Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 2

Use Green's second identity in the source coordinate:Take $u({\mathbf{x}}^{\prime })=\psi ({\mathbf{x}}^{\prime })$ and $v({\mathbf{x}}^{\prime })=-1/[4\pi |\mathbf{x}-{\mathbf{x}}^{\prime }|]$, the fundamental solution of the Laplace equation satisfying ${\Delta }^{\prime }v=\delta ({\mathbf{x}}^{\prime }-\mathbf{x})$. With localized charge and the decaying potential, the boundary integral tends to zero on expanding spheres. The volume identity then gives $\psi (\mathbf{x})+(1/{ϵ}_0)\int vqd{V}^{\prime }=0$, hence the Coulomb potentialConvergence of this integral and a condition fixing the potential at infinity are needed: mere pointwise decay of $q$ is not enough, and a harmonic function could otherwise be added. A bounded, compactly supported charge distribution is a sufficient setting for this derivation.

For the disk put $\sigma =Q/(\pi a^2)$. On its axis, every ring of radius $\rho$ is at distance $\sqrt{{\rho }^2+z^2}$, soFor $z>0$ subtract $z$; for $z<0$ add $z$. This is the electrostatic potential of a uniformly charged disk with zero potential at infinity. The absolute value is essential for reflection symmetry and the correct behavior on the lower axis.

Write $A=Q/(2\pi {ϵ}_0{a}^2)$. The Taylor expansion $\sqrt{a^2+z^2}=a+z^2/(2a)+O(z^4/a^3)$ gives the near-axis resultTo extend off the axis, work separately above and below the charged plane. In either source-free half-space the regular axisymmetric solid harmonics are $r^n{P}_n(\cos \theta )$. Matching the positive-axis coefficients, using $P_n(1)=1$, and then reflecting in the disk givesHere $P_2(t)=(3t^2-1)/2$ is a Legendre polynomial. The quadratic term is $A(z^2-{\rho }^2/2)/(2a)$. The linear term is $-Az$ above and $+Az$ below; it represents the discontinuity of the normal electric field. A single regular spherical expansion across the entire charged sheet would incorrectly remove that jump.

For large $|z|$, expand in $a^2/z^2$ to obtainThe axisymmetric exterior multipole expansion replaces each $|z{|}^{-n-1}$ on the positive axis by $r^{-n-1}{P}_n(\cos \theta )$. Reflection symmetry eliminates the odd multipoles, soThe leading term is the potential of total charge $Q$ concentrated at the center; the next is its electric quadrupole correction. On either axis $P_2(\pm 1)=1$, recovering the same even axis potential.