilon$-identity $[(d\vec s\times\vec\nabla)\times\vec A]_k =\sum_i ds_i\partial_kA_i-ds_k\sum_i\partial_iA_i$ (expand $\varepsilon_{kij}\varepsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$ — the operator BAC–CAB rule, with $\vec\nabla$ acting only on $\vec A$).

(a) Superpose loops: the turns in $dz_0$ at height $z_0$ carry $nI\,dz_0$, so replace $I\to nI\,dz_0$, $z\to z-z_0$ in (6.83)–(6.84) and integrate $z_0\in(-d/2,d/2)$. Both integrals factor through

For $|z|<d/2$ split at $z_0=z$:

The $\sinh$ line inserted into (6.83) gives $B_\rho$ directly. For $B_z$, (6.84) produces

and the first ($d$-independent) integral is exactly the step function identity (6.74): $a\int_0^\infty J_1(ka)J_0(k\rho)\,dk=H(\rho)$ — the infinite-solenoid term $\frac{4\pi nI}{c}H(\rho)$. ✓ (b) At $\rho=0$, $z=0$ the correction is $\Delta=\frac{4\pi anI}{c}\int_0^\infty dk\,J_1(ka)e^{-kd/2}$. For $d\gg a$ the exponential confines $k\lesssim2/d\ll1/a$, where $J_1(ka)\simeq ka/2$:

So the outside return field near the middle of a long solenoid dies as $1/d^2$ (each end looks like a monopole-pair aperture; cf. P16 for the $|z|\gg d$ dipole regime).

The annulus $(\rho_0,\rho_0+d\rho_0)$ is a loop of current $dI=\sigma\,(\omega\rho_0)\,d\rho_0$ (charge per length $\sigma d\rho_0$ passing at speed $\omega\rho_0$). Insert $a\to\rho_0$, $I\to dI$ in (6.83)–(6.84) and integrate. Both components need

which follows from the Bessel recursion $\frac{d}{dx}[x^2J_2(x)]=x^2J_1(x)$ (the $m=2$ member of the family that gave (6.72)). Then, e.g.,

and identically for $B_\rho$ with $J_1(k\rho)$ and the loop field's $\mathrm{sign}(z)$. ✓ Far field. On axis ($\rho=0$, $J_0=1$), $e^{-k|z|}$ confines $k\lesssim1/|z|\ll1/R$ where $J_2(kR)\simeq(kR)^2/8$:

Matching the on-axis dipole field $B_z=2m_z/|z|^3$ (6.154):

Slice the cylinder into disks of thickness $dz_0$: each is a spinning disk (P7) of surface density $\rho_0\,dz_0$. Integrate P7's $B_z$ with $z\to z-z_0$:

The inner integral (P6's split) is $\frac2k[1-e^{-kd/2}\cosh kz]$, so

The first, $d$-independent integral is the Weber–Schafheitlin closed form

(at $\rho=0$ it is $\int J_2(x)dx/x=1/2$ ✓), turning the leading term into $\frac{2\pi\rho_0\omega}{c}(R^2-\rho^2)H(R-\rho)$ — precisely the infinite spinning cylinder: nested solenoid sheets $K_\phi=\rho_0\omega\rho'\,d\rho'$ give $B_z(\rho)=\frac{4\pi}{c}\int_\rho^R\rho_0\omega\rho'\,d\rho' =\frac{2\pi\rho_0\omega}{c}(R^2-\rho^2)$ inside and $0$ outside. ✓

(The given $\vec J$ integrates over a half-plane cross-section to $I$ ✓.) By symmetry $\vec A=A_\phi(\rho,z)\hat e_\phi$, and evaluating at $\phi=0$ (where $\hat e_\phi=\hat y$) as in (6.100)–(6.101),

Now insert the other cylindrical expansion of the free Green function — Fourier in $z$, modified Bessel in $\rho$ (the §4.6 companion of (6.70)):

The $\phi'$ integral $\int_0^{2\pi}\cos\phi'\,e^{-im\phi'}d\phi'=\pi(\delta_{m,1}+\delta_{m,-1})$ keeps only $m=\pm1$, each contributing equally ($I_{-1}K_{-1}=I_1K_1$), so with $z'=0$ and $\phi=0$:

Here $\rho_<=\min(\rho,a)$, $\rho_>=\max(\rho,a)$ because the source sits at $\rho'=a$. This form is manifestly regular on the axis ($I_1\to0$) and decays for $\rho\to\infty$ ($K_1\to0$); the numeric check confirms it agrees with the $e^{-k|z|}J_1J_1$ representation (6.227) everywhere off the wire.

Slice the cone into rings by axial position $z\in(L_1,L_2)$. The ring at $z$ has radius $\rho(z)=z\tan\alpha$; its width measured along the slant is $d\ell=dz/\cos\alpha$, so it carries

The on-axis loop formula (6.62), with the field point a distance $z$ below the ring's plane, gives $dB_z=\frac{2\pi\,dI}{c}\frac{\rho^2}{(\rho^2+z^2)^{3/2}}$. With $\rho^2+z^2=z^2/\cos^2\alpha$:

which is independent of $z$ — every slice contributes equally (the $1/z^2$ geometric dilution exactly cancels the growth of ring radius and current). Hence

Limits: $\alpha\to0$ kills the field as $\alpha^3$ (thin needle); $\alpha\to\pi/2$ diverges (the cone flattens into an infinite sheet-like annulus whose inner edge approaches the origin).

Exterior. Start from (6.69), $\Omega=\partial_z\int_{S'}da'/|\vec x-\vec x'|$. Every source point has $r'=\rho'\le a<r$ and sits at polar angle $\pi/2$, so the $m=0$ (azimuthally averaged) Coulomb expansion gives

$P_\ell(0)$ kills odd $\ell$; for $\ell=2n$, $P_{2n}(0)=(-1)^n\frac{(2n-1)!!}{(2n)!!}$. Now apply the hinted derivative. Acting on $f_\ell(r)P_\ell(\cos\theta)$ with $f_\ell=r^{-(\ell+1)}$, the two pieces combine via the derivative identities (4.181)–(4.183) into the pure $P_{\ell+1}$ ladder

(the textbook identity $\partial_z[r^{-(\ell+1)}P_\ell]$: both $\cos\theta\,\partial_r$ and the angular term produce $P_{\ell+1}$-content only, by $(\ell+1)P_{\ell+1}=(2\ell+1)xP_\ell-\ell P_{\ell-1}$ and $(1-x^2)P'_\ell=\ell(P_{\ell-1}-xP_\ell)$; Ex. 4.12.1). Hence

using $(2n+1)(2n-1)!!=(2n+1)!!$ and $(2n+2)(2n)!!=(2n+2)!!$. ✓ Interior. For $r\le a$ split the radial integral at $\rho'=r$:

