Derive an expression for the electric potential at a point due to a short electric dipole. Hence find the potential on its axial and equatorial lines.

1. 1 · Draw the figure. [−q at A, +q at B, centre O, AB = 2a; point P at distance r, at angle θ to p; r₁ = BP, r₂ = AP]
2. 2 · Add the potentials. V = (q/4πε₀)(1/r₁ − 1/r₂).
3. 3 · Use the geometry. r₁² = r² + a² − 2ar cos θ; r₂² = r² + a² + 2ar cos θ.
4. 4 · Approximate for r ≫ a. By the binomial approximation, 1/r₁ − 1/r₂ ≈ 2a cos θ/r², so V = p cos θ/(4πε₀r²).
5. 5 · Special cases. Axial (θ = 0°): V = p/(4πε₀r²). Equatorial (θ = 90°): V = 0.
6. 6 · Close in one line. Dipole potential falls as 1/r² and depends on direction through cos θ.

Scoring tip: State 'r ≫ a' before you drop terms. Without it, the approximation step loses its mark.