Static fields: electrostatics and magnetostatics, the same vector calculus twice

Idea. Flux conservation through spheres of area \(\propto r^{n-1}\) fixes the field as \(r^{1-n}\); the only freedom is the constant, fixed by unit flux.

1. \(\partial_if = f'x_i/r\), and \(\partial_i(f'x_i/r) = f''x_i^2/r^2 + f'(1/r - x_i^2/r^3)\). Summing with \(\sum x_i^2 = r^2\): \(\nabla^2f = f'' + (n-1)f'/r = r^{1-n}(r^{n-1}f')'\). So harmonic iff \(r^{n-1}f' = \beta\), i.e. \(f' = \beta r^{1-n}\):

• \(n = 2\): \(f = \alpha + \beta\log r\);
• \(n = 3\): \(f = \alpha - \beta/r\);
• \(n = 4\): \(f = \alpha - \beta/(2r^2)\).

2. \(-\Gamma_4' \cdot 2\pi^2r^3 = 1\) gives \(\Gamma_4' = -\frac{1}{2\pi^2r^3}\), so \[ \Gamma_4 = \frac{1}{4\pi^2r^2} . \] The field of a point charge is \(-\nabla\Gamma_4 \propto r^{-3}\): in four dimensions Coulomb's law would be an inverse-cube law.

3. Between the conductors the field is radial and \(z\)-independent, and by Gauss on a coaxial cylinder of radius \(s\) it is that of the enclosed line charge: \(\varphi = -\frac{\lambda}{2\pi\varepsilon_0}\log s + C\). The potential difference is \[ \varphi(a) - \varphi(b) = \frac{\lambda}{2\pi\varepsilon_0}\log\frac ba = \frac{10^{-8}\times\log5}{2\pi\times8.8542\times10^{-12}} = \mathbf{289.3\ V}. \] The field \(E = \lambda/2\pi\varepsilon_0s\) is \(3.595\times10^5\ \mathrm{V\,m^{-1}}\) at the inner conductor and \(7.190\times10^4\ \mathrm{V\,m^{-1}}\) at the outer; outside the shell the enclosed charge is zero and the field vanishes.

Check. Dimensions: \([\lambda/\varepsilon_0] = \mathrm{C\,m^{-1}}/\mathrm{F\,m^{-1}} = \mathrm V\) ✓. Limit \(b\to a\): the gap closes and the potential difference \(\to0\) ✓. The two fields are in ratio \(b/a = 5\) ✓. And \(\Gamma_4\) fits the pattern \(\Gamma_n \propto r^{2-n}\) with constant \(1/((n-2)\omega_{n-1}) = 1/(2\cdot2\pi^2)\) ✓. The logarithm's argument is the ratio \(b/a\): the arbitrary scale inside \(\log s\) cancels, which is the only way a two-dimensional potential difference can be meaningful.

Idea. The image makes the plane an equipotential; uniqueness makes the image potential correct above the plane; everything else is reading it off.

1. \(\varphi = \frac{q}{4\pi\varepsilon_0}\Big(\frac{1}{|\vec x - d\hat z|} - \frac1{|\vec x + d\hat z|}\Big)\). In \(z \gt 0\) it is harmonic except at the charge, where it has the right singularity; it vanishes on \(z = 0\) (equal distances) and tends to \(0\) at infinity. By the uniqueness theorem (Dirichlet data on the plane, decay at infinity, applied to the difference of two solutions, which is harmonic everywhere in \(z \gt 0\)) it is the answer. The half-space is neither bounded nor the exterior of a bounded set, but the deck's proof of case 1 works on it verbatim: apply the maximum principle on the half-ball \(\{z \gt 0\}\cap B_R\) and let \(R\to\infty\).

2. From rung 3, \[ \sigma(\rho) = \varepsilon_0E_z = -\frac{qd}{2\pi(\rho^2+d^2)^{3/2}} . \] Total: \(\int_0^\infty\sigma\,2\pi\rho\,\mathrm d\rho = -qd\int_0^\infty\frac{\rho\,\mathrm d\rho}{(\rho^2+d^2)^{3/2}} = -qd\Big[-\frac1{\sqrt{\rho^2+d^2}}\Big]_0^\infty = -q\). Within \(\rho \lt d\) the same integral up to \(d\) gives \(-q\big(1 - 1/\sqrt2\big)\): \(\mathbf{29.3\,\%}\) of the induced charge.

3. The field at \(q\) is that of the image alone: \(F = \frac{q^2}{4\pi\varepsilon_0(2d)^2} = \frac{q^2}{16\pi\varepsilon_0d^2}\), towards the plane. The work to pull it from \(d\) to infinity is \(\int_d^\infty\frac{q^2}{16\pi\varepsilon_0z^2}\mathrm dz = \frac{q^2}{16\pi\varepsilon_0d}\) — half of \(q^2/(4\pi\varepsilon_0\cdot2d)\), because the image is not a fixed charge: it recedes as \(q\) does, and only the region \(z \gt 0\) holds real field energy.

4. \(U = -\frac{e^2}{16\pi\varepsilon_0d} = -\frac{(1.602\times10^{-19})^2}{16\pi\times8.854\times10^{-12}\times10^{-9}}\ \mathrm J = -5.77\times10^{-20}\ \mathrm J = \mathbf{-0.360\ eV}\). The force there is \(5.77\times10^{-11}\) N.

Check. Dimensions of \(\sigma\): \([q d/\rho^3] = \mathrm{C\,m^{-2}}\) ✓. Limit \(d\to0\): \(\sigma\) concentrates at \(\rho = 0\) with total \(-q\), cancelling the charge ✓. Limit \(\rho\gg d\): \(\sigma\sim-qd/2\pi\rho^3\), the charge density that a dipole \(2qd\) (charge plus image) would induce ✓. The \(-0.36\) eV is a fraction of a typical metal work function of 4–5 eV, the right size for the correction it is known to be.