(for $\ell\ge2$; the $\ell=0$ term is $a-r/2$, and only its $-r/2$ piece survives $\partial_z$). Applying the same $\partial_z$ ladder — now also to the growing solutions, via $\partial_z[r^{\ell}P_\ell]=\ell\,r^{\ell-1}P_{\ell-1}$ — and collecting coefficients of $P_{2n+1}$ produces the stated interior branch; at $r=a$ it matches the exterior branch termwise because $\frac{(2n-1)!!}{(2n)!!}\bigl[\frac{4n+3}{2n+2}-1\bigr] =\frac{(2n-1)!!}{(2n)!!}\frac{2n+1}{2n+2}=\frac{(2n+1)!!}{(2n+2)!!}$. Sign split. The $r$-independent part of the interior series is $-2\pi\sum_n(-1)^n\frac{(2n-1)!!}{(2n)!!}\frac{4n+3}{2n+2}P_{2n+1}(\cos\theta)$. Since $\int_0^1P_{2n+1}(x)\,dx=(-1)^n\frac{(2n-1)!!}{(2n+2)!!}$, the Legendre coefficients of $\mathrm{sign}(x)$ are $c_{2n+1}=(4n+3)\int_0^1P_{2n+1}=(-1)^n\frac{(2n-1)!!(4n+3)}{(2n+2)!!}$ — exactly the numbers above. So that constant part is $-2\pi\,\mathrm{sign}(\cos\theta)$, and pulling it out leaves the stated geometric-convergent power series. The sign term carries the whole $\Omega=\mp2\pi$ jump across the disk ((6.80)–(6.81)); the remaining series is continuous through $z=0$.

$B_r=\frac{I}{c}\partial_r\Omega$ and $B_\theta=\frac{I}{c}\frac1r\partial_\theta\Omega$ (6.82-analog in spherical form). Exterior $B_r$: $\partial_r(a/r)^{2n+2}=-(2n+2)(a/r)^{2n+2}/r$, so

i.e. (6.119) with $r_<=a$, $r_>=r$. Interior $B_r$: use the sign-split form (the $\mathrm{sign}(z)$ term is $r$-independent on either side, so $\partial_r$ kills it):

the same formula with $r_<=r$, $r_>=a$: $B_r$ is one continuous expression. ✓ $B_\theta$: $\frac{\partial}{\partial\theta}P_\ell(\cos\theta) =-\sin\theta\,P'_\ell(\cos\theta)=P^1_\ell(\cos\theta)$ in the Condon–Shortley convention. Exterior:

since $(2n+2)!!=2(n+1)(2n)!!$. Interior (split form again):

matching the stated $r<a$ branch. The apparent jump of $B_\theta$ across $r=a$ is the book's point: on the sphere $r=a$ the two branches disagree termwise, but the sums only actually differ at $\theta=\pi/2$ — the wire itself — because the discarded $\mathrm{sign}(z)$ term contributes the compensating $\delta$-sheet on the spanning disk ($\frac1r\partial_\theta$ of $-2\pi\,\mathrm{sign}(\cos\theta)$), cf. (6.85)–(6.94). Off the wire the full field is continuous — the numeric check hits points on both sides against the elliptic closed form.

The surface current is $\vec K=\sigma\vec v=\sigma\omega a\sin\theta'\,\hat e_{\phi'}$. In $\vec A=\frac1c\oint K\,da'/|\vec x-\vec x'|$ expand the Green function in spherical harmonics (6.102). Because $K\propto\sin\theta'=-P^1_1(\cos\theta')$, the $\hat e_{\phi'}$ projection picks out only the $\ell=1$, $m=\pm1$ terms (the same bookkeeping as (6.101)–(6.113), where now the $\theta'$ integral $\int P^1_\ell P^1_1\sin\theta'\,d\theta'\propto\delta_{\ell1}$). The result must therefore be of the pure $\ell=1$ form

(the coefficient fixed by the $\ell=1$ radial factor $\frac{4\pi}{3}\frac{r_<}{r_>^2}\,a^2\cdot\frac{\sigma\omega a}{c}$, continuous at $r=a$). Then $\vec B=\vec\nabla\times(A_\phi\hat e_\phi)$:

Inside ($A_\phi\propto r\sin\theta$, i.e. $\vec A=\frac12\vec B_0\times\vec x$): $B_r=\frac{8\pi\sigma\omega a}{3c}\cos\theta$, $B_\theta=-\frac{8\pi\sigma\omega a}{3c}\sin\theta$, which is the uniform field $\frac{8\pi a\sigma\omega}{3c}\hat z$. ✓ Outside ($A_\phi=\frac{m\sin\theta}{r^2}$ with $m=\frac{4\pi\sigma\omega a^4}{3c}$): $B_r=\frac{2m\cos\theta}{r^3}$, $B_\theta=\frac{m\sin\theta}{r^3}$ — exactly the point dipole (6.154)–(6.155). The magnetization-sphere identification: $K_{\rm eff}=cM_0\sin\theta\,\hat e_\phi$ (6.268) matches $K=\sigma\omega a\sin\theta\,\hat e_\phi$ at $M_0=\sigma\omega a/c$, and then $B_{\rm in}=\frac{8\pi}{3}M_0$ ✓, $m=\frac{4\pi a^3}{3}M_0$ ✓ ((6.259), (6.267)).

Slice into shells of radius $r'$, surface density $\sigma_{\rm eff}=\rho\,dr'$, and superpose P13. Outside ($r>a$): every shell is a dipole,

Inside ($r<a$): shells with $r'<r$ act as dipoles, accumulating $m(r)=4\pi r^5\omega\rho/(15c)$ — the same integral cut at $r$; shells with $r'>r$ each contribute their uniform interior field $\frac{8\pi r'(\rho\,dr')\omega}{3c}\hat z$, summing to

which indeed vanishes at $r=a$, where the total field joins the exterior dipole continuously (no surface current at the outer boundary: $\vec K$ is a volume effect here). So $\vec B_{\rm in}=\vec B_{\rm dip}\bigl(m(r)\bigr)+\frac{4\pi\rho\omega}{3c}(a^2-r^2)\hat z$. ✓

Now $K_\phi=\omega a\sigma_0\cos\theta\sin\theta =-\frac{\omega a\sigma_0}{3}P^1_2(\cos\theta)$ (Condon–Shortley $P^1_2=-3\cos\theta\sin\theta$): a pure $\ell=2$ current, so the same projection as P13 leaves only the $\ell=2$ radial pair. The regular/decaying combination continuous at $r=a$ is

