Paper 2, Section II, C

Methods | Part IB, 2015

(i) The Laplace operator in spherical coordinates is

2=1r2r(r2r)+1r2sinθθ(sinθθ)+1r2sin2θ2ϕ2\vec{\nabla}^{2}=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \frac{\partial}{\partial r}\right)+\frac{1}{r^{2} \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial}{\partial \theta}\right)+\frac{1}{r^{2} \sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}}

Show that general, regular axisymmetric solutions ψ(r,θ)\psi(r, \theta) to the equation 2ψ=0\vec{\nabla}^{2} \psi=0 are given by

ψ(r,θ)=n=0(Anrn+Bnr(n+1))Pn(cosθ)\psi(r, \theta)=\sum_{n=0}^{\infty}\left(A_{n} r^{n}+B_{n} r^{-(n+1)}\right) P_{n}(\cos \theta)

where An,BnA_{n}, B_{n} are constants and PnP_{n} are the Legendre polynomials. [You may use without proof that regular solutions to Legendre's equation ddx[(1x2)ddxy(x)]=λy(x)-\frac{d}{d x}\left[\left(1-x^{2}\right) \frac{d}{d x} y(x)\right]=\lambda y(x) are given by Pn(x)P_{n}(x) with λ=n(n+1)\lambda=n(n+1) and non-negative integer nn.]

(ii) Consider a uniformly charged wire in the form of a ring of infinitesimal width with radius r0=1r_{0}=1 and a constant charge per unit length σ\sigma. By Coulomb's law, the electric potential due to a point charge qq at a point a distance dd from the charge is

U=q4πϵ0dU=\frac{q}{4 \pi \epsilon_{0} d}

where ϵ0\epsilon_{0} is a constant. Let the zz-axis be perpendicular to the circle and pass through the circle's centre (see figure). Show that the potential due to the charged ring at a point on the zz-axis at location zz is given by

V=σ2ϵ01+z2.V=\frac{\sigma}{2 \epsilon_{0} \sqrt{1+z^{2}}} .

(iii) The potential VV generated by the charged ring of (ii) at arbitrary points (excluding points directly on the ring which can be ignored for this question) is determined by Laplace's equation 2V=0\vec{\nabla}^{2} V=0. Calculate this potential with the boundary condition limrV=0\lim _{r \rightarrow \infty} V=0, where r=x2+y2+z2r=\sqrt{x^{2}+y^{2}+z^{2}}. [You may use without proof that

11+x2=m=0x2m(1)m(2m)!22m(m!)2\frac{1}{\sqrt{1+x^{2}}}=\sum_{m=0}^{\infty} x^{2 m}(-1)^{m} \frac{(2 m) !}{2^{2 m}(m !)^{2}}

for x<1|x|<1. Furthermore, the Legendre polynomials are normalized such that Pn(1)=1.]\left.P_{n}(1)=1 .\right]

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