Electric Dipole Field

An electric dipole consists of two equal and opposite charges, $+q$ and $-q$, separated by a small distance $d$. Its field has a beautiful, distinctive shape: field lines start on the positive charge, curve outward through space, and end on the negative charge. There are no "free ends" — every line connects $+q$ to $-q$ (or extends to infinity).

The Dipole Moment

The dipole's strength is measured by its dipole moment vector:

$\vec p = q\vec d$

where $\vec d$ points from $-q$ to $+q$. The magnitude is $p = qd$.

The Field at a Point

At a general point $(x, y)$, the total field is the vector sum of the contributions from the two charges:

$\vec E = \vec E_+ + \vec E_-$

Each individual field is a Coulomb $1/r^{2}$ field. The two fields add as vectors, producing the curved pattern.

Far-Field Limit

Far from the dipole (where $r \gg d$), the field magnitude on the axis (along the line through both charges) is:

$E_{\text{axial}}(r) = \dfrac{2kp}{r^{3}}$

And on the perpendicular bisector:

$E_{\text{perp}}(r) = \dfrac{kp}{r^{3}}$

Notice both drop as $1/r^{3}$ — faster than a point charge ($1/r^{2}$) because the two opposite charges partially cancel each other at large distances.

Step-by-Step Solution

Given: Two charges $+q$ and $-q$ with $q = 1\;\text{nC}$, separated by $d = 0.02\;\text{m}$ along the $x$-axis.

Find: The field magnitude on the dipole axis at $r = 0.1\;\text{m}$.

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Step 1 — Compute the dipole moment.

$p = qd = (10^{-9})(0.02) = 2 \times 10^{-11}\;\text{C}\cdot\text{m}$

Step 2 — Apply the axial-field formula.

$E = \dfrac{2 k p}{r^{3}} = \dfrac{2(9 \times 10^{9})(2 \times 10^{-11})}{(0.1)^{3}}$

Step 3 — Compute step by step.

Numerator: $2 \times 9 \times 10^{9} \times 2 \times 10^{-11} = 36 \times 10^{-2} = 0.36$.

Denominator: $0.001$.

$E = \dfrac{0.36}{0.001} = 360\;\text{V/m}$

Step 4 — Compare to a single point charge at the same distance.

A single $+q = 1\;\text{nC}$ at $0.1\;\text{m}$ would produce:

$E_{\text{single}} = \dfrac{(9 \times 10^{9})(10^{-9})}{0.01} = 900\;\text{V/m}$

The dipole gives a much weaker field (360 vs 900) because the negative charge is partially canceling the positive. And the falloff is faster: at $r = 0.2\;\text{m}$ the dipole gives $E \propto 1/8$ of its previous value, while a single charge gives $1/4$.

Step 5 — The visual signature.

The famous dipole field-line pattern shows:

• Lines emerging from $+q$ in all directions
• Curving around toward $-q$
• Most lines end on $-q$, but some go far out to infinity
• The pattern is symmetric under $180°$ rotation about the center
• The pattern looks like a figure-8 if you look at the magnetic-field analog

This pattern is the same shape as a magnetic field around a small bar magnet — and that's not a coincidence: a tiny current loop also has a "magnetic dipole moment" with the same field pattern.

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Answer: The field 0.1 m from a 1 nC dipole with separation 0.02 m, on the dipole axis, is

$\boxed{\;E = \dfrac{2kp}{r^{3}} = 360\;\text{V/m}\;}$

Less than a single charge of the same magnitude, because of partial cancellation. Field lines connect the $+q$ to the $-q$ in graceful curves, with stronger lines along the dipole axis.

Try It

• Adjust the separation distance $d$ between the charges.
• Watch the field lines curve from positive to negative.
• Toggle the strength to see the lines spread out more or cluster more tightly.