(coefficient from the $\ell=2$ kernel $\frac{4\pi}{5}\frac{r_<^2}{r_>^3}$ and the projection integral $\int_0^\pi P^1_2P^1_2\sin\theta\,d\theta=\frac{2\cdot4!}{5\cdot2!\cdot... } =\frac{12}{5}$, folded with the $a^2$ area factor). Taking $B_r=\frac{1}{r\sin\theta}\partial_\theta(\sin\theta A_\phi)$, $B_\theta=-\frac1r\partial_r(rA_\phi)$: inside $A_\phi\propto r^2\sin\theta\cos\theta$ gives $B_r=\frac{4\pi\sigma_0\omega}{5c}\,r\,(3\cos^2\theta-1)$, $B_\theta=-\frac{4\pi\sigma_0\omega}{5c}\,3r\sin\theta\cos\theta$; converting with $x=r\sin\theta\cos\phi$ etc. gives exactly $\frac{4\pi\sigma_0\omega}{5c}(-x,-y,2z)$ — divergence-free and curl-free (a harmonic $\ell=2$ interior field, the magnetostatic analog of a uniform field gradient). Outside the same derivatives of $a^4/r^3$ produce the quoted Cartesian quadrupole, falling as $1/r^4$ — no net moment: the shell's dipole $\propto\int\sigma_0\cos\theta\cdot\sin^2\theta\,(\dots)$ integrates to zero, so the leading multipole is $\ell=2$. Both branches agree at $r=a$ except for the $K$-sheet jump in $B_\theta$, as they must.

(a) Same superposition as P6, but with $|z|>d/2$ the field point is outside the $z_0$ range and $|z-z_0|=|z|-z_0\,\mathrm{sign}(z)$ never changes branch:

which in (6.84) gives the stated $B_z$ (and the matching $B_\rho\propto J_1(k\rho)$ form with the same kernel). ✓ (b) On axis, $e^{-k|z|}$ confines $k\lesssim1/|z|$, small enough that $J_1(ka)\simeq ka/2$, $J_0\simeq1$, $\sinh(kd/2)\simeq kd/2$:

(c) Each turn is a loop of moment $I\pi a^2/c$ (6.157); $N$ turns give $m=NI\pi a^2/c$ along $\hat z$. On the axis of a dipole (6.154), $B_z=2m/|z|^3=2\pi a^2IN/(c|z|^3)$ — the multipole expansion reproduces the $k\to0$ limit of the exact kernel: the three factors $ka/2$, $kd/2$, $k$ in (b) are precisely the loop area, the turn count and the dipole differentiation.

Route 1 (effective currents). Uniform $\vec M$ has $\vec J_{\rm eff}=c\vec\nabla\times\vec M=0$ in the bulk. On the two faces $\hat n'=\pm\hat n\parallel\vec M$, so $\vec K_{\rm eff}=c\vec M\times\hat n'=0$ there; only the rim (outward normal $\hat\nu\perp\hat n$) carries $\vec K_{\rm eff}=cM_0\,\hat n\times\hat\nu$ — a current of surface density $cM_0$ circulating around $C$ (right-handed about $\vec M$). Integrated over the thickness it is a line current $I=|\vec K|\,d=cM_0d$. Hence, by (6.42) restricted to a line current (6.22), $\vec A=\frac{I}{c}\oint_C d\vec\ell'/|\vec x-\vec x'|=M_0d\oint_Cd\vec\ell'/g$. ✓ Route 2 (dipole superposition, the Ex. 6.13.1 result). From P33, $\vec A=\int(\vec\nabla'\times\vec M)/g\,-\oint(\hat n'\times\vec M)/g$: the volume term dies, the face terms die ($\hat n'\parallel\vec M$), and the rim term is the same line integral. Route 3 (Stokes). For the thin wafer, $\vec A=\int d^3x'\,\vec M\times\vec\nabla'(1/g) = M_0d\int_S da'\,\hat n\times\vec\nabla'(1/g)$, and the surface-curl corollary of Stokes' theorem, $\int_S(\hat n\times\vec\nabla')f\,da'=\oint_Cf\,d\vec\ell'$, lands on the same answer — this route shows the curved wafer works too, since only $\vec M\parallel\hat n'$ (locally) was used. The check compares fields, not potentials: outside the material $\vec B=\vec H$ of the two $\pm M_0$ charged faces, which converges to the rim-loop Biot–Savart field quadratically in $d$.

For a line current (6.131), $\vec m=\frac{I}{2c}\oint\vec x\times d\vec\ell$, equivalently the directed area $\vec m=\frac Ic\vec S$ (6.133): $S_i$ = signed area of the circuit's projection onto the plane $\perp\hat e_i$. (a) Projection on the $x$–$y$ plane ($S_z$): the arcs project onto the two segments $y\in[0,2R]$ at $x=0,-L$; the circuit projects to the $2R\times L$ rectangle traversed once; with the parametrization above the traversal is counterclockwise seen from $+z$… tracking the order (origin $\to(0,2R)\to(-L,2R)\to(-L,0)\to$ origin) the loop runs clockwise in the conventional $(x,y)$ orientation? Compute honestly: $\frac12\oint(x\,dy-y\,dx)$ over that rectangle path gives $+2RL$ — the quadrature settles the sign: $S_z=+2RL$. Projection on the $y$–$z$ plane ($S_x$): the legs project to points; arc 1 (at $x=0$, bulging $+z$) traverses $(0,0)\to(R,R)_{\;(y,z)}\to(2R,0)$, arc 2 (at $x=-L$, bulging $-z$) traverses $(2R,0)\to(R,-R)\to(0,0)$: together a full circle of radius $R$ traversed clockwise in the $(y,z)$ plane, i.e. $S_x=-\pi R^2$. Projection on $x$–$z$: the two arcs trace the same half-disk boundary in opposite senses — cancels; the legs are collinear segments: $S_y=0$. Hence $\vec m=\frac Ic(-\pi R^2,0,2RL)$. (The saddle geometry is exactly why the arcs add in $S_x$: one bulges up traversed forward, the other bulges down traversed backward.) (b) Each half is a half-disk of area $\pi R^2/2$; the unbent half (in the $x$–$y$ plane, traversed counterclockwise from $+z$) contributes $\frac{I\pi R^2}{2c}\hat z$; the folded half lies in the $x$–$z$ plane and, for the fold that lifts the $y<0$ half to $z>0$, its traversal $(-R,0)\to(0,R)\to(R,0)$ in $(x,z)$ has directed area $+\frac{\pi R^2}{2}\hat y$ (verified by the quadrature). Total $\vec m=\frac{I\pi R^2}{2c}(\hat y+\hat z)$, magnitude $\pi IR^2/(\sqrt2c)$ — bending a loop by $90^\circ$ cuts its moment by $1/\sqrt2$ and rotates it to bisect the fold.

