Field of a dipole

Take two equal and opposite charges, hold them a short fixed distance apart, and you have built an electric dipole — the same arrangement nature uses inside every polar molecule and every antenna. On its own axis, the two fields nearly cancel but not quite; on the line perpendicular to the axis, they cancel differently again. The result is a field that dies away as $1/r^3$ , far faster than the $1/r^2$ of a lone charge, because a dipole carries zero net charge and "leaks" field only through the tiny separation between its charges. ISC Class 12 asks you to derive this field along two special lines — axial and equatorial — compare their strengths, and use both in numerical problems. This section builds the two formulas from first principles, works a full numerical pair, and flags the sign and direction traps that cost the most marks.

The core insight

An electric dipole's field falls off as $1/r^3$ and is twice as strong on the axial line as on the equatorial line at the same distance, because the two charges' contributions add along the axis but only partially survive off it.

The core idea

An electric dipole is a pair of charges $+q$ and $-q$ separated by a small distance $2a$ . Its dipole moment is the vector

\vec p = q(2a)\,\hat n

where $q$ is the magnitude of either charge, $2a$ is the separation, and $\hat n$ points from $-q$ to $+q$ ; the SI unit of $p$ is coulomb-metre (C·m). Two special points matter for ISC: a point P on the axial line (the line through both charges, at distance $r > a$ from the centre O), and a point P on the equatorial line (the perpendicular bisector of that line, at distance $r$ from O). At each point the total field is the vector sum of the fields due to $+q$ and $-q$ separately — the same superposition idea used for any point-charge system, just applied to two charges that almost cancel. On the axial line the two fields point along the same line, so the algebra reduces to a difference of two inverse-square terms:

E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{(r-a)^2} - \frac{q}{(r+a)^2}\right] = \frac{1}{4\pi\varepsilon_0}\frac{2pr}{(r^2-a^2)^2}

where: $\varepsilon_0$ is the permittivity of free space, $r$ is the distance of P from the centre O, and $a$ is the half-separation. For a short dipole viewed far away ( $r \gg a$ ), $a^2$ is negligible next to $r^2$ , giving the far-field form

E_{\text{axial}} \approx \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}

directed along $\vec p$ — away from the dipole on the side nearer $+q$ . On the equatorial line, both charges sit at the same distance $\sqrt{r^2+a^2}$ from P, so their field magnitudes are equal; by symmetry the components perpendicular to the axis cancel while the components parallel to the axis (both pointing from $+q$ toward $-q$ ) add:

E_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0}\frac{p}{(r^2+a^2)^{3/2}} \;\approx\; \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3} \quad (r \gg a)

directed antiparallel to $\vec p$ — the reverse of the axial-field direction. Comparing the two far-field forms at equal $r$ gives a compact and frequently tested relation:

E_{\text{axial}} = 2\,E_{\text{equatorial}}

Both fields fall as $1/r^3$ , not $1/r^2$ , because a dipole is net-neutral: at large distance the $+q$ and $-q$ contributions almost cancel, and what survives depends on the separation $2a$ , not on any leftover charge.

Figure

Axial line field twice the equatorial line field at equal distance from a dipole centre

Derivation AField on the axial line of a short dipole

The goal is to show, starting only from Coulomb's law applied to the two point charges $+q$ and $-q$ separately, that the net field at an axial point P a distance $r$ from the dipole's centre reduces — after algebraic simplification and the far-field approximation $r \gg a$ — to the compact form $E_{\text{axial}} \approx \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$ , directed along the dipole moment $\vec p$ .

Step 1. Place the dipole with $-q$ at distance $a$ to the left of centre O and $+q$ at distance $a$ to the right, and let P lie on the axis beyond $+q$ , a distance $r$ from O with $r > a$ . The distance from $+q$ to P is $(r-a)$ , and the distance from $-q$ to P is $(r+a)$ .

Step 2. Write each charge's field at P using Coulomb's law: the field due to $+q$ has magnitude $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r-a)^2}$ and points away from $+q$ (toward increasing $r$ ); the field due to $-q$ has magnitude $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r+a)^2}$ and points toward $-q$ (also toward increasing $r$ , since $-q$ is farther away). Both fields point the same way at P, so the net field is their sum with a relative minus sign only because the $-q$ term is smaller and subtracted when both are written as "field along the axis":

E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{(r-a)^2} - \frac{q}{(r+a)^2}\right]

Step 3. Combine the two fractions over a common denominator. The numerator becomes $q\left[(r+a)^2-(r-a)^2\right] = q(4ar)$ , using the identity $(r+a)^2-(r-a)^2 = 4ar$ , and the denominator is $(r-a)^2(r+a)^2 = (r^2-a^2)^2$ . So

E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{4aqr}{(r^2-a^2)^2} = \frac{1}{4\pi\varepsilon_0}\frac{2pr}{(r^2-a^2)^2}

using $p = q(2a)$ , i.e. $4aq = 2p$ .