Idea. One image makes the sphere a zero equipotential; a second, at the centre, changes the sphere's potential without changing its shape as an equipotential, and fixes its charge.

1. Images. Put the axis through \(q\). The deck's Apollonius computation shows that \(q\) at \(R\hat z\) and \(q' = -qa/R\) at \(b\hat z\), \(b = a^2/R\), give \(\varphi = 0\) on \(r = a\). Outside the sphere the pair is harmonic except at \(q\), and decays; by uniqueness it is the answer there.

Legendre. In \(a \lt r \lt R\), \(\varphi\) is axisymmetric and harmonic, so \(4\pi\varepsilon_0\varphi = \sum(A_\ell r^\ell + B_\ell r^{-\ell-1})P_\ell\). The growing part is the source's: \(A_\ell = q/R^{\ell+1}\). The sphere demands \(A_\ell a^\ell + B_\ell a^{-\ell-1} = 0\), so \(B_\ell = -qa^{2\ell+1}/R^{\ell+1}\).

Reconciliation. The image's coefficients are \(q'b^\ell = -\frac{qa}{R}\frac{a^{2\ell}}{R^\ell} = -\frac{qa^{2\ell+1}}{R^{\ell+1}}\), equal to \(B_\ell\) for every \(\ell\). The total induced charge is the \(\ell = 0\) coefficient, \(B_0 = -qa/R = q'\) (Gauss on a sphere just outside \(r = a\)).

2. Add \(q'' = +qa/R\) at the centre. It adds the constant \(q''/4\pi\varepsilon_0a\) on the sphere, which stays an equipotential, and it makes the total charge inside zero. The configuration has the right singularity at \(q\), makes the sphere a conductor with total charge \(0\), and decays. The uniqueness theorem for conductor data (the deck's proof 2) then says its field is the field. The sphere's potential is \(\frac{q}{4\pi\varepsilon_0R}\) — exactly the potential \(q\) alone would produce at the centre, which is the mean value property in disguise.

3. Here \(q' = -0.500\) nC at \(b = 5.00\) cm, so \(R - b = 0.150\) m. Grounded: \[ F = \frac{kqq'}{(R-b)^2} = \frac{8.98755\times10^9\times10^{-9}\times(-0.5\times10^{-9})}{0.150^2} = \mathbf{-2.00\times10^{-7}\ N}, \] attractive. Neutral: add \(kqq''/R^2 = 8.98755\times10^9\times10^{-9}\times0.5\times10^{-9}/0.04 = 1.12\times10^{-7}\) N of repulsion: total \(\mathbf{-8.74\times10^{-8}\ N}\), still attractive — a neutral sphere always attracts a charge, because the near side's induced charge is closer. (For a neutral conductor of general shape this is not a theorem: suitably shaped conductors can repel a nearby charge — cited: Levin and Johnson, Am. J. Phys. 79 (2011) 843.)

4. \(q' = -qa/(a+d)\to-q\). The image's distance from the surface is \(a - a^2/R = ad/R \to d\). So: image \(-q\) at depth \(d\) behind a surface that has become flat — P2 ✓.

Check. Dimensions: \([q'] = [q]\), \([b] = \mathrm L\) ✓. Limit \(R\to\infty\) at fixed \(q\): induced charge \(-qa/R\to0\) ✓. Limit \(R\to a^+\): \(q'\to-q\) and \(b\to a\), the charge and image meet on the surface ✓. Large-\(R\) force on the neutral sphere: expanding, \(F \approx -2kq^2a^3/R^5\) — a charge polarising a sphere of polarisability \(4\pi\varepsilon_0a^3\) and attracting the induced dipole, which is the running example's \(p = 4\pi\varepsilon_0a^3E\) with \(E = kq/R^2\) ✓.

Idea. Near a corner of angle \(\alpha\) the true potential behaves like \(s^{\pi/\alpha}\). Point sources away from the corner can only produce integer powers of \(s\). When \(\pi/\alpha\) is not an integer the two are incompatible.

1. Reflection in a grounded wall sends a charge to minus itself on the other side. The group generated by reflections in the two walls is dihedral of order \(2\pi/\alpha\) when \(\pi/\alpha\) is an integer \(n\); its \(2n - 1\) non-identity elements give the images, with charge \(-\lambda\) for reflections and \(+\lambda\) for rotations. \(\alpha = \pi/2\): \(-\lambda\) at \((s_0, -\phi_0)\) and at \((s_0, \pi - \phi_0)\), and \(+\lambda\) at \((s_0, \pi + \phi_0)\) — three images. \(\alpha = \pi/3\): \(-\lambda\) at \(\phi = -\phi_0,\ 2\pi/3 - \phi_0,\ 4\pi/3 - \phi_0\) and \(+\lambda\) at \(\phi = 2\pi/3 + \phi_0,\ 4\pi/3 + \phi_0\) — five images, all at radius \(s_0\), all outside the wedge. Each wall is a mirror line of the whole configuration with charges reversed, so \(\varphi = 0\) on it; decay holds because the total charge is zero; by uniqueness this is the solution.

