Potential of an Electric Dipole
An electric dipole consists of two equal and opposite charges, +q and −q, separated by a small distance 2a. While the dipole as a whole is electrically neutral, it creates a non-zero electric potential at surrounding points because the two charges are not at the same location. The potential contributions from the two charges do not perfectly cancel except at points equidistant from both — and that happens only on the perpendicular bisector plane. Think of the dipole like two people pulling a rope from opposite ends: individually they are strong, but together their net pull depends on where you stand. Stand exactly between them and you feel equal tension from both sides. Move closer to one and that person dominates. The electric potential of a dipole works the same way — it depends critically on both the distance from the dipole and the direction (angle) from the dipole axis. The key result is that the dipole potential falls off as 1/r², faster than the 1/r fall-off of a single point charge. This makes sense physically: at large distances the positive and negative charges increasingly 'cancel', leaving only a small residual effect that encodes the geometry of the pair. Understanding dipole potential is foundational for molecular chemistry, antenna theory, and the physics of dielectric materials.
- Set Up Geometry
- Write the Exact Potential
- Apply the Far-Field Approximation (r ≫ a)
- Introduce the Dipole Moment
- Interpret the Angular Dependence
- Special Cases on Axial and Equatorial Lines
Set Up Geometry
Place the dipole along the z-axis with −q at position −a and +q at position +a, so the total separation is 2a. Let P be an observation point at distance r from the centre, making angle θ with the positive z-axis (the dipole axis). Define the distances from P to +q and −q as r₊ and r₋ respectively.
Write the Exact Potential
The potential is a scalar, so we simply add the contributions: V = (kq/r₊) + (k(−q)/r₋) = kq(1/r₊ − 1/r₋). No vector addition is needed — this is the great advantage of working with potential rather than field.
Apply the Far-Field Approximation (r ≫ a)
Using the law of cosines, r₊² = r² + a² − 2ar cosθ and r₋² = r² + a² + 2ar cosθ. For r ≫ a, we can expand using the binomial approximation: 1/r₊ ≈ (1/r)(1 + a cosθ/r) and 1/r₋ ≈ (1/r)(1 − a cosθ/r). Substituting: 1/r₊ − 1/r₋ ≈ 2a cosθ/r².
Introduce the Dipole Moment
We define the electric dipole moment p = q(2a), pointing from −q to +q. Substituting into the expression gives V = kp cosθ / r². This is the standard dipole potential formula valid for r ≫ a. In vector form, p⃗ · r̂ = p cosθ, so V = k(p⃗ · r̂)/r².
Interpret the Angular Dependence
At θ = 0° (along the positive axis, toward +q): V is maximum positive. At θ = 90° (perpendicular bisector): cosθ = 0, so V = 0. At θ = 180° (toward −q): V is maximum negative. These zero-potential surfaces (equipotentials at V = 0) are not points but the entire plane perpendicular to the dipole axis through its centre.
Special Cases on Axial and Equatorial Lines
On the axial line (θ = 0° or 180°): V_axial = ±kp/r². On the equatorial line (θ = 90°): V_equatorial = 0. Note that even though the potential is zero on the equatorial plane, the electric field there is NOT zero — it points anti-parallel to p⃗. This illustrates that zero potential does not mean zero field.
The Key Equation
- V — Electric Potential (V (volts)): The electric potential at the observation point P due to the dipole.
- k — Coulomb's Constant (N·m²/C²): Equal to 1/(4πε₀) ≈ 8.99 × 10⁹ N·m²/C².
- p — Dipole Moment Magnitude (C·m (coulomb-metre)): Product of charge magnitude q and separation 2a: p = q(2a). Measures the strength of the dipole.
- \theta — Polar Angle (degrees or radians (dimensionless)): Angle between the dipole axis (pointing from −q to +q) and the line from the dipole centre to point P.
- r — Distance from Dipole Centre (m (metres)): Distance from the midpoint of the dipole to the field point P. Must satisfy r ≫ a for the approximation to hold.
Worked Example
Problem. A water molecule can be modelled as an electric dipole with dipole moment p = 6.17 × 10⁻³⁰ C·m. Calculate the electric potential at a point P located 1.0 nm from the centre of the molecule at an angle of 60° from the dipole axis. (k = 8.99 × 10⁹ N·m²/C²)
Answer. The electric potential at P is approximately 27.7 mV (millivolts).
Origins · 1 minCharles-Augustin de Coulomb and how this was discovered
Coulomb established the quantitative law of electrostatic force between point charges (Coulomb's Law, 1785), which is the direct foundation upon which the dipole potential formula is built. Without the 1/r² force law, the elegant 1/r² dipole potential would not exist in its present form. Coulomb invented the torsion balance — a device using the twisting of a thin wire — to measure the incredibly tiny forces between charged spheres. This same instrument design was later used by Henry Cavendish to measure the gravitational constant G.
Related Concepts
- Electric Dipole Moment — The dipole moment p = q(2a) is the central quantity in the dipole potential formula. Understanding what p represents and how it is defined is essential before computing V.
- Electric Potential of a Point Charge — The dipole potential is derived by superposing the 1/r potentials of two point charges. Mastery of V = kq/r is the prerequisite step.
- Electric Field of a Dipole — The electric field of the dipole can be obtained from the potential by E = −∇V. Comparing the field and potential illuminates why V = 0 on the equatorial plane does not mean E = 0 there.
- Equipotential Surfaces — The dipole potential produces distinctive equipotential surfaces shaped like nested shells distorted along the dipole axis. Visualising these deepens understanding of the cosθ dependence.