(a) Write $(\vec m\times\vec r/r^3)_k =-\varepsilon_{klm}m_l\partial_m\frac1r$ (since $\vec r/r^3=-\vec\nabla\frac1r$). Then

The first term is $+4\pi m_i\,\delta(\vec x)$. For the second use (6.139), $\partial_i\partial_j\frac1r=\frac{3x_ix_j-r^2\delta_{ij}}{r^5} -\frac{4\pi}{3}\delta_{ij}\delta(\vec x)$ (its diagonal delta weight $-4\pi/3$ is the box-flux computation checked numerically; off-diagonal terms carry no delta):

(The $\frac{8\pi}{3}\vec m$ weight — versus the electric dipole's $-\frac{4\pi}{3}\vec p$ (6.147) — is what the ball-integral check measures: $\int_{\rm ball}\vec\nabla\times\vec A=\oint\hat n\times\vec A\,da =\frac{8\pi}{3}\vec m$ for any ball centered on the dipole, since the traceless dipole term integrates to zero.) (b) With $\vec\nabla\cdot\vec A_m=0$ (it is $-\vec m\cdot\vec\nabla\times\vec\nabla\frac1r$-type, i.e. the divergence of a curl-like object: explicitly $\vec\nabla\cdot(\vec m\times\vec F)=-\vec m\cdot\vec\nabla\times\vec F=0$ for the gradient field $\vec F=-\vec\nabla\frac1r$),

identifying $\vec J_m=-c\,\vec m\times\vec\nabla\delta$. ✓ Its moment closes the loop: $\frac1{2c}\int\vec x\times\vec J_m =-\frac12\int\vec x\times(\vec m\times\vec\nabla\delta) =\vec m$ after one integration by parts (checked with a Gaussian-smeared delta). This is the magnetic twin of $\rho=-\vec p\cdot\vec\nabla\delta$ (6.150).

(a) For static currents $\vec\nabla'\cdot\vec J=0$ (6.34), so

where the total-divergence integral vanishes for localized $\vec J$ (surface at infinity) and $\vec J\cdot\vec\nabla'(x'_ix'_jx'_k) =J_ix'_jx'_k+x'_iJ_jx'_k+x'_ix'_jJ_k$. ✓ (This generalizes (6.124) ($x'_i$ alone: $\int J_i=0$) and (6.126) (two coordinates: symmetric part).) (b) Use (a) to remove the totally symmetric part:

Now $x'_jJ_k-x'_kJ_j=\sum_r\varepsilon_{jkr}(\vec x'\times\vec J)_r$ (contract two epsilons), so

using the definition of $m_{ij}$ ($\int(\vec x'\times\vec J)_rx'_i=\frac{3c}{2}m_{ri}$). Flipping $\varepsilon_{jkr}=-\varepsilon_{kjr}$, $\varepsilon_{ikr}=-\varepsilon_{kir}$ gives the stated minus-sign form. ✓ Note $\mathrm{tr}\,m=\frac2{3c}\int(\vec x'\times\vec J)\cdot\vec x'=0$ identically. (c) The quadratic Taylor term of (6.123) is

The two $\varepsilon$-terms contribute equally under the symmetric $\sum_{ij}Q_{ij}$ (swap $i\leftrightarrow j$ in the second), so

the stated formula (with the dimensional caveat on the printed $1/c$: the dipole term is $\vec m\times\vec x/r^3$ with no extra $c$, and $m_{ij}$ has dimensions [dipole]$\times$[length], so no further $1/c$ can appear).

(a) Take the curl of P20's $A^{(q)}$. Two structural facts do the work. First, split $m_{rj}=s_{rj}+a_{rj}$: the antisymmetric part $a_{rj}=\frac12\varepsilon_{rjl}v_l$ turns $-\frac{1}{2r^5}\varepsilon_{kir}Q_{ij}a_{rj}$ into $\vec\nabla\times$-exact-free content — explicitly it reduces to a term $\propto\vec\nabla(x_lv_l/r^3$-type$)$, i.e. a gradient, whose curl vanishes: only $s_{ij}$ enters $\vec B$. Second, with $\partial_l(x_ix_j/r^5)=\frac{\delta_{li}x_j+\delta_{lj}x_i}{r^5}-\frac{5x_ix_jx_l}{r^7}$ and $\partial_l r^{-3}=-3x_l/r^5$, the curl $B^{(q)}_k=\varepsilon_{klm}\partial_lA^{(q)}_m$ contracts (two $\varepsilon$'s again) to

after using $\mathrm{tr}\,s=0$ to drop the $\delta_{ij}$ contractions. (The numeric identity check against $\vec\nabla\times A^{(q)}$, with a generic non-symmetric $m_{ij}$, certifies both facts.) (b) $s_{ij}=s_{ji}$ gives 6 components; $\mathrm{tr}\,s=\mathrm{tr}\,m=0$ (P20) removes one: 5, matching the electric quadrupole's count — the $\ell=2$ multiplicity $2\ell+1$. (c) For the $\cos\theta$ shell, $\vec K=\omega a\sigma_0\cos\theta\sin\theta\,\hat e_\phi$ and $\vec x\times\vec K=-a|K|\hat e_\theta$ on the surface, so

and by azimuthal symmetry $+$ tracelessness $s_{11}=s_{22}=-\frac12s_{33}=-\frac{8\pi\sigma_0a^5\omega}{45c}$ (off-diagonals vanish by the $\phi$ integrals). Plugging into (a): e.g. on the axis $B^{(q)}_z=\frac{3}{2r^7}s_{33}\cdot3r^3=\frac{9s_{33}}{2r^4} =\frac{8\pi\sigma_0a^5\omega}{5c\,r^4}$, exactly P15's exterior $B_z(0,0,r)$. The full tensor match is the numeric check — P15's shell is a pure quadrupole (no dipole, and the $\ell\ge3$ moments vanish for this $K\propto P^1_2$ current).

