The ideal electric dipole field is one of the first places in electromagnetism where a “simple-looking” formula hides a lot of structure. It encodes both a directional pattern and a different distance-scaling than the Coulomb field of a single charge. Rather than starting from the final answer, let us derive it directly from the superposition of two point-charge fields in Cartesian form.

Our goal is to connect the exact two-charge field to the familiar far-field dipole expression,

\[\mathbf E(\mathbf r)\approx \frac{1}{4\pi\varepsilon_0}\frac{1}{r^3}\left[3(\mathbf p\cdot\hat{\mathbf r})\hat{\mathbf r}-\mathbf p\right],\qquad r\gg s.\]

We will first examine two especially useful directions:

  • On-axis (parallel to the dipole moment)
  • Equatorial line (perpendicular to the dipole moment)

These two limits already reveal the characteristic \(1/r^3\) behavior and make the angular dependence much easier to interpret.

Setup: a physical dipole and Coulomb superposition

Consider a physical dipole made from two point charges \(+q\) and \(-q\) separated by a distance \(s\), centered at the origin, and aligned with the \(z\)-axis. Their positions are

\[\mathbf r_{+}=\frac{s}{2}\hat{\mathbf z},\qquad \mathbf r_{-}=-\frac{s}{2}\hat{\mathbf z}.\]

Let the observation point be

\[\mathbf r=(x,y,z),\qquad r=\sqrt{x^2+y^2+z^2}.\]

We are interested in the far-field regime,

\[r\gg s,\]

meaning the observation distance is much larger than the charge separation. In that limit, the pair of charges begins to look like a single “effective object” (an ideal dipole) rather than two individually resolved charges.

Coulomb field of one point charge

For a point charge \(q\) located at \(\mathbf r_0\), the electric field at \(\mathbf r\) is

\[\mathbf E(\mathbf r)=\frac{1}{4\pi\varepsilon_0}q\frac{\mathbf r-\mathbf r_0}{|{\mathbf r-\mathbf r_0}|^3}.\]

Exact dipole field from superposition

The field of the dipole is just the vector sum of the fields from the two charges. Writing the contribution from \(-q\) explicitly as a subtraction gives

\[\mathbf E(\mathbf r)=\frac{1}{4\pi\varepsilon_0}q\left(\frac{\mathbf r-\mathbf r_{+}}{|{\mathbf r-\mathbf r_{+}}|^3}-\frac{\mathbf r-\mathbf r_{-}}{|{\mathbf r-\mathbf r_{-}}|^3}\right).\]

This expression is exact. Everything that follows is just careful simplification in special directions, followed by a far-field expansion.

Field on the dipole axis (parallel case)

To study the field along the dipole axis, place the observation point on the \(z\)-axis:

\[\mathbf r=z\hat{\mathbf z},\qquad z\gg s.\]

By symmetry, the field must point purely along \(\hat{\mathbf z}\). The exact axial field is

\[E_{\parallel}(z)=\frac{1}{4\pi\varepsilon_0}q\left[\frac{1}{(z-\tfrac{s}{2})^2}-\frac{1}{(z+\tfrac{s}{2})^2}\right].\]

This already contains the essential physics: the nearer charge contributes slightly more strongly than the farther charge, and that mismatch is what survives at large distance.

Combine the two terms

To expose the scaling more clearly, combine the fractions:

\[\frac{1}{(z-\tfrac{s}{2})^2}-\frac{1}{(z+\tfrac{s}{2})^2} =\frac{2zs}{\left(z^2-\tfrac{s^2}{4}\right)^2}.\]

So the exact on-axis field becomes

\[E_{\parallel}(z)=\frac{1}{4\pi\varepsilon_0}q\frac{2zs}{\left(z^2-\tfrac{s^2}{4}\right)^2}.\]

On-axis far-field expansion

Now use the assumption \(z\gg s\). Rewrite the denominator in a form suitable for expansion:

\[\left(z^2-\tfrac{s^2}{4}\right)^{-2}=z^{-4}\left(1-\frac{s^2}{4z^2}\right)^{-2}.\]

Since \(s^2/z^2\ll 1\), the leading term is

\[\left(z^2-\tfrac{s^2}{4}\right)^{-2}\approx z^{-4}+\mathcal O(z^{-6}).\]

Substituting back into the field gives

\[\mathbf E_{\parallel}(z)\approx\frac{1}{4\pi\varepsilon_0}q(2zs)z^{-4}\hat{\mathbf z} =\frac{1}{4\pi\varepsilon_0}\frac{2qs}{z^3}\hat{\mathbf z}.\]

Define the dipole moment magnitude by

\[p\equiv qs.\]

Then the leading on-axis far-field result is

\[\boxed{\mathbf E_{\parallel}\approx\frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}\hat{\mathbf z}\qquad (r\gg s,\ \text{on-axis})}.\]

This is the first clean appearance of the dipole scaling law: the field falls as \(1/r^3\) rather than \(1/r^2\). The reason is that the net charge is zero, so the monopole (Coulomb) contribution cancels.