2. On \(0 \lt s \lt s_0\) the potential is harmonic. Differentiating \(b_k\) under the integral (Leibniz) and using Laplace's equation in polar form, \(s(sb_k')' = -\frac2\alpha\int_0^\alpha\varphi_{\phi\phi}\sin\frac{k\pi\phi}\alpha\,\mathrm d\phi\). Two integrations by parts, with \(\varphi = 0\) and \(\sin = 0\) at \(\phi = 0, \alpha\), turn this into \((k\pi/\alpha)^2b_k\). So \(b_k = c_ks^{k\pi/\alpha} + d_ks^{-k\pi/\alpha}\); \(\varphi\) is bounded near the edge (it is continuous and zero there), so \(d_k = 0\). Completeness of the sine series on \((0,\alpha)\) (unit 19) gives the expansion.

Sign. Unit 05's maximum principle was proved in three dimensions; it applies here because a \(z\)-independent function harmonic in the plane is harmonic on the three-dimensional region wedge \(\times\,\mathbb R\), and an interior extremum in the plane is an interior extremum there. \(\varphi \ge 0\): on the region between a small circle round the charge (where \(\varphi\) is large and positive, the logarithm dominating), the walls (\(\varphi = 0\)) and a large arc (\(|\varphi|\) small), the minimum principle bounds \(\varphi\) below by the boundary minimum, which tends to \(0\) as the circle shrinks and the arc grows. So \(\varphi \ge 0\), and \(\varphi \gt 0\) inside, since a harmonic function attaining an interior minimum is constant. Then \(b_1(s) = c_1s^{\pi/\alpha}\) is the integral of a positive function against \(\sin(\pi\phi/\alpha) \gt 0\): \(c_1 \gt 0\).

3. Suppose finitely many line charges \(\mu_i\) at points \(\vec x_i\) outside the open wedge reproduce \(\varphi\) there. Then \(\varphi = -\frac{1}{2\pi\varepsilon_0}\big(\lambda\log|\vec x-\vec x_0| + \sum\mu_i\log|\vec x-\vec x_i|\big) + C\) in the wedge. Split off any charge \(\mu_*\) sitting on the edge itself: the rest, \(h\), is harmonic on a disc \(|\vec x| \lt \delta\). A harmonic function on that disc has \(h = \sum_{j\ge0}s^j(a_j\cos j\phi + \beta_j\sin j\phi)\) with \(|a_j|, |\beta_j| \le 2M\delta^{-j}\) (the full-circle version of part 2's argument; \(M = \max|h|\) on the circle of radius \(\delta\)), so the series converges uniformly for \(s \le \delta/2\) and may be integrated term by term. Hence, for small \(s\), \[ b_1(s) = -\frac{\mu_*}{2\pi\varepsilon_0}\gamma\log s + \sum_{j\ge0}\kappa_js^j , \qquad \gamma = \frac2\alpha\int_0^\alpha\sin\frac{\pi\phi}\alpha\,\mathrm d\phi = \frac4\pi , \] a logarithm plus a convergent power series in integer powers of \(s\).

But part 2 says \(b_1(s) = c_1s^{3/2}\) with \(c_1 \gt 0\). As \(s\to0\), \(c_1s^{3/2}\to0\) while \(\log s\to-\infty\), so \(\mu_* = 0\). Then \(c_1s^{3/2} = \sum\kappa_js^j\), a function that is infinitely differentiable at \(s = 0\). Differentiate twice: the right side stays bounded as \(s\to0^+\), the left is \(\tfrac34c_1s^{-1/2}\to\infty\). Contradiction. So no finite image set exists for \(\alpha = 2\pi/3\). \(\square\)

Check. The argument must fail when it should: for \(\alpha = \pi/n\), \(\pi/\alpha = n\) is an integer and \(s^n\sin n\phi\) is a harmonic polynomial — no contradiction, and indeed part 1 found images ✓. Dimensionally \(c_k\) carries \(\mathrm V\,\mathrm m^{-k\pi/\alpha}\), a non-integer power of length for the \(120°\) corner, which is already a hint that no finite arrangement of charges at definite distances produces it. Physically, the field goes as \(s^{\pi/\alpha - 1}\): at an opening of \(120° \lt 180°\) that is \(s^{1/2}\), which vanishes at the edge, as in any hollow corner ✓; for \(\alpha \gt \pi\) (a protruding edge) the exponent is negative and the field diverges — why sharp points spark first.

Another way

A look ahead, not needed for anything above: unit 20's complex analysis maps the wedge onto a half-plane by \(w = \zeta^{\pi/\alpha}\), \(\zeta = x + \mathrm iy\), and solves the problem there with one image. Pulled back, the solution involves \(\zeta^{3/2}\), which is not single-valued round the edge — the same obstruction, seen as a branch point.

Idea. Expand the boundary data in \(P_\ell\); attach \((r/a)^\ell\) inside and \((a/r)^{\ell+1}\) outside so that each term matches at \(r = a\) and behaves at the centre and at infinity.

1. By the expansion theorem, inside \(\varphi = \sum A_\ell r^\ell P_\ell\) (boundedness at the centre kills \(r^{-\ell-1}\)) and outside \(\varphi = \sum B_\ell r^{-\ell-1}P_\ell\) (decay kills \(r^\ell\)). At \(r = a\) both must equal \(\sum V_\ell P_\ell\), so \[ \varphi_{\text{in}} = \sum_\ell V_\ell\Big(\frac ra\Big)^\ell P_\ell(\cos\theta),\qquad \varphi_{\text{out}} = \sum_\ell V_\ell\Big(\frac ar\Big)^{\ell+1}P_\ell(\cos\theta). \] Oddness leaves odd \(\ell\) only. \(V_1 = \tfrac32V_0\); \(V_3 = 7V_0(\tfrac58 - \tfrac34) = -\tfrac78V_0\); \(V_5 = 11V_0\int_0^1\tfrac18(63x^5 - 70x^3 + 15x)\,\mathrm dx = 11V_0\cdot\tfrac18\big(\tfrac{63}6 - \tfrac{70}4 + \tfrac{15}2\big) = \tfrac{11}{16}V_0\).