The wire field is $\vec B=\frac{2I}{c\rho^2}(-y,\,x,\,0)$, so

Force (6.170): $\vec F=\vec\nabla(\vec m\cdot\vec B)$ at $(x_0,0,0)$:

and $\partial_z=0$: $\vec F=-\frac{2I}{cx_0^2}(m_y,m_x,0)$. ✓ (Note the pattern: the $m_x$ part is pulled sideways, the $m_y$ part radially; an $\vec m\parallel\hat z$ dipole feels no force — $\vec m\cdot\vec B=0$ everywhere — but does feel a torque.) Torque (6.177): with $\vec B(x_0,0,0)=\frac{2I}{cx_0}\hat y$,

The torque tries to align $\vec m$ with the local $\hat e_\phi=\hat y$; the force then pulls the aligned dipole toward the wire ($\vec m=m\hat y$: $\vec F=-\frac{2Im}{cx_0^2}\hat x$, attraction — a small co-circulating loop is drawn in, "likes attract").

Insert the expansion:

Around a closed loop $\oint d(x_lx_j)=0$ gives $\oint x_ldx_j=-\oint x_jdx_l$: the object is antisymmetric in $(l,j)$, so it is dual to the directed area,

(Contract: $\sum_{lj}\varepsilon_{ljm}\oint x_ldx_j=2S_m$ ✓.) Then

— care with the contraction: $\varepsilon_{ijk}\varepsilon_{ljm}=\varepsilon_{jki}\varepsilon_{jml} =\delta_{km}\delta_{il}-\delta_{kl}\delta_{im}$, so

using $\vec\nabla\cdot\vec B=0$ (6.27) and constant $\vec m$. ✓ This is (6.170) re-derived at the level of a single rigid loop; no assumption $\vec\nabla\times\vec B=0$ was needed (that would further convert it to $(\vec m\cdot\vec\nabla)\vec B$, (6.171)).

(a) With $\vec A\equiv\vec M$ (constant), $\hat n$ the outward normal, expand the double cross product $(\vec M\times\hat n)\times\vec B_0=\hat n\,(\vec M\cdot\vec B_0)-\vec M\,(\hat n\cdot\vec B_0)$:

First integral: $\oint\vec x\times\hat n\,da=-\int\vec\nabla\times\vec x\,d^3x=0$. Second: for constant $\vec B_0$ and $\vec f(\vec x)=\vec x\times\vec M$, the divergence theorem componentwise gives $\oint da\,(\hat n\cdot\vec B_0)f_i=\int d^3x\,(\vec B_0\cdot\vec\nabla)f_i$, and $(\vec B_0\cdot\vec\nabla)(\vec x\times\vec M)=\vec B_0\times\vec M$. Hence

So although the formula is the familiar dipole torque, it is delivered entirely by the $\vec K_{\rm eff}\times\vec B_0$ stress on the surface (for uniform $\vec M$ there is no volume current to push on). (b) $\vec m=\vec M\,V=\frac{2\pi a^3}{3}M_0\hat k$. On the flat face $\hat n=-\hat k\parallel\vec M$, so $\vec K_{\rm eff}=0$ there — the entire torque comes from the dome, where $\vec K_{\rm eff}=cM_0\sin\theta\,\hat e_\phi$. The quadrature over the dome alone indeed returns

(a) $\vec\nabla\times(\vec m\times\vec r)=2\vec m$, so the ansatz fields are

with $\vec B_{\rm dip}$ the point-dipole field (6.152, $r\ne0$). For $\vec m=m\hat z$: $B_r=\frac{2m\cos\theta}{r^3}(\cdots)$, $B_\theta=\frac{m\sin\theta}{r^3}(\cdots)$, and the uniform piece adds $\frac{2C_1m}{a^3}(\cos\theta,-\sin\theta)$ in $(r,\theta)$. Boundary conditions at $r=a$: $B_r$ continuous (6.204) and $H_\theta$ continuous (6.205, no free surface current), with $\vec H_{\rm in}=\vec B_{\rm in}$ (vacuum), $\vec H_{\rm out}=\vec B_{\rm out}/\mu$:

Solving: $C_1=\frac{\mu-1}{2\mu+1}$, $C_2=\frac{3\mu}{2\mu+1}$. ✓ For $\mu>1$ the cavity field is boosted along $\vec m$ (the medium's magnetization adds a uniform field), and the far dipole is enhanced by $C_2>1$; $\mu\to\infty$: $C_1\to\frac12$, $C_2\to\frac32$. (b) Swapping which side is vacuum exchanges the roles of $\vec B$ and $\mu\vec H$: dividing both media's $\mu$'s by $\mu$ (allowed — only the ratio enters the BCs) turns (medium $\mu$ outside, vacuum inside) into (vacuum outside, $1/\mu$ inside)… run the same two conditions with $\vec H_{\rm in}=\vec B_{\rm in}/\mu$, $\vec H_{\rm out}=\vec B_{\rm out}$: $1+C_1'=C_2'$ and $\frac{1}{\mu}(1-2C_1')=C_2'$, whose solution is exactly (a) with $\mu\to1/\mu$: $C_1'=\frac{1-\mu}{2+\mu}$, $C_2'=\frac{3}{\mu+2}$. For $\mu>1$ the sphere now screens: $C_2'<1$, and the interior uniform correction opposes $\vec m$. (c) The medium's magnetization at $r=a^+$ is $\vec M=\chi_m\vec H_{\rm out}=\frac{\mu-1}{4\pi}\frac{C_2}{\mu}\vec B_{\rm dip}(\vec m)/1$… keeping the $\theta$ component (which is what survives in $\vec M\times\hat n$): $M_\theta=\frac{\mu-1}{4\pi}\frac{C_2}{\mu}\frac{m\sin\theta}{a^3}$. The bound current on the cavity wall (outward normal of the material is $-\hat r$) is

using $\frac{\mu-1}{4\pi}\frac{C_2}{\mu}=\frac{3(\mu-1)}{4\pi(2\mu+1)}$. This is precisely the rotating shell's $K_f=\sigma\omega a\sin\theta\,\hat e_\phi$ under $\sigma\omega a\leftrightarrow\frac{3cm}{4\pi a^3}\frac{\mu-1}{2\mu+1}$ — the cavity problem is Ex. 6.7.3's shell superposed on the bare dipole.

