 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015/9/a/solution

p = 2 a q \hat{z}

V (r) = \frac{q}{4 π ϵ _{0}} [\frac{1}{r ^{2} + a ^{2} - 2 a r c o s θ} - \frac{1}{r ^{2} + a ^{2} + 2 a r c o s θ}] .

For

a / r ≪ 1

, the binomial expansion gives each reciprocal distance as

r^{- 1} (1 \pm (a / r) cos θ + O (a^{2} / r^{2}))

. The leading

1/ r

terms cancel because the total electric charge is zero. The next even terms also cancel between the two signs. Consequently

V (r) = \frac{2 a q}{4 π ϵ _{0}} \frac{c o s θ}{r ^{2}} + O (\frac{q a ^{3}}{4 π ϵ _{0} r ^{4}}), α = \frac{2 a q}{4 π ϵ _{0}} = \frac{p}{4 π ϵ _{0}} .

This is the leading electric dipole term of the multipole expansion; it describes the far field, not the electric potential arbitrarily close to either point charge.

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