2. Near the centre \(\varphi = \frac{3V_0}{2a}z + O(r^3)\), so \(\vec E(0) = -\frac{3V_0}{2a}\hat z\): from the positive hemisphere towards the negative one. Far away \(\varphi \approx \frac32V_0\frac{a^2}{r^2}\cos\theta = \frac{p\cos\theta}{4\pi\varepsilon_0r^2}\) with \(p = 6\pi\varepsilon_0a^2V_0\).

3. \(|\vec E(0)| = 1.5\times100/0.1 = \mathbf{1500\ V\,m^{-1}}\), along \(-\hat z\). On the axis \(P_\ell(1) = 1\), so \(\varphi(a/2) = V_0\sum V_\ell/V_0\cdot2^{-\ell}\):

• one term: \(\tfrac32\cdot\tfrac12 = 0.7500\,V_0\);
• two: \(0.7500 - \tfrac78\cdot\tfrac18 = 0.6406\,V_0\);
• three: \(+\tfrac{11}{16}\cdot\tfrac1{32} \Rightarrow 0.6621\,V_0\);
• four: \(V_7 = -\tfrac{75}{128}V_0\), \(-\tfrac{75}{128}\cdot\tfrac1{128} \Rightarrow 0.6575\,V_0\).

Exact: \(2 - \frac{3/4}{\frac12\sqrt{5/4}} = 2 - 1.3416 = \mathbf{0.6584\,V_0} = 65.84\) V. The partial sums alternate about it and close in geometrically, the factor \(2^{-2}\) per step at \(r = a/2\) beating the slow \(\ell^{-1/2}\) decay of \(V_\ell\) that a jump in the data forces.

Check. Dimensions: \([V_0/a] = \mathrm{V\,m^{-1}}\) ✓. On the equatorial plane \(\varphi = 0\) by antisymmetry, and every odd \(P_\ell(0) = 0\) ✓. As \(r\to a^-\) on the axis the exact formula gives \(V_0[1 - 0] = V_0\) ✓. The closed form comes from Poisson's integral formula (the deck's notes on the images slide): on the axis the kernel integrates in elementary terms.

Idea. The sphere's problem, one dimension down: \(r^{-2}\cos\theta\) becomes \(s^{-1}\cos\phi\), and the peak field factor \(3\) becomes \(2\).

1. \(\varphi = -E_0(s - a^2/s)\cos\phi\) for \(s \ge a\) (from rung 3, \(B = E_0a^2\)). It is harmonic, zero on \(s = a\), and \(\varphi + E_0x\to0\). Then \(E_s(a) = -\partial_s\varphi|_{a} = E_0(1 + a^2/s^2)\cos\phi|_a = 2E_0\cos\phi\) and \[ \sigma = 2\varepsilon_0E_0\cos\phi . \] The induced term \(E_0a^2\cos\phi/s\) is a 2D dipole: a dipole \(p'\) per unit length (charges \(\pm\lambda'\) separated by \(\delta\), \(p' = \lambda'\delta\)) has potential \(p'\cos\phi/2\pi\varepsilon_0s\), so \(p' = 2\pi\varepsilon_0a^2E_0\).

2. On the circle \(s = a\), with \(b = a^2/D\), \(|\vec x - b\hat e|^2 = a^2 + b^2 - 2ab\cos\gamma = \frac{a^2}{D^2}(D^2 + a^2 - 2aD\cos\gamma) = \frac{a^2}{D^2}|\vec x - D\hat e|^2\). So \(\lambda\) at \(D\) and \(-\lambda\) at \(b\) give \(-\frac{\lambda}{2\pi\varepsilon_0}\log\frac{|\vec x - D\hat e|}{|\vec x - b\hat e|} = -\frac{\lambda}{2\pi\varepsilon_0}\log\frac Da\), a constant on the cylinder — not zero, unlike 3D, because the 2D image has the same magnitude as the source. For the pair \(+\lambda\) at \(-D\) and \(-\lambda\) at \(+D\) the two constants are equal and opposite, so the grounded condition holds exactly. The images are \(-\lambda\) at \(-b\) and \(+\lambda\) at \(+b\): a dipole per length \(\lambda\cdot2b = 2\lambda a^2/D\). The source pair makes \(E_0 = 2\cdot\frac{\lambda}{2\pi\varepsilon_0D} = \frac{\lambda}{\pi\varepsilon_0D}\) at the axis, so \(p' = 2\pi\varepsilon_0a^2E_0\), independent of \(D\) — part 1's value ✓.

3. Peak \(\sigma = 2\varepsilon_0E_0 = \mathbf{1.771\ \mu C\,m^{-2}}\), peak field \(2E_0 = \mathbf{2.00\times10^5\ V\,m^{-1}}\). The sphere of the same field had \(2.656\ \mu\mathrm{C\,m^{-2}}\) and \(3E_0\): ratio \(2/3\), the ratio of the dimensions of the space the field lines crowd in from.

Check. Dimensions: \([\varepsilon_0E_0a^2] = \mathrm{C\,m^{-2}}\cdot\mathrm m^2 = \mathrm C\), which is charge × length per length, as a dipole per metre should be ✓. \(\int_0^{2\pi}\sigma\,a\,\mathrm d\phi = 0\): neutral ✓. Limit \(a\to0\): the uniform field ✓. The field meets the surface normally: \(E_\phi(a) = -\frac1s\partial_\phi\varphi|_a = -E_0(1 - a^2/s^2)\sin\phi|_a = 0\) ✓.