In the source-free region $\vec H=-\vec\nabla\Phi_m$ with $\vec\nabla\cdot\vec B=0$, $\vec B=\mu\vec H$: mathematically identical to the dielectric problem under $\vec H\leftrightarrow\vec E$, $\vec B\leftrightarrow\vec D$, $\mu\leftrightarrow\epsilon$ (§6.11, (6.210)–(6.211)). The dielectric sphere in $\vec E_0$ (Ex. 5.7.3) has interior field $\frac{3}{\epsilon+2}E_0$ and exterior dipole $\frac{\epsilon-1}{\epsilon+2}a^3E_0$; transcribing gives the stated $\Phi_m$. Verify directly: both branches are harmonic; at $r=a$, $\Phi_m$ continuous ($-\frac{3}{\mu+2}=-1+\frac{\mu-1}{\mu+2}$ ✓ — this is $H_\theta$ continuity), and $B_r=\mu H_r$ inside vs $H_r$ outside: $\mu\frac{3}{\mu+2}B_0\cos\theta =\bigl[1+2\frac{\mu-1}{\mu+2}\bigr]B_0\cos\theta=\frac{3\mu}{\mu+2}B_0\cos\theta$ ✓. (b) $\vec M=\frac{\vec B-\vec H}{4\pi}$ inside:

Sanity: $\mu\to\infty$ gives $B_{\rm in}\to3B_0$, $M\to\frac{3B_0}{4\pi}$ (the "perfectly permeable" saturation of the sphere geometry, demagnetizing factor $4\pi/3$); $\mu<1$ gives $M$ opposing $B_0$ (diamagnetic screening).

Ansatz (mirroring the electrostatic (6.214)–(6.221)): $\vec B_{z>0}={\rm BS}[\vec J]+{\rm BS}[\vec J^]$ and $\vec B_{z<0}={\rm BS}[c'\vec J]$, with BS the vacuum Biot–Savart operator (6.24). The mirrored current with coefficients $(a,a,b)$, $b=-a$, is geometrically the mirror image of the circuit (current path reflected in $z=0$, which flips $dl_z$) scaled by $a$: for a planar horizontal loop this is a coaxial loop at $-z_0$ circulating in the same sense. Key mirror-symmetry facts at the plane $z=0$: the field of the mirror image of a current satisfies $B^_z(x,y,0)=+a\,B^J_z(x,y,0)$ and $\vec B^*_\parallel(x,y,0)=-a\,\vec B^J_\parallel(x,y,0)$ (reflect the Biot–Savart integrand; a mirror current's field is the mirror of the field with the pseudovector flip). Then the conditions: normal $\vec B$ continuous, and tangential $\vec H$ continuous with $\vec H_{z<0}=\vec B_{z<0}/\mu$:

and $b=-a$ is forced by consistency for currents with vertical segments (the mirror flip of $J_z$ is what keeps $\vec\nabla\cdot\vec J^*=0$ and makes the parallel/normal field symmetries hold simultaneously). Limits: $\mu\to1$: no image ($a\to0$, $c'\to1$); $\mu\to\infty$: $a\to1$ (the "magnetic conductor" image, field lines forced normal to the interface); $\mu\to0$: $a\to-1$ (superconductor limit — see P31's boundary condition).

In an infinite medium $\mu$, a free current's field is $\mu$ times its vacuum field ($\vec\nabla\times\vec B=\frac{4\pi\mu}{c}\vec J_{\rm free}$ from (6.203)), so write $\vec B_{z>0}=\mu\bigl({\rm BS}[\vec J]+{\rm BS}[\vec J^*]\bigr)$, $\vec B_{z<0}={\rm BS}[c''\vec J]$. The same mirror relations as P27 now give, at $z=0$:

Solving: $a'=\frac{1-\mu}{1+\mu}=-\frac{\mu-1}{\mu+1}$, $c''=\frac{2\mu}{\mu+1}$. ✓ So the embedded source sees an image of the opposite sign relative to P27 (a co-circulating loop buried in iron is repelled from the surface toward the bulk), while the transmitted-side strength happens to be the same $2\mu/(\mu+1)$ — though evaluated with the vacuum kernel rather than the medium's. Both P27 and P28 reduce to the same physical field when $\mu\to1$.

Potential of a uniform dipole layer. A layer of moment density $D\hat n'$ on $S'$ has $\Phi(\vec x)=\int_{S'}da'\,D\,\hat n'\cdot\frac{\vec x-\vec x'}{|\vec x-\vec x'|^3} =-D\,\Omega(\vec x)$ in the book's sign convention (6.67) — e.g. just above a large layer $\Omega=-2\pi$ and $\Phi=+2\pi D$ ✓ (the standard layer jump $4\pi D$). Images. The dielectric half-space images a charge $q$ at height $h$ as $q^=-\lambda q$ at $-h$ ($\lambda=\frac{\epsilon-1}{\epsilon+1}$), and transmits $\frac{2}{\epsilon+1}q$. Image the layer sheet by sheet: the $+$ sheet (upper) maps to a $-\lambda\sigma$ sheet at the mirrored (lower) position, the $-$ sheet to $+\lambda\sigma$ above it — mirroring swaps the stacking, so the image layer has moment $+\lambda D$ along $+\hat z$ (two sign flips cancel). Hence for $z>0$, $\Phi=-D\Omega_{\rm src}-\lambda D\,\Omega_{\rm img}$, and below, $\Phi=-\frac{2D}{\epsilon+1}\Omega_{\rm src}$. Gradients give the stated $\vec E$; multiplying by $\epsilon$ below gives $\vec D=\frac{2\epsilon}{\epsilon+1}D\,\vec\nabla\Omega_{\rm src}$. Magnetic transcription.* In the source-free regions $\vec H=-\vec\nabla\Phi_m$ with the same boundary mathematics ($\vec H\leftrightarrow\vec E$, $\vec B\leftrightarrow\vec D$, $\mu\leftrightarrow\epsilon$). A current loop $I$ is the magnetic twin of a dipole layer of strength $D=I/c$ on any spanning surface (§6.5: $\vec B=\frac Ic\vec\nabla\Omega$ off the surface). Substituting:

which is exactly P27's image system (image loop $\lambda I$, same circulation sense; transmitted loop $\frac{2\mu}{\mu+1}I$) — the numeric check closes that triangle.

A point dipole $\vec p$ at $\vec x'$ is the charge density $\rho=-\vec p\cdot\vec\nabla'\delta(\vec x-\vec x')$ (6.150). Feeding it through the linear relation $\Phi=\int G\rho$ and integrating by parts moves the gradient onto $G$:

A dipole surface of density $D\hat n'$ then superposes to

the general-$G$ version of P29's $-D\Omega$ (for free space $G=1/|\vec x-\vec x'|$ and $\int\partial_{n'}(1/g)\,da'=-\Omega$ by (6.68) — signs consistent). Now transcribe: in regions with no free current the magnetostatic problem is the $\mu\leftrightarrow\epsilon$ electrostatic one (§6.11), a current loop is a dipole layer of strength $I/c$ on any spanning surface $S'$ (P29), so

with $\hat n'$ right-handed about the current. The payoff: any dielectric image/Green construction (plane, sphere, layered media, …) instantly generates the magnetostatic loop solution in matched permeable geometry — one scalar quadrature instead of three vector ones. The representation fails only on $S'$ itself, where the layer discontinuity ($\Phi_m$ jumps by $4\pi I/c$) encodes the multivaluedness of the loop's scalar potential — crossing $S'$ is how Ampère's law survives $\vec H=-\vec\nabla\Phi_m$.

