How this instrument works
A dipole is nothing more exotic than two equal and opposite charges held a fixed distance apart — a positive pole and a negative pole, separated by d. Multiply the pole charge by that separation and you get the dipole moment, p = q·d, measured in coulomb-metres. It stands in for the whole arrangement as a single number: double either the charge or the gap and the moment doubles with it, but a large charge crammed onto a tiny separation and a modest charge held further apart can share the identical moment, and from a distance their fields look the same.
The on-axis field, E = 2kp ⁄ r³, comes from adding the two poles' individual point-charge fields and watching most of each cancel the other. Stand on the line running through both charges: the near pole's field is slightly stronger than the far pole's, simply because it sits slightly closer, and that leftover difference, not either field alone, is what survives at long range. Expanding the difference for r much larger than d turns two 1/r² terms into one 1/r³ term — a full power of r weaker than either pole would produce by itself, which is why dipole fields die out so much faster with distance than a lone charge's does.
That shortcut assumes r is large next to d and that you are standing precisely on the axis through both poles; step off that line, or move in close enough that r and d are comparable, and it collapses back into two separate inverse-square calculations at the individual pole distances. Chemists lean on the same p = q·d relationship to describe polar molecules like water, whose bent shape leaves a permanent moment near 6.2 × 10⁻³⁰ C·m; atmospheric physicists model a thundercloud the same way, treating its separated layers of positive and negative charge as one large dipole to predict the field a storm produces at ground level.
- Enter Charge of each pole — the magnitude shared by both the positive and negative charge, in nC or µC; the two poles are assumed equal and opposite.
- Set Separation between charges to the physical gap between the two poles, in mm or cm — this is d in the moment formula, not the distance to your observation point.
- Set Distance from dipole center, along axis to how far out, along the line through both poles, you want the field evaluated; keep it well beyond Separation for the far-field formula to hold.
- Read Dipole moment, C·m for p = q·d, a fixed property of the charge pair that does not depend on where you measure it.
- Read On-axis electric field strength, N ⁄ C for the field at the distance you set, falling off as the cube of that distance.
Worked example — two 10 nC charges, 2 mm apart, at 1 m
Set Charge of each pole to 10 nC (1 × 10⁻⁸ C) and Separation between charges to 2 mm (0.002 m) — roughly the pole spacing inside a small electret element. The dipole moment comes straight from the first formula: p = q·d = (1 × 10⁻⁸ C)(0.002 m) = 2 × 10⁻¹¹ C·m, a fixed number that stays the same no matter where you go on to measure the field.
Set Distance from dipole center, along axis to 1 m — five hundred times the pole separation, comfortably inside the far-field regime — and E = 2kp ⁄ r³ = 2(8.9875518 × 10⁹)(2 × 10⁻¹¹) ⁄ 1³ ≈ 0.3595 N/C. A single 10 nC point charge on its own would produce about 89.9 N/C at that same 1 m distance; pairing it with an equal and opposite charge only 2 mm away cuts the field by a factor of roughly 250, the dipole's steep 1/r³ falloff at work.
Questions
Why does the dipole field fall off as 1/r³ instead of 1/r² like a single charge?
Because the two poles' fields nearly cancel. Each pole alone would create a field that falls as 1/r², but on the axis the near pole's slightly larger field only partly outweighs the far pole's slightly smaller one; expanding that difference for r much bigger than d turns the leading term into 1/r³ — one extra power of r weaker, which is why dipole fields fade faster with distance than a lone charge's does.
Does the E = 2kp ⁄ r³ formula work anywhere around the dipole, or only on the axis?
Only on the axis running through both poles. Off that line the field has a different magnitude and a different direction — along the perpendicular bisector, for instance, the field is half this size and points the opposite way. The general dipole field needs an angular term that only vanishes for points on the axis, which is the special case this sheet computes.
How much bigger than the separation does the distance need to be?
As a rule of thumb, at least ten times bigger for accuracy within a few percent, and a hundred times bigger for the far-field formula to be essentially exact. At r equal to ten times the separation, the exact two-charge calculation and the 2kp/r³ shortcut already agree to within about 1 percent; push r down toward d and the individual pole fields need to be added directly instead.
What is Coulomb's constant k, and why does it show up here?
k = 1/(4πε₀) ≈ 8.9875518 × 10⁹ N·m²/C² is the same proportionality constant from Coulomb's law, because the dipole field is built by adding two point-charge fields together and simplifying. Nothing new enters the physics — only algebra that combines the pair of 1/r² terms into one 1/r³ term, while k rides along unchanged.
Is 2 × 10⁻¹¹ C·m, the worked example's moment, a realistic figure?
It sits in the range of small engineered or lab dipoles rather than everyday charged objects. A single water molecule's permanent dipole moment is about 6.2 × 10⁻³⁰ C·m, vastly smaller because its charges sit only atomic distances apart, while the 10 nC pair 2 mm apart used here is closer to what a bench electrostatics setup or a small electret element might produce.
Does the same 2kp ⁄ r³ shape describe a magnetic dipole too?
The exponent matches — an ideal magnetic dipole's axial field also falls as 1/r³ — but the formula does not transfer directly: magnetic dipole moment is measured in ampere-square-metres, not coulomb-metres, and the leading constant becomes μ₀/(2π) instead of 2k. Treat the resemblance as a family likeness between two different quantities, not an interchangeable formula.