Idea. The energy of the difference field is a sum over conductors of (potential difference) × (charge difference), and every term has one factor zero.

1. Let \(\varphi_1, \varphi_2\) both solve the problem and put \(w = \varphi_1 - \varphi_2\). Then \(\nabla^2w = 0\) in \(D\) (same \(\rho\)), \(w = 0\) on \(S_0\), and \(w\) is constant, say \(W_i\), on each \(S_i\) (each \(\varphi_j\) is constant on a conductor). Green's identity (I) with \(u = v = w\): \[ \int_D|\nabla w|^2\,\mathrm dV = \oint_{\partial D}w\,\partial_nw\,\mathrm dA = 0 + \sum_{i=1}^mW_i\oint_{S_i}\partial_nw\,\mathrm dA . \] By Gauss's law on a surface enclosing conductor \(i\) and hugging \(S_i\), \(\oint_{S_i}\partial_nw = \pm(Q_i^{(1)} - Q_i^{(2)})/\varepsilon_0\), the sign set by the normal's direction. For each \(i\): if \(V_i\) is given, \(W_i = 0\); if \(Q_i\) is given, the charge difference is \(0\). So every term vanishes, \(\int|\nabla w|^2 = 0\), and since \(|\nabla w|^2\) is continuous and nonnegative, \(\nabla w \equiv 0\) — the fields agree. Then \(w\) is constant on the connected region \(D\), and it is \(0\) on \(S_0\) and continuous up to it, so \(w \equiv 0\): \(\varphi\) is unique too. \(\square\)

2. A single conducting sphere of radius \(a\) centred inside a large grounded sphere of radius \(b\), no other charge. Potential \(V\) given: the solution \(\varphi = \frac{Q}{4\pi\varepsilon_0}(\frac1r - \frac1b)\) forces \(Q = 4\pi\varepsilon_0V\big/\big(\frac1a - \frac1b\big)\). So prescribing \(Q\) as well is consistent for exactly one value and contradictory for all others: over-determined.

3. The same geometry with nothing prescribed on the inner sphere: every \(Q\) gives a solution, a one-parameter family. Under-determined.

Check. The count is one number per conductor, which matches the physics: a conductor has one unknown constant (its potential) and one unknown integral (its charge), related by the capacitance, so exactly one of them is free data. Limit \(b\to\infty\) in part 2: \(Q = 4\pi\varepsilon_0aV\), the capacitance of an isolated sphere, \(1.11\times10^{-11}\) F for \(a = 10\) cm ✓.

Idea. The lemma's formula needs no ingenuity; a better-adapted gauge comes from asking for a potential with one component.

1. \(\nabla\cdot\vec B = c(\partial_xy + \partial_yx) = 0\). \(\nabla\times\vec B = c\,(0 - 0,\ 0 - 0,\ \partial_xx - \partial_yy) = \vec 0\). So \(\mu_0\vec J = \vec 0\): there is no current near the axis — the field is made by coils further out.

2. \((y,x,0)\times(x,y,z) = (xz,\ -yz,\ y^2 - x^2)\), so \[ \vec A_1 = \frac c3\big(xz,\ -yz,\ y^2 - x^2\big). \] Curl: \(x\)-component \(\partial_y\big(\tfrac c3(y^2 - x^2)\big) - \partial_z\big(-\tfrac c3yz\big) = \tfrac c3(2y + y) = cy\) ✓; \(y\)-component \(\partial_z(\tfrac c3xz) - \partial_x\big(\tfrac c3(y^2 - x^2)\big) = \tfrac c3(x + 2x) = cx\) ✓; \(z\)-component \(\partial_x(-\tfrac c3yz) - \partial_y(\tfrac c3xz) = 0\) ✓. Divergence: \(\tfrac c3(z - z + 0) = 0\): Coulomb gauge.

3. \(\partial_yf = cy\) and \(-\partial_xf = cx\) give \(f = \tfrac c2(y^2 - x^2)\): \(\vec A_2 = \big(0,\ 0,\ \tfrac c2(y^2 - x^2)\big)\), also divergence-free. The difference \(\vec A_2 - \vec A_1 = \big(-\tfrac c3xz,\ \tfrac c3yz,\ \tfrac c6(y^2 - x^2)\big) = \nabla\chi\) with \[ \chi = \frac c6\,z\,(y^2 - x^2) . \] (Check: \(\partial_x\chi = -\tfrac c3xz\), \(\partial_y\chi = \tfrac c3yz\), \(\partial_z\chi = \tfrac c6(y^2-x^2)\) ✓.) \(\nabla^2\chi = \tfrac c6z(2 - 2) = 0\): harmonic, as it must be, since both potentials are in Coulomb gauge and \(\nabla\cdot\nabla\chi = 0\).

4. \(|\vec B| = c\sqrt{x^2+y^2} = cs = 10\times0.01 = \mathbf{0.100\ T}\).

Check. Dimensions: \([A] = [c]\,\mathrm m^2 = \mathrm{T\,m}\) ✓. \(\vec A_2\) is the natural one for the lens: a single component along the beam, the analogue of the wire's \(A_z\propto\log s\). In a current-free region the field also has a scalar potential, \(\vec B = -\nabla\psi\) with \(\psi = -cxy\), harmonic ✓.

Idea. Symmetry and the two integral laws fix \(\vec B\); Stokes fixes \(\vec A\); the outside region has a hole, and \(\vec A\) there is the vortex.