Ring reduction. Decompose $\vec J$ into coaxial rings (the $\phi$-components; poloidal pieces follow the same scaling — see below). A ring at $(r_0,\theta_0)$ has, in spherical coordinates, only $A_\phi(r,\theta)$, and

so $\vec B\cdot\hat r|_{r=a}=0\iff A_\phi(a,\theta)$ is $\theta$-independent $\iff A_\phi(a,\theta)=0$ (it must vanish at the poles). The ring potential expands as

(the (6.113) machinery with the source at $\theta_0$). Add an interior image ring at $r_1<a$, same $\theta_0$, current $I^$: for $r=a$ (outside both), each $\ell$ term of the sum* must vanish:

An $\ell$-independent solution requires $r_1=a^2/r_0$ (then $r_1^{\ell+1}=a^{2\ell+2}/r_0^{\ell+1}$), giving

— one image ring kills every multipole at once, exactly like the electrostatic sphere image. ✓ Density form. Write the ring as $\vec J=\frac{I}{r_0}\delta(r-r_0)\delta(\theta-\theta_0)\hat e_\phi$ and demand $\vec J^(r)=-(a/r)^5\vec J(a^2/r)$: the delta transforms as $\delta(a^2/r-r_0)=\frac{r^2}{a^2}\delta(r-a^2/r_0)$, and collecting factors returns precisely $I^=-(r_0/a)I$ at $r_1=a^2/r_0$ — the $(a/r)^5$ is the Jacobian bookkeeping $(a/r)^{2}$ (delta) $\times(a/r)^{3}$ (Kelvin scaling of a current density) that makes the rule ring-consistent and, applied componentwise, extends it to arbitrary $\vec J$ (same fifth power as the electrostatic image charge density $\rho^*=-(a/r)^5\rho(a^2/r)$ of Ex. 3.3.3). Physically: $\mu\to0$ expels flux (P27's $a\to-1$ image in the plane limit), and $\vec B\cdot\hat n=0$ is the superconductor (Type-1, Meissner) boundary condition.

By P27, the field above the slab due to the slab is that of an image dipole: mirror the small current loop ($z\to-z$ flips the moment's parallel components — a pseudovector — and the image coefficient $a=\lambda$ scales it):

located at $(0,0,-d)$. The force on $\vec m$ is exact (no factor-$\frac12$ subtlety for forces: differentiate the field of the fixed image): $\vec F=\vec\nabla(\vec m\cdot\vec B^)$, with $\vec B^$ the image's dipole field. Along the axis joining them ($\hat z$, separation $s=z+d$):

With $\vec m=m(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$: $\vec m\cdot\vec m^=\lambda m^2(\cos^2\theta-\sin^2\theta)$ and $(\vec m\cdot\hat z)(\vec m^\cdot\hat z)=\lambda m^2\cos^2\theta$, so the numerator is $\lambda m^2(3\cos^2\theta-\cos^2\theta+\sin^2\theta) =\lambda m^2(1+\cos^2\theta)$ — $\phi$ drops out (rotational symmetry about the normal). Then

and the transverse components vanish on axis by symmetry (checked numerically). For $\mu>1$, $F_z<0$: attraction — likes attract, and the image co-circulates (P27). The pull is strongest for $\vec m\perp$ surface ($\theta=0$, factor 2) and weakest, but still attractive, for $\vec m\parallel$ surface (factor 1); a magnet never hangs repelled above iron, but does above a superconductor ($\mu\to0$: $F_z=+3m^2(1+\cos^2\theta)/16d^4$).

Superpose:

since $\vec\nabla'(1/g)=+(\vec x-\vec x')/g^3$. Use the product rule $\vec\nabla'\times\bigl(\vec M/g\bigr) =\frac{1}{g}\vec\nabla'\times\vec M+\vec\nabla'(1/g)\times\vec M$, i.e.

Integrate; the second term is a total curl, converted by the curl theorem $\int_V\vec\nabla\times\vec F\,d^3x=\oint_S\hat n\times\vec F\,da$:

Reading through (6.42): the first term is $\frac1c\int\vec J_{\rm eff}/g$ with $\vec J_{\rm eff}=c\vec\nabla\times\vec M$; the second is $\frac1c\oint\vec K_{\rm eff}/g$ with $\vec K_{\rm eff}=c\vec M\times\hat n=-c\,\hat n\times\vec M$ ✓ — the dipole picture and the bound-current picture are the same object, split by one integration by parts. The two numeric regimes probe each piece separately: a smooth interior $\vec M$ (only the volume term) and the uniform sphere (only the surface term, $\vec K_{\rm eff}=cM_0\sin\theta\,\hat e_\phi$, whose field P13 already established).

Sources. $\vec\nabla\cdot\vec M=0$ within each hemisphere, but (6.255) needs all of $\vec M$'s divergence: across the equatorial plane $M_z$ jumps from $-M_0$ to $+M_0$, a magnetic surface charge $\sigma_m^{\rm disk}=-(M_z^+-M_z^-)=-2M_0$ on the disk $\rho'\le a$, $z'=0$; on the outer shell $\sigma_m^{\rm shell}=\vec M\cdot\hat n=M_0|\cos\theta'|$ (positive at both poles). Total magnetic charge: $2\pi a^2M_0-2M_0\pi a^2=0$ ✓. (a) Multipoles: both sources are even in $\cos\theta'$ → only even $\ell$; $\ell=0$ vanishes (zero net charge); the first survivor is $\ell=2$:

so $\Phi_m\simeq q_2P_2(\cos\theta)/r^3$ and $\vec B=-\vec\nabla\Phi_m$: $B_r=3\pi M_0a^4P_2/r^4$, $B_\theta=-\frac{\pi M_0a^4}{r^4}\partial_\theta P_2$ — a pure quadrupole falloff $1/r^4$ (compare P15/P21: same multipole order, different source). (b) Exterior. Expand each source with (4.225) ($r_>=r$, $r_<=r'$). Shell: $\int_0^1xP_{2n}(x)dx=(-1)^{n+1}\frac{(2n-3)!!}{(2n+2)!!}$ (Ex. 4.9.1 values) gives the coefficient $4\pi M_0a^2\,(-1)^{n+1}\frac{(2n-3)!!}{(2n+2)!!}\,\frac{a^{2n}}{r^{2n+1}}$. Disk: with $P_{2n}(0)=(-1)^n\frac{(2n-1)!!}{(2n)!!}$,

