 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2/2/ii/solution

For large

∣ z ∣

, expand in

a^{2} / z^{2}

to obtain

ψ (0, z) = \frac{Q}{4 π ϵ _{0}} [\frac{1}{∣ z ∣} - \frac{a ^{2}}{4∣ z ∣ ^{3}} + O (a^{4} /∣ z ∣^{5})] .

The axisymmetric exterior multipole expansion replaces each

∣ z ∣^{- n - 1}

on the positive axis by

r^{- n - 1} P_{n} (cos θ)

. Reflection symmetry eliminates the odd multipoles, so

ψ (r, θ) = \frac{Q}{4 π ϵ _{0}} [\frac{1}{r} - \frac{a ^{2}}{4 r ^{3}} P_{2} (cos θ) + O (a^{4} / r^{5})], r ≫ a .

The 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.

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