--- title: "Dipole field on axial line" subject: "Engineering Physics" chapter: 7 key: "semester_1/engineering_physics:dipole-axial-field-inverse-cube" source: https://www.bctengineer.com/q/semester_1/engineering_physics/dipole-axial-field-inverse-cube site: BCT Engineering --- # Dipole field on axial line Engineering Physics · Chapter 7: Electrostatics - Asked in: 2074 Asoj, 5 marks - Video: [Electric Field at a Point on Axial Line of Dipole | Class 12 Physics Chapter 1 Derivations](https://www.youtube.com/watch?v=r1l8vqxY7ds) — Mandeep Education Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/dipole-axial-field-inverse-cube - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7 ## Question What is an electric dipole and dipole moment? Show that the electric field for a short dipole drops inversely to the cube of the distance at any point from the dipole on an axial line. ## Short answer An electric dipole consists of equal and opposite charges ±q separated by 2a; dipole moment p = 2qa from −q to +q. On the axial line at distance r ≫ a, E = (1/4πε₀)·2p/r³, so E ∝ 1/r³. ## Answer An **electric dipole** is a pair of equal and opposite point charges +q and −q separated by a small distance 2a. The **dipole moment** is the vector p = q(2a), of magnitude 2aq, directed from −q to +q. Its unit is C·m. #### Field on the axial line Let O be the centre of the dipole and P a point on the axis at distance r from O (r > a). ```text -q O +q P --(-)---------+---------(+)------------------.--> <--- a ---> <--- a ---> <------------------ r -------------------> ``` 1. Distance of P from +q is (r − a); from −q is (r + a). 2. Field of +q at P (away from +q, along p): E₁ = (1/4πε0) q/(r − a)² 3. Field of −q at P (towards −q, opposite to p): E₂ = (1/4πε0) q/(r + a)² 4. E₁ is larger and the fields are collinear, so the net field is along p: ```text E = E₁ − E₂ = (q/4πε0) [ 1/(r−a)² − 1/(r+a)² ] = (q/4πε0) · 4ar / (r² − a²)² = (1/4πε0) · 2pr / (r² − a²)² (p = 2aq) ``` 5. For a short dipole, a ≪ r, so r² − a² ≈ r²: **E = (1/4πε0) · 2p / r³** Hence E ∝ 1/r³: the field of a short dipole falls off inversely as the cube of the distance, faster than the 1/r² of a single charge, because the fields of +q and −q nearly cancel at large distance. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which is the dipole moment and its unit? A. p = q/(2a), in C/m B. p = q(2a), in N/C C. p = 2q, in C D. p = q(2a), in C·m, from -q to +q Answer: D. Dipole moment is charge times separation. (Key point: Dipole moment) 2. On the axial line, why is the net field directed along p? A. The farther -q gives the larger field B. The two fields are perpendicular C. The nearer +q gives the larger field, E1 > E2, and the fields are collinear and opposite D. The two fields are equal Answer: C. E = E1 - E2 with E1 (from +q) greater. (Key point: Net field on axis) 3. For a << r, the exact E = (1/4πε0)·2pr/(r^2 - a^2)^2 reduces to what? A. E = (1/4πε0)·p/r^2 B. E = (1/4πε0)·2p/r^3 C. E = (1/4πε0)·2p/r^2 D. E = (1/4πε0)·p/r^3 Answer: B. r^2 - a^2 ≈ r^2 so (r^2)^2 = r^4 and 2pr/r^4 = 2p/r^3. (Key point: Short dipole result) 4. Why does a dipole's field fall off faster (1/r^3) than a single charge's (1/r^2)? A. The fields of +q and -q nearly cancel at large distance B. The charges move closer C. The net charge increases D. The charge is spread over a larger area Answer: A. Only a small difference remains between the two nearly equal fields. 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