How fast a dipole's field falls off.

The field of a dipole probed on its axis and equator, arrows to scale, with a log-log plot of E against r beside a single charge's.

Find the equatorial distance where the field equals the axial field at 20 cm (≈15.9 cm), and confirm doubling r cuts E to one-eighth.

Using a 1/r² law for a dipole, or pointing the equatorial field along p.

• An electric dipole is two charges +q and −q, equal in size, held a distance 2a apart. Its dipole moment is p = q × 2a, a vector pointing from −q to +q. The SI unit is C m.
• Its total charge is zero, but its field is not: the fields of +q and −q do not cancel exactly, and far away the dipole field falls off as 1/r³, faster than the 1/r² of a single charge.
• On the axis at distance r from the centre: E = 2kp/r³ (r >> a), directed along p. The exact expression is E = k·4qar/(r² − a²)².
• On the equatorial plane at distance r: E = kp/r³ (r >> a), directed opposite to p. The exact expression is E = k·2qa/(r² + a²)^(3/2).
• At the same large distance, the axial field is twice the equatorial field.
• Far fields depend on q and a only through the product p. A point dipole is the limit 2a → 0 with p kept finite; for it the 1/r³ formulas are exact.
• Polar molecules such as H₂O have a permanent dipole moment because their centres of positive and negative charge do not coincide; CO₂ and CH₄ have zero dipole moment but acquire an induced one in an applied field.

Teaching this idea? Put the simulation on your own page or LMS with this code.