Sum: $(2n-3)!!+(2n-1)!!=(2n-3)!!\,[1+(2n-1)]=2n\,(2n-3)!!$, hence

zero at $n=0$ and $C_1=2\cdot1/4!!\cdot(-1)^2\cdot(-1)!!=2/8=\frac14$ — matching (a): $4\pi M_0a^2C_1a^2/r^3=\pi M_0a^4/r^3$ ✓. Interior (corrected). The shell is still a pure $r^{2n}$ series inside (all its sources at $r'=a$), coefficient $2\pi M_0a^2\,I_{2n}\,r^{2n}/a^{2n+1}$ with $I_0=1$, $I_{2n}=C_n/n$ ($n\ge1$); but the disk potential at $r<a$ must be split at $\rho'=r$:

At the center only $n=0$ survives: $\Phi_m(0)=2\pi M_0a-4\pi M_0a=-2\pi M_0a$ (shell pulls up, the stronger disk pulls down) — the direct-quadrature check confirms the corrected interior everywhere, and confirms that the book's $r\leftrightarrow a$ swap of the exterior form does not reproduce it.

(a) In (6.255) the volume term dies ($\vec\nabla\cdot\vec M=0$ for uniform $\vec M$), leaving surface charges $\sigma_m=\vec M\cdot\hat n$: zero on the sides ($\hat n\perp\hat k$), $+M_0$ on the end face $S$, $-M_0$ on the far face. Near (but a few face-diameters away from) the near end, the far face's contribution is negligible, and

the monopole term of the face's multipole expansion about any interior point $\vec x''$ (corrections are the face's dipole/quadrupole moments about $\vec x''$, down by powers of $\sqrt S/|\vec x-\vec x''|$). Hence $\vec H=-\vec\nabla\Phi_m=M_0S\,(\vec x-\vec x'')/|\vec x-\vec x''|^3$. ✓ This is how "magnetic poles" emerge from a pole-free theory: the end of a uniformly magnetized rod carries an effective surface charge $M_0$ whose far field is Coulombic in $\vec H$ (not in $\vec B$: inside the rod $\vec B=\vec H+4\pi\vec M$ carries the return flux). (b) Both ends visible: superpose $q_m=\pm M_0\pi R^2$ at $\vec x_\pm=(0,0,\pm L/2)$:

valid whenever the distance to each face $\gg R$. At $r\gg L$ this pair merges into a dipole $m=q_mL=M_0\pi R^2L=M_0V$ ✓ (the total moment).

(a) As in P35, the only sources are $\sigma_m=+M_0$ on the top face (the sides carry none; the bottom is at $-\infty$), so $\Phi_m(\vec x)=M_0\int_S da'/|\vec x-\vec x'|$ and

by exactly the identity (6.69) that defined $\Omega$. Nothing in the step distinguished inside from outside — the formula holds in both regions, with $\Omega$ jumping from $-2\pi$ (just above the face) to $+2\pi$ (just below, inside the material): $H_z$ jumps from $+2\pi M_0$ to $-2\pi M_0$ across the charged sheet, while $B_z=H_z+4\pi M_0$ (inside) stays continuous ✓. (b) The cube's two faces carry $\sigma_m=\pm M_0$ at $z=\pm b/2$; superposing two copies of (a):

From the center, each of the six faces subtends the same magnitude, and a closed surface seen from inside subtends $4\pi$ total: $|\Omega|=4\pi/6=2\pi/3$ per face (the arctan closed form gives $4\arctan\frac{1}{\sqrt3}=\frac{2\pi}{3}$ ✓). Signs: viewed from below, $\Omega_{\rm top}=+2\pi/3$; from above, $\Omega_{\rm bot}=-2\pi/3$; so

The sphere gives the same $-\frac{4\pi}{3}\vec M$ everywhere inside ((6.263)); the cube only at special points — away from the center its $H_z=-M_0[\Omega_t-\Omega_b]$ varies (the checked quadrature shows it).

(a) Sources: $\sigma_m=\pm M_0$ on the end disks at $z=\pm d/2$ (sides clean, volume clean). The charged-disk potential in Bessel form follows from the cylindrical Green expansion (6.70) and the disk integral (6.72):

so

For $|z|<d/2$: $e^{-k(d/2-z)}-e^{-k(d/2+z)}=2e^{-kd/2}\sinh(kz)$, giving

Then $H_\rho=-\partial_\rho\Phi_m$ with $\partial_\rho J_0(k\rho)=-kJ_1(k\rho)$ gives the first stated integral, and $H_z=-\partial_z\Phi_m$ the second. ✓ (b) On the curved side, $\vec K_{\rm eff}=c\vec M\times\hat\rho =cM_0\,\hat e_\phi$: a solenoid sheet of strength $nI\leftrightarrow cM_0$. Ex. 6.6.1's field with that substitution:

— the two computations are the same field seen through the two decompositions: bound currents give $\vec B$ directly; magnetic charges give $\vec H$; they differ inside the material by exactly $4\pi\vec M$ (6.199), here visible as the Weber–Schafheitlin step term.

Route 1 (scalar potential). In (6.254) replace $\vec M\,d^3x'\to M_A\hat n'\,da'$:

This is Gauss's-law geometry: the integrand is (minus) the closed-surface solid-angle density, and for a closed surface with outward normals

(outside, the pierced cone contributions cancel in pairs; inside, the surface wraps the observation point once; the sign follows the book's (6.67) orientation — from inside, $\hat n'\cdot(\vec x-\vec x')<0$ over the whole shell). Hence $\Phi_m=0$ outside, $-4\pi M_A$ inside — constant in each region, so $\vec H=0$ in both. ✓ (Equivalently: the shell is a closed dipole layer, across which $\Phi_m$ drops by $4\pi M_A$; a closed constant-strength layer shifts the interior potential and nothing else.) Route 2 (vector potential). $\vec M\parallel\hat n'$ makes $\vec K_{\rm eff}=c\vec M\times\hat n'=0$ on both wall faces, and within the thin wall $\vec\nabla\times\vec M$ integrates to zero across the thickness for normal magnetization of a uniform shell (no net tangential circulation — all the "dipole needles" stand on end, and side-by-side needles carry counter-circulating molecular currents that cancel). With no effective currents anywhere, $\vec A=0$ and $\vec B=\vec\nabla\times\vec A=0$ everywhere; then $\vec H=\vec B-4\pi\vec M$ is $0$ outside and in the cavity, and $-4\pi M_A/d\cdot\hat n\cdot d$-worth across the wall — integrating $-\vec H\cdot d\vec\ell$ through the wall reproduces the $-4\pi M_A$ potential step. Both routes agree with the quadrature on the sphere and the cube — the shape never entered.