1. The configuration is invariant under rotations about and translations along \(z\), so the cylindrical components depend on \(s\) only. Azimuthal: Ampère on the circle of radius \(s\) in a plane \(z = \) const: the current is azimuthal, so none crosses the disc; \(2\pi sB_\phi = 0\). Radial: \(\nabla\cdot\vec B = 0\), so the flux out of a coaxial cylinder of radius \(s\), length \(L\) is zero; the end faces cancel (same \(B_z(s)\) on both, opposite normals), leaving \(2\pi sLB_s = 0\). Axial: a rectangle in a plane through the axis, vertical sides of length \(\ell\) at \(s_1 \lt s_2\) (the horizontal sides contribute nothing since \(B_s = 0\)): \((B(s_1) - B(s_2))\ell = \mu_0nI\ell\) if \(s_1 \lt R \lt s_2\), and \(0\) if both are on the same side. So \(B\) is constant outside, hence \(0\) by the hypothesis at infinity; and constant inside, equal to \(\mu_0nI\).

2. Try \(\vec A = A(s)\hat\phi\). Stokes on the disc of radius \(s\): \(2\pi sA(s) = \) flux through it, which is \(\mu_0nI\pi s^2\) inside and \(\Phi = \mu_0nI\pi R^2\) outside: \[ \vec A = \tfrac12\mu_0nI\,s\,\hat\phi\ \ (s \le R),\qquad \vec A = \frac{\Phi}{2\pi s}\hat\phi\ \ (s \ge R). \] Inside this is \(\tfrac12\vec B\times\vec x\), whose curl is \(\vec B\) (worked example 4). Outside, in Cartesian form \(\vec A = \frac{\Phi}{2\pi}\frac{(-y, x, 0)}{x^2+y^2}\), whose curl is zero (unit 04) — and \(\vec B = \vec 0\) there ✓. Divergence: inside, \(\nabla\cdot\tfrac12\vec B\times\vec x = 0\) (worked example 4); outside, in Cartesian form, \(\partial_x\frac{-y}{x^2+y^2} + \partial_y\frac{x}{x^2+y^2} = \frac{2xy - 2xy}{(x^2+y^2)^2} = 0\). Coulomb gauge.

3. Outside, \(\vec A = \frac{\Phi}{2\pi}\vec F_{\mathrm V}\) with \(\vec F_{\mathrm V}\) unit 04's vortex: closed, with circulation \(\frac{\Phi}{2\pi}\cdot2\pi = \Phi\) round any loop encircling the solenoid once. If \(\vec A = \nabla\chi\) on the outside region, every circulation there would vanish — so no such \(\chi\) exists. On the half-space \(x \gt R\) (a box, unit 04), \(\vec F_{\mathrm V} = \nabla\arctan(y/x)\), so with \(\chi = \frac{\Phi}{2\pi}\arctan(y/x)\) the gauge-transformed \(\vec A - \nabla\chi\) vanishes there. It cannot be extended round the solenoid: \(\chi\) would have to increase by \(\Phi\) on a circuit.

4. \(B = 4\pi\times10^{-7}\times1000\times2.00 = \mathbf{2.513\ mT}\); \(\Phi = B\pi(0.015)^2 = \mathbf{1.777\ \mu Wb}\); at \(s = 0.03\) m, \(A = \Phi/(2\pi\times0.03) = \mathbf{9.42\times10^{-6}\ T\,m}\), where \(\vec B = \vec 0\).

Check. Dimensions: \([\mu_0nI] = \mathrm{N\,A^{-2}\cdot m^{-1}\cdot A} = \mathrm T\) ✓; \([\Phi/s] = \mathrm{T\,m}\) ✓. At \(s = R\) both formulas give \(\tfrac12\mu_0nIR = 1.885\times10^{-5}\) T m: \(\vec A\) is continuous ✓ (its derivative is not — the surface current). Limit \(R\to0\) at fixed \(\Phi\): an ideal flux line, \(\vec A = \frac{\Phi}{2\pi}\vec F_{\mathrm V}\) everywhere off the axis and \(\vec B\) zero everywhere except on it — a magnetic field entirely hidden, with a potential that remembers it.

Idea. On the axis every element is at the same distance, and symmetry cancels the transverse parts.

1. \(\mathrm d\vec\ell\times(\vec x - \vec y) = a(-\sin\phi,\cos\phi,0)\times(-a\cos\phi, -a\sin\phi, z)\,\mathrm d\phi = \big(az\cos\phi,\ az\sin\phi,\ a^2\big)\,\mathrm d\phi\). The first two integrate to zero over \([0,2\pi]\). So \[ \vec B(z\hat z) = \frac{\mu_0I}{4\pi}\frac{2\pi a^2}{(a^2+z^2)^{3/2}}\hat z = \frac{\mu_0Ia^2}{2(a^2+z^2)^{3/2}}\hat z . \]

2. For \(|z|\gg a\), \((a^2+z^2)^{3/2}\approx|z|^3\), so \(B\approx\frac{\mu_0Ia^2}{2|z|^3} = \frac{\mu_0(I\pi a^2)}{2\pi|z|^3}\). It has the electric dipole's form with \(p/\varepsilon_0 \leftrightarrow \mu_0m\): a small loop is a magnetic dipole of moment current × area.

3. Centre: \(B = \mu_0I/2a = 4\pi\times10^{-7}\times10/0.1 = \mathbf{1.257\times10^{-4}\ T}\). At \(z = 0.1\) m: \(\frac{\mu_0\times10\times0.0025}{2(0.0125)^{3/2}} = \mathbf{1.124\times10^{-5}\ T}\). Dipole approximation: \(\mu_0\times10\times\pi\times0.0025/(2\pi\times0.001) = 1.571\times10^{-5}\) T. Ratio \(\big(z^2/(a^2+z^2)\big)^{3/2} = 0.8^{3/2} = \mathbf{0.716}\) — at two radii the dipole approximation is still 40 % high.

