--- title: "Dipole potential; no work along bisector" subject: "Engineering Physics" chapter: 7 key: "semester_1/engineering_physics:dipole-potential-no-work-perpendicular" source: https://www.bctengineer.com/q/semester_1/engineering_physics/dipole-potential-no-work-perpendicular site: BCT Engineering --- # Dipole potential; no work along bisector Engineering Physics · Chapter 7: Electrostatics - Asked in: 2073 Shrawan, 5 marks - Video: [Potential due to dipole (logical derivation) | Electric potential & cap. | Physics | Khan Academy](https://www.youtube.com/watch?v=utY112E_NIc) — Khan Academy India - English - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/dipole-potential-no-work-perpendicular - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7 ## Question Derive the relation for potential at any point due to an electric dipole and show that no work is done in bringing a charge from infinity to the dipole along the perpendicular bisector of the dipole. ## Short answer Potential due to a dipole is V = (1/4πε₀) p cosθ/r². On the perpendicular bisector θ = 90°, so V = 0 everywhere; work W = q₀(V_P − V_∞) = 0, so no work is done. ## Answer #### Potential due to a dipole Dipole: charges +q and −q, separation 2a, moment p = 2aq. Take point P at distance r from the centre O, where OP makes angle θ with the dipole axis, and r ≫ a. ```text P .' | r .' | .' | r1 (from +q) .' θ | -q ---------O-----------+q ------ axis <---- a ---><---- a ---> r1 ≈ r − a cosθ (from +q) r2 ≈ r + a cosθ (from −q) ``` 1. V = (q/4πε0)(1/r₁ − 1/r₂) = (q/4πε0)(r₂ − r₁)/(r₁r₂) 2. From the geometry (for r ≫ a): r₂ − r₁ ≈ 2a cosθ and r₁r₂ ≈ r². 3. V = (q/4πε0)(2a cosθ)/r² **V = (1/4πε0) · p cosθ / r²** Special cases: on the axis (θ = 0), V = p/(4πε0 r²); on the equatorial line (θ = 90°), V = 0. #### No work along the perpendicular bisector 1. Every point on the perpendicular bisector has θ = 90°, so cos θ = 0 and V = 0 at all those points, including at infinity (V = 0 there). 2. Work done in bringing a test charge q₀ from infinity to a point P on the bisector: W = q₀ (V_P − V_∞) = q₀ (0 − 0) = 0 Equivalently, the forces from +q and −q on the test charge have equal magnitudes, and their components along the bisector cancel; only a component parallel to the axis remains, which is perpendicular to the path, so it does no work. **Hence no work is done in bringing a charge from infinity to the dipole along its perpendicular bisector; the bisector is an equipotential at zero potential.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What approximations give V = p cos θ/(4πε0 r^2)? A. r2 - r1 ≈ 2a cos θ and r1 r2 ≈ r^2 for r >> a B. r2 - r1 ≈ 2a and r1 r2 ≈ a^2 C. r2 - r1 ≈ 2a sin θ and r1 r2 ≈ r^2 D. r1 = r2 = r Answer: A. V = (q/4πε0)(r2 - r1)/(r1 r2) with those substitutions. (Key point: Path difference) 2. What is the potential at every point on the perpendicular bisector? A. Maximum, since cos 90° = 1 B. p/(4πε0 r^2) C. Undefined D. Zero, since θ = 90° makes cos θ = 0 Answer: D. The bisector is an equipotential at V = 0. (Key point: Potential on bisector) 3. Why is the work to bring a charge from infinity along the bisector zero? A. The force from +q and -q is zero everywhere B. The path length is zero C. W = q0 (VP - V∞) = q0 (0 - 0) = 0 D. Work depends on the path and the path is perpendicular Answer: C. Both endpoints are at zero potential, so the potential difference is zero. (Key point: Work) 4. How does the force argument explain zero work along the bisector? A. The net force is along the path but zero in magnitude B. The net force is parallel to the dipole axis, perpendicular to the path, so it does no work C. The net force opposes the path and cancels D. The force from -q exceeds that from +q Answer: B. Force components along the bisector cancel; the remaining force is perpendicular to the displacement. 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