Field on the equatorial line (perpendicular case)

Next, place the observation point on the \(x\)-axis, perpendicular to the dipole axis:

\[\mathbf r=x\hat{\mathbf x},\qquad x\gg s.\]

For the two charges, the displacement vectors are

\[\mathbf r-\mathbf r_{\pm}=x\hat{\mathbf x}\mp \frac{s}{2}\hat{\mathbf z},\]

and both have the same squared magnitude,

\[|{\mathbf r-\mathbf r_{\pm}}|^2=x^2+\frac{s^2}{4}.\]

It is convenient to define

\[R^2=x^2+\frac{s^2}{4},\]

so that both denominators are the same.

Superpose the fields

Substituting into the exact dipole field,

\[\mathbf E(x)=\frac{1}{4\pi\varepsilon_0}q\left(\frac{x\hat{\mathbf x}-\tfrac{s}{2}\hat{\mathbf z}}{R^3}-\frac{x\hat{\mathbf x}+\tfrac{s}{2}\hat{\mathbf z}}{R^3}\right).\]

The \(x\)-components cancel exactly (as they must by symmetry), leaving only a \(z\)-component:

\[\mathbf E(x)=-\frac{1}{4\pi\varepsilon_0}\frac{qs}{R^3}\hat{\mathbf z}.\]

The minus sign is physically meaningful: on the equatorial line, the field points opposite to the dipole moment direction.

Equatorial far-field expansion

Write the exact result explicitly as

\[\mathbf E(x)=-\frac{1}{4\pi\varepsilon_0}\frac{qs}{\left(x^2+\tfrac{s^2}{4}\right)^{3/2}}\hat{\mathbf z}.\]

Now factor out \(x^2\) from the denominator:

\[\left(x^2+\frac{s^2}{4}\right)^{-3/2}=x^{-3}\left(1+\frac{s^2}{4x^2}\right)^{-3/2}.\]

For \(x\gg s\), the leading term is

\[\left(x^2+\frac{s^2}{4}\right)^{-3/2}\approx x^{-3}+\mathcal O(x^{-5}).\]

So the far-field equatorial result is

\[\boxed{\mathbf E_{\perp}\approx-\frac{1}{4\pi\varepsilon_0}\frac{qs}{x^3}\hat{\mathbf z} =-\frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}\hat{\mathbf z}\qquad (r\gg s,\ \text{equatorial})}.\]

What changed relative to the on-axis case?

Both directions have the same distance scaling \(1/r^3\), but the coefficient differs:

  • On-axis magnitude: \(\displaystyle \frac{2p}{4\pi\varepsilon_0 r^3}\)
  • Equatorial magnitude: \(\displaystyle \frac{p}{4\pi\varepsilon_0 r^3}\)

So at the same distance, the on-axis field is twice the equatorial field in magnitude.

The general far-field dipole field

We can now summarize the result in its standard vector form. Define the dipole moment vector (pointing from \(-q\) to \(+q\)) as

\[\mathbf p = qs\hat{\mathbf z}.\]

Then for a general observation direction \(\hat{\mathbf r}=\mathbf r/r\), the leading far-field term is

\[\boxed{\mathbf E(\mathbf r)\approx\frac{1}{4\pi\varepsilon_0}\frac{1}{r^3}\left[3(\mathbf p\cdot\hat{\mathbf r})\hat{\mathbf r}-\mathbf p\right],\qquad r\gg s.}\]

This compact expression reproduces both special cases and makes the angular structure of the dipole field manifest.

Quick consistency checks

It is worth checking that the general expression reduces to the results derived above.

On-axis: \(\hat{\mathbf r}\parallel \mathbf p\)

If \(\hat{\mathbf r}\) is parallel to \(\mathbf p\), then \(\mathbf p\cdot\hat{\mathbf r}=p\), and

\[\mathbf E=\frac{1}{4\pi\varepsilon_0}\frac{1}{r^3}\left(3p\hat{\mathbf z}-p\hat{\mathbf z}\right) =\frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}\hat{\mathbf z},\]

which matches the on-axis result.

Equatorial: \(\mathbf p\cdot\hat{\mathbf r}=0\)

If the observation direction is perpendicular to \(\mathbf p\), then \(\mathbf p\cdot\hat{\mathbf r}=0\), so

\[\mathbf E=-\frac{1}{4\pi\varepsilon_0}\frac{\mathbf p}{r^3}.\]

For \(\mathbf p=p\hat{\mathbf z}\), this becomes

\[\mathbf E=-\frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}\hat{\mathbf z},\]

which matches the equatorial result.

Takeaways

A few points are worth keeping in mind:

  • A physical dipole has zero net charge, so the leading \(1/r^2\) monopole field cancels.
  • The first nonzero far-field term is the dipole field, with characteristic scaling \(\boxed{E\sim \frac{1}{r^3}}.\)
  • The field is strongly directional: it is largest on-axis, smaller on the equatorial line, and in general determined by the projection \(\mathbf p\cdot\hat{\mathbf r}\).

A note for students: this is your first multipole expansion

This derivation is also a preview of a much broader idea in electromagnetism and mathematical physics: multipole expansions. At large distance, fields are organized into a hierarchy of terms,

  • monopole (\(1/r^2\) field),
  • dipole (\(1/r^3\) field),
  • quadrupole (\(1/r^4\) field),
  • and so on.

That hierarchy is powerful because it tells us, almost immediately, which physical features of a source dominate at long range.