Check. Dimensions: \([\mu_0I/a] = \mathrm T\) ✓. Ampère check along the whole axis, a line that closes through infinity and links the loop once: \(\int_{-\infty}^{\infty}\frac{\mu_0Ia^2\,\mathrm dz}{2(a^2+z^2)^{3/2}} = \frac{\mu_0I}{2}\Big[\frac{z}{\sqrt{a^2+z^2}}\Big]_{-\infty}^{\infty} = \mu_0I\) ✓. Direction: anticlockwise current, field along \(+\hat z\) by the right-hand rule ✓. Symmetric in \(z\to-z\), as the loop is ✓.

Idea. Either gauge gives the Lorentz force; each makes a different conservation law obvious.

1. \(L = \tfrac12m(\dot x^2 + \dot y^2 + \dot z^2) + \tfrac{qB}2(x\dot y - y\dot x)\). Euler–Lagrange: \(m\ddot x = qB\dot y\), \(m\ddot y = -qB\dot x\), \(m\ddot z = 0\) — the components of \(q\dot{\vec x}\times B\hat z\) ✓. With \(w = \dot x + \mathrm i\dot y\): \(m\dot w = qB(\dot y - \mathrm i\dot x) = -\mathrm iqBw\), so \(w = w_0\mathrm e^{-\mathrm i\omega_ct}\): the velocity rotates at constant speed with \(\omega_c = qB/m\), clockwise seen from \(+z\) for \(q \gt 0\). Constant speed and constant turning rate: a circle of radius \(v_\perp/\omega_c\).

2. \(L = \tfrac12m|\dot{\vec x}|^2 + qBx\dot y\). No \(y\): \(p_y = m\dot y + qBx\) is conserved. Integrating \(m\ddot y = -qB\dot x\) once gives \(m\dot y = -qB(x - x_c)\) for a constant \(x_c\), and on the circle \(x_c\) is the \(x\)-coordinate of its centre (where \(\dot y\) changes sign). So \(p_y = qBx_c\): the Landau gauge makes the position of the guiding centre a conserved momentum, where the symmetric gauge made angular momentum about the origin one.

3. \(\omega_c = eB/m_e = 1.602\times10^{-19}/9.109\times10^{-31} = \mathbf{1.759\times10^{11}\ s^{-1}}\); \(\omega_c/2\pi = \mathbf{27.99\ GHz}\); radius \(v/\omega_c = 10^6/1.759\times10^{11} = \mathbf{5.69\ \mu m}\).

Check. Dimensions: \([qB/m] = \mathrm{C\,T\,kg^{-1}} = \mathrm{s^{-1}}\) ✓. The frequency is independent of speed — why the cyclotron works — and the energy \(\tfrac12m|\dot{\vec x}|^2\) is constant, as the no-work theorem requires ✓. The two Lagrangians differ by \(\frac{\mathrm d}{\mathrm dt}\big(\tfrac{qB}2xy\big)\), the total derivative of \(q\chi\) with worked example 4's \(\chi\) ✓.

Idea. A sphere round the origin cannot be shrunk in the domain, and the flux through it is the obstruction; remove a half-line to make the domain star-shaped (about any point of the remaining half-axis), and the obstruction reappears on the overlap as unit 04's vortex.

1. \(\nabla\cdot(\vec x/r^3) = 3/r^3 - 3r^2/r^5 = 0\) (unit 05). Flux through the unit sphere: \(g\cdot4\pi\). If \(\vec B = \nabla\times\vec A\) with \(\vec A\in C^1\), Stokes on the northern hemisphere gives \(\oint_{\text{equator}}\vec A\cdot\mathrm d\vec r\), anticlockwise; on the southern hemisphere, whose outward normal induces the opposite orientation of the equator, it gives minus that. The total flux is \(0 \ne 4\pi g\). Contradiction.

2. \(r + z = 0\) exactly when \(x = y = 0\), \(z \le 0\); elsewhere \(f = g/(r(r+z))\) is \(C^\infty\), hence so is \(\vec A_N\). With \(r_i = x_i/r\): \(\partial_x(r^2 + rz) = 2x + zx/r = x(2r+z)/r\), similarly for \(y\), and \(\partial_z(r^2 + rz) = 2z + r + z^2/r = (r+z)^2/r\). So \(f_z = -g/r^3\) and \(xf_x + yf_y = -g\frac{s^2(2r+z)}{r^3(r+z)^2} = -g\frac{(r-z)(2r+z)}{r^3(r+z)}\). Then \[ 2f + xf_x + yf_y = \frac{g\big[2r^2 - (r-z)(2r+z)\big]}{r^3(r+z)} = \frac{g\big[rz + z^2\big]}{r^3(r+z)} = \frac{gz}{r^3}. \] With the first two components \(-xf_z = gx/r^3\), \(-yf_z = gy/r^3\): \(\nabla\times\vec A_N = g\vec x/r^3\) ✓.

3. \(\vec A_S\) is \(\vec A_N\) under \(z\to-z\) with an extra sign, and the same computation gives \(\nabla\times\vec A_S = \vec B\). On the overlap (off the whole \(z\)-axis), \[ \vec A_N - \vec A_S = g(-y,x,0)\Big[\frac1{r(r+z)} + \frac1{r(r-z)}\Big] = g(-y,x,0)\frac{2r}{r(r^2 - z^2)} = 2g\,\frac{(-y,\ x,\ 0)}{x^2+y^2} = 2g\,\vec F_{\mathrm V}, \] unit 04's vortex: closed (as it must be — both have curl \(\vec B\)) and not exact, circulation \(4\pi g\) round the axis. Locally it is \(\nabla(2g\phi)\).