Step 4. Apply the short-dipole limit $r \gg a$ : since $a^2 \ll r^2$ , the denominator $(r^2-a^2)^2 \approx r^4$ , so $E_{\text{axial}} \approx \dfrac{1}{4\pi\varepsilon_0}\dfrac{2pr}{r^4} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$ , directed from the $-q$ side toward the $+q$ side — the same sense as $\vec p$ .

Worked example

An electric dipole has charges $+2\ \text{nC}$ and $-2\ \text{nC}$ separated by $1\ \text{cm}$ . Find the field at a point $10\ \text{cm}$ from the centre, both on the axial line and on the equatorial line, and compare them.

Step 1. Dipole moment: $p = q\times2a = (2\times10^{-9})\times(1\times10^{-2}) = 2\times10^{-11}\ \text{C}\,\cdot\,\text{m}$ . Here $r = 0.10\ \text{m}$ and $a = 0.005\ \text{m}$ , and $r/a = 20$ , so the short-dipole approximation is safe to use.

Step 2. Axial field: $E_{\text{axial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3} = (9\times10^9)\times\dfrac{2\times(2\times10^{-11})}{(0.10)^3} = (9\times10^9)\times(4\times10^{-8}) = 360\ \text{N/C}$ , directed along $\vec p$ .

Step 3. Equatorial field: $E_{\text{equatorial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3} = (9\times10^9)\times\dfrac{2\times10^{-11}}{(0.10)^3} = (9\times10^9)\times(2\times10^{-8}) = 180\ \text{N/C}$ , directed antiparallel to $\vec p$ .

Step 4. Check the ratio: $360/180 = 2$ , confirming $E_{\text{axial}} = 2E_{\text{equatorial}}$ at equal $r$ . The axial field is 360 N/C (along p) and the equatorial field is 180 N/C (opposite to p).

Common traps

Treating the equatorial field as parallel to $\vec p$ — it is always antiparallel, while the axial field is always parallel.
Reaching for the point-charge formula $E = kq/r^2$ for a dipole; a dipole has zero net charge, so only the dipole-specific $1/r^3$ formulas apply.
Using the far-field approximate formulas without first checking $r \gg a$ — if $r$ is only a few times $a$ , the exact $(r^2-a^2)^2$ or $(r^2+a^2)^{3/2}$ forms must be used instead.
Forgetting the factor of 2: at the same distance $r$ , the axial field is always the larger of the two, never equal or smaller.

Ask the tutor: want a walkthrough of the exact (non-approximated) equatorial-field derivation or a mixed axial-versus-equatorial practice set?

Why does the dipole field fall off faster than a single charge's field?

A lone charge's field falls as $1/r^2$ because its source, the net charge, never disappears. A dipole's net charge is exactly zero, so at large distance the $+q$ and $-q$ fields almost cancel. What survives is proportional to the tiny separation $2a$ between them, which brings in an extra factor of $1/r$ beyond the usual $1/r^2$ , giving $1/r^3$ overall. The farther you go, the more completely the two opposite charges look like a single point with no charge at all — only the "twist" of their moment $p$ remains visible.

Why is the equatorial field exactly opposite to the axial field in direction, not just different in magnitude?

On the equatorial line, a point P is equidistant from both charges, so the field vectors from $+q$ and $-q$ have equal magnitude but point in different directions — one toward P from $+q$ , one toward $-q$ from P. Their components perpendicular to the dipole axis are mirror images and cancel exactly by symmetry. Their components along the axis both point from the $+q$ side toward the $-q$ side, so they add up pointing that way — which is defined as antiparallel to $\vec p$ , since $\vec p$ points from $-q$ to $+q$ . On the axial line, by contrast, both fields already lie along the same line as $\vec p$ , and the near charge dominates, so the net field points along $\vec p$ .

Do I need to memorise both derivations, or just the final formulas?

ISC exams commonly ask for the full derivation of at least one of the two (axial or equatorial), so both should be practised at least once, but the fastest route to marks is knowing the two boxed results, $E_{\text{axial}} \approx \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$ and $E_{\text{equatorial}} \approx \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3}$ , their directions, and the relation $E_{\text{axial}} = 2E_{\text{equatorial}}$ . For numericals, plug in directly; for derivation questions, reproduce the Coulomb's-law setup and the algebraic simplification shown above.

Field of a dipole

The core idea

Worked example

Common traps

Before this

Where this leads