Solution

Let

θ

be the angle between

r

and the positive

z

-axis. By superposition of the electric potential of point charges,

Φ (r) = \frac{Q}{4 π ϵ _{0}} (\frac{N}{r} - \frac{1}{∣ r - d z ∣} - \frac{M}{∣ r + d z ∣}) .

(20)

For

r > d

, the Legendre polynomial expansion gives

\frac{1}{∣ r \mp d z ∣} = \frac{1}{r} [1 \pm \frac{d}{r} cos θ + \frac{d ^{2}}{2 r ^{2}} (3 cos^{2} θ - 1) + O (r^{- 3})] .

(21)

Therefore

Φ (r, θ) = \frac{Q}{4 π ϵ _{0}} [\frac{N - M - 1}{r} + \frac{( M - 1 ) d cos θ}{r ^{2}} - \frac{( M + 1 ) d ^{2} ( 3 cos ^{2} θ - 1 )}{2 r ^{3}} + O (r^{- 4})] .

(22)

The three displayed terms are respectively the monopole, dipole, and quadrupole terms of the electric multipole expansion. In particular, the total charge is

Q (N - M - 1)

and the electric dipole moment is

Q d (M - 1) z

. This is the multipole expansion of three collinear charges.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 5D / a / Solution

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