 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-4/10a/b/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Let

ρ = a / 3

be the distance from the origin to each fixed charge, and denote their position vectors by

R_{j}

. For a small planar displacement

r

\frac{1}{∣ R _{j} - r ∣} = \frac{1}{ρ} + \frac{R _{j} \cdot r}{ρ ^{3}} + \frac{3 ( R _{j} \cdot r ) ^{2} - ρ ^{2} ∣ r ∣ ^{2}}{2 ρ ^{5}} + O (∣ r ∣^{3})

can be summed using the threefold symmetry relations

\sum_{j} R_{j} = 0, \sum_{j} R_{j} R_{j}^{T} = \frac{3 ρ ^{2}}{2} I .

The electric potential due to the fixed charges is consequently

Φ (r) = \frac{q}{4 π ϵ _{0}} (\frac{3}{ρ} + \frac{3∣ r ∣ ^{2}}{4 ρ ^{3}} + O (∣ r ∣^{3})) .

U = q Φ

. Because the quadratic coefficient is positive, the origin is a strict local minimum of

U

, and the equilibrium is stable.

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