In the previous notes we considered the far-field expression for the electric field of an ideal dipole

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

along the dipole and equitorial axes. In many practical problems it is useful to evaluate this expression in Cartesian components at a specific point. In this note we compute the electric field at the point

\[(x,y)=(a,b), \qquad a,b>0,\]

for a dipole whose moment points along the \(x\) direction.

Setup: dipole aligned with \(\hat{\mathbf x}\)

Consider an ideal dipole at the origin with dipole moment

\[\mathbf p = p\,\hat{\mathbf x}.\]

We want the electric field at the point

\[\mathbf r = a\,\hat{\mathbf x}+b\,\hat{\mathbf y}.\]

The distance from the origin is therefore

\[r=\sqrt{a^2+b^2}.\]

The general dipole field formula will now be evaluated at this location.

The dipole field formula

The electrostatic field of an ideal dipole at the origin can be rewritten1

\[\boxed{\mathbf E(\mathbf r)=\frac{1}{4\pi\varepsilon_0}\left[\frac{3(\mathbf p\cdot\mathbf r)\mathbf r}{r^5}-\frac{\mathbf p}{r^3}\right].}\]

To compute the field we therefore need

  • the dot product \(\mathbf p\cdot\mathbf r\),
  • the vector \(\mathbf r\),
  • and the powers \(r^3\) and \(r^5\).

Computing the dot product \(\mathbf p\cdot\mathbf r\)

With

\[\mathbf p=p\,\hat{\mathbf x}, \qquad \mathbf r=a\,\hat{\mathbf x}+b\,\hat{\mathbf y},\]

the dot product becomes

\[\mathbf p\cdot\mathbf r=(p\,\hat{\mathbf x})\cdot(a\,\hat{\mathbf x}+b\,\hat{\mathbf y})=pa.\]

The distance from the origin is

\[r=\sqrt{a^2+b^2},\]

so that

\[r^3=(a^2+b^2)^{3/2}, \qquad r^5=(a^2+b^2)^{5/2}.\]

The \(x\) and \(y\) components

Insert \(\mathbf p\cdot\mathbf r = pa\) and \(\mathbf r=a\hat{\mathbf x}+b\hat{\mathbf y}\) into the general formula:

\[\mathbf E(a,b)=\frac{1}{4\pi\varepsilon_0}\left[\frac{3(pa)(a\hat{\mathbf x}+b\hat{\mathbf y})}{(a^2+b^2)^{5/2}}-\frac{p\hat{\mathbf x}}{(a^2+b^2)^{3/2}}\right].\]

We now extract the Cartesian components. The \(\hat{\mathbf y}\) contribution comes only from the first term

\[\boxed{E_y(a,b)=\frac{p}{4\pi\varepsilon_0}\frac{3ab}{(a^2+b^2)^{5/2}}}.\]

The \(\hat{\mathbf x}\) component receives contributions from both terms

\[E_x(a,b)=\frac{1}{4\pi\varepsilon_0}\left[\frac{3pa\,a}{(a^2+b^2)^{5/2}}-\frac{p}{(a^2+b^2)^{3/2}}\right].\]

Factor out \(p/(4\pi\varepsilon_0)\) yeilding

\[E_x(a,b)=\frac{p}{4\pi\varepsilon_0}\left[\frac{3a^2}{(a^2+b^2)^{5/2}}-\frac{1}{(a^2+b^2)^{3/2}}\right].\]

To combine the terms, we rewrite the second denominator which gives

\[\frac{1}{(a^2+b^2)^{3/2}}=\frac{a^2+b^2}{(a^2+b^2)^{5/2}}.\]

Substitute this into the expression

\[E_x(a,b)=\frac{p}{4\pi\varepsilon_0}\frac{3a^2-(a^2+b^2)}{(a^2+b^2)^{5/2}}.\]

Simplifying the numerator gives

\[\boxed{E_x(a,b)=\frac{p}{4\pi\varepsilon_0}\frac{2a^2-b^2}{(a^2+b^2)^{5/2}}}.\]

Quick consistency checks

Two useful limits confirm that the result behaves as expected.

On the dipole axis (\(b=0\))

\[\mathbf E=\frac{p}{4\pi\varepsilon_0}\frac{2}{a^3}\hat{\mathbf x}.\]

This matches the familiar on-axis dipole field.

On the equatorial axis (\(a=0\))

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

Here the field points opposite to the dipole moment, as expected for the equatorial line.

The same result in polar form

It is helpful to rewrite the geometry in terms of plane polar coordinates. Let

\[a=r\cos\theta,\qquad b=r\sin\theta,\]

where

\[r=\sqrt{a^2+b^2},\qquad \theta=\tan^{-1}\!\left(\frac{b}{a}\right).\]

Substitute these into the Cartesian result:

\[\mathbf E(a,b)=\frac{p}{4\pi\varepsilon_0}\frac{(2a^2-b^2)\hat{\mathbf x}+(3ab)\hat{\mathbf y}}{(a^2+b^2)^{5/2}}.\]

Since \(a^2+b^2=r^2\), the denominator becomes \(r^5\), while the numerator becomes

\[2a^2-b^2=2r^2\cos^2\theta-r^2\sin^2\theta=r^2(2\cos^2\theta-\sin^2\theta),\]

and

\[3ab=3r^2\sin\theta\cos\theta.\]

Therefore

\[\mathbf E(r,\theta)=\frac{p}{4\pi\varepsilon_0 r^3}\left[(2\cos^2\theta-\sin^2\theta)\hat{\mathbf x}+(3\sin\theta\cos\theta)\hat{\mathbf y}\right].\]

Using the identities

\[\hat{\mathbf r}=\cos\theta\,\hat{\mathbf x}+\sin\theta\,\hat{\mathbf y}, \qquad \hat{\boldsymbol\theta}=-\sin\theta\,\hat{\mathbf x}+\cos\theta\,\hat{\mathbf y},\]

this can be rewritten in the standard polar-component form

\[\boxed{\mathbf E(r,\theta)=\frac{p}{4\pi\varepsilon_0 r^3}\left(2\cos\theta\,\hat{\mathbf r}+\sin\theta\,\hat{\boldsymbol\theta}\right)}.\]

This is the familiar dipole field in a plane containing the dipole axis. It makes the angular structure very clear:

  • the radial component is proportional to \(2\cos\theta\),
  • the angular component is proportional to \(\sin\theta\),
  • and the overall magnitude still falls as \(1/r^3\).

  1. The two expressions are identical because \(\mathbf r=r\hat{\mathbf r}\), so \(\frac{3(\mathbf p\cdot\mathbf r)\mathbf r}{r^5}=\frac{3(\mathbf p\cdot r\hat{\mathbf r})(r\hat{\mathbf r})}{r^5}=\frac{3(\mathbf p\cdot\hat{\mathbf r})\hat{\mathbf r}}{r^3}\). ↩