4. The monopole is a closed 2-form-like field (zero divergence) that is not a curl on \(\mathbb R^3\setminus\{0\}\), the degree-two failure of the Poincaré lemma. Removing a half-line makes a region on which a vector potential exists (\(\vec A_N\) or \(\vec A_S\)). The two potentials differ on the overlap by a closed field that is not a gradient, the degree-one failure; and the circulation \(4\pi g\) of that difference is exactly the flux that part 1 said could not be produced by a single potential. The obstruction has not disappeared; it has moved down a degree, onto the overlap. Unit 34 turns this into the statement that \(H^2\) of the sphere is built from \(H^1\) of the overlap of its two hemispheres.

Check. Dimensions: \([g] = \mathrm{T\,m^2}\) (flux), \([A] = [g]/\mathrm m\) ✓. On the upper axis \(z \gt 0\), \(r(r+z)\approx2z^2\) and \(\vec A_N\to0\) smoothly as \(x, y\to0\) ✓; on the lower axis it blows up as \(1/s\) — the "Dirac string", carrying the flux \(4\pi g\) back up to the origin. Physically: if a monopole existed, \(\vec B\) would be the whole field and the string an artefact of the gauge; Dirac showed in 1931 that quantum mechanics hides it only if \(qg\) is quantised.

Idea. A decaying field with no divergence and no curl is harmonic and decaying, hence zero. Everything else follows from applying that to \(\vec E\) and \(\vec B\).

1. Let \(\vec F\in C^2(\mathbb R^3)\) with \(\vec F(\vec x)\to\vec 0\) as \(|\vec x|\to\infty\), and suppose \(D = \nabla\cdot\vec F\) and \(\vec C = \nabla\times\vec F\) are \(C^1\) with compact support. Then \(\vec F = -\nabla\Phi + \nabla\times\vec A\) with \(\Phi = \frac1{4\pi}\int\frac{D(\vec y)}{|\vec x-\vec y|}\mathrm dV\), \(\vec A = \frac1{4\pi}\int\frac{\vec C(\vec y)}{|\vec x-\vec y|}\mathrm dV\), and \(\vec F\) is the only field tending to \(\vec 0\) with divergence \(D\) and curl \(\vec C\).

2. Let \(\vec F_1, \vec F_2\) both qualify; \(\vec W = \vec F_1 - \vec F_2\) is \(C^2\), tends to \(\vec 0\), and has zero divergence and curl. By the curl-curl identity (proved in the deck's Biot–Savart notes by Clairaut), \(\nabla^2W_i = 0\) for each Cartesian component. By the maximum principle (unit 05) on the ball \(B_R\), \(|W_i(\vec x)| \le \max_{|\vec y| = R}|W_i(\vec y)|\) for \(|\vec x| \le R\) (apply it to \(W_i\) and \(-W_i\)). Fix \(\vec x\) and let \(R\to\infty\): the right side tends to \(0\). So \(\vec W = \vec 0\). \(\square\)

3. Decay: \(\vec F = \hat z\) has zero divergence and curl and is not \(\vec 0\); so is \(\nabla h\) for any harmonic polynomial \(h\), e.g. \(\nabla(xy) = (y, x, 0)\). Without decay, the divergence and curl determine \(\vec F\) only up to such fields. Independence: the curl of anything has zero divergence, so a prescribed \(\vec C\) with \(\nabla\cdot\vec C \ne 0\) — say \(\vec C = (x,0,0)\eta(\vec x)\) with \(\eta\) a bump equal to \(1\) near the origin — is the curl of no field at all.

4. By part 2 each of \(\vec E\), \(\vec B\) is fixed by its divergence and curl, and nothing less fixes it (part 3's non-uniqueness returns if one is dropped: add \(\nabla\times\vec G\) or \(\nabla h\) with decay). So the theory must specify exactly four things, \(\nabla\cdot\vec E\), \(\nabla\times\vec E\), \(\nabla\cdot\vec B\), \(\nabla\times\vec B\), in terms of the sources: four equations. Count: the divergences are one function each, the curls — divergence-free by necessity — two functions' worth each, so \(3 + 3 = 6\) functions determined, matching the six components. In statics these are Gauss (\(\rho/\varepsilon_0\)), \(\vec 0\), \(0\), and \(\mu_0\vec J\). The constraint of part 3 applied to \(\nabla\times\vec B = \mu_0\vec J\) forces \[ \nabla\cdot\vec J = 0 , \] steady currents. When charge accumulates, continuity gives \(\nabla\cdot\vec J = -\partial_t\rho \ne 0\), the static curl equation becomes inconsistent, and the right side must gain a term whose divergence is \(\mu_0\partial_t\rho\). Since \(\partial_t\rho = \varepsilon_0\partial_t\nabla\cdot\vec E\), the term \(\mu_0\varepsilon_0\partial_t\vec E\) does it — Maxwell's displacement current, which unit 24 derives exactly this way.

Check. Test the statement on the running example: the total field \(-\nabla\varphi\) does not decay and Helmholtz does not apply to it; the induced part does, and is fixed by \(\sigma = 3\varepsilon_0E_0\cos\theta\) alone — the deck's third derivation ✓. Dimensions of the predicted term: \([\mu_0\varepsilon_0\partial_t E] = \mathrm{s^2m^{-2}}\cdot\mathrm{V\,m^{-1}s^{-1}} = \mathrm{T\,m^{-1}} = [\nabla\times\vec B]\) ✓, using \(\mu_0\varepsilon_0 = 1/c^2\) and \(\mathrm V = \mathrm{T\,m^2\,s^{-1}}\).