Electricity-And-Magnetism · Unit 3 · Video 4 · Interactive Practice

The Electric Dipole and Its 1/r³ Field

IKey Formulas

Formula	Name	When it applies
$\vec{p} = 2qa\,\hat{j}$ , $\vec{p} = \sum_{i} q_i\,\vec{r}_i$	Dipole moment	Points from $-q$ to $+q$ ; the sum needs $\sum q_i = 0$
$E_x = \frac{q}{4\pi\varepsilon_0}\left(\frac{x}{[x^2+(y-a)^2]^{3/2}} - \frac{x}{[x^2+(y+a)^2]^{3/2}}\right)$ $E_y = \frac{q}{4\pi\varepsilon_0}\left(\frac{y-a}{[x^2+(y-a)^2]^{3/2}} - \frac{y+a}{[x^2+(y+a)^2]^{3/2}}\right)$	Exact field, $\vec{E} = \vec{E}_+ + \vec{E}_-$	Every point off the charges
$E_x = \frac{3p}{4\pi\varepsilon_0 r^3}\sin\theta\cos\theta$ , $E_y = \frac{p}{4\pi\varepsilon_0 r^3}(3\cos^2\theta - 1)$	Point-dipole field	$r \gg a$ , with $\theta$ measured from the dipole axis
$\vec{E}(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\left(-\frac{\vec{p}}{r^3} + \frac{3(\vec{p}\cdot\vec{r})\,\vec{r}}{r^5}\right)$	Vector form	$r \gg a$ , any orientation of $\vec{p}$ , in 3D

Key Insight: With zero net charge, the two $1/r^2$ fields cancel at leading order. The remainder is smaller by a factor of about $a/r$ , so $E \propto p/r^3$ .

IIVisualization 1 — The Exact Field Is a Vector Sum

At every point $P$ , $\vec{E}$ is $\vec{E}_+$ (away from $+q$ ) plus $\vec{E}_-$ (toward $-q$ ).

Field point P

IIIVisualization 2 — One Power of r Faster

Far from the pair the two $1/r^2$ fields nearly cancel, and the remainder falls off as $1/r^3$ .

Direction Distance r 4.0a

💡 Challenge: on the bisector, find the smallest distance at which the point-dipole formula is within $1\%$ of the exact field.

IVVisualization 3 — The Point-Dipole Field Around a Circle

At fixed $r$ , the field on the axis is twice as strong as on the bisector, and points the opposite way.

Angle θ from p 30°

VQuiz Questions

Problem 1 · Dipole Moment of a Pair

Given: a charge $+3.0\ \text{nC}$ at $y = +1.5\ \text{mm}$ and a charge $-3.0\ \text{nC}$ at $y = -1.5\ \text{mm}$ . Find the dipole moment $\vec{p}$ .

✅ Correct! The magnitude is the charge times the full separation

2a = 3.0\ \text{mm}

, and

\vec{p}

points from

-q

up to

+q

❌ Not quite.

1.5\ \text{mm}

a

, the distance from the origin to one charge. The separation is

2a = 3.0\ \text{mm}

❌ Check the direction. The field in the gap runs from

+q

down to

-q

, but

\vec{p}

runs the other way, from

-q

+q

❌ Not quite. Use

p = q\cdot 2a

with

q = 3.0\ \text{nC}

and

2a = 3.0\ \text{mm}

Show solution

Here $q = 3.0\times10^{-9}\ \text{C}$ and $a = 1.5\times10^{-3}\ \text{m}$ , so the separation is $2a = 3.0\times10^{-3}\ \text{m}$ .

p = 2qa = (3.0\times10^{-9})(3.0\times10^{-3}) = 9.0\times10^{-12}\ \text{C}\cdot\text{m}

The sum definition gives the same vector:

\vec{p} = q(a\hat{j}) + (-q)(-a\hat{j}) = 2qa\,\hat{j} = 9.0\times10^{-12}\ \text{C}\cdot\text{m}\ \hat{j}

It points along $+\hat{j}$ , from $-q$ toward $+q$ .

Problem 2 · Axis Versus Bisector

Given: a small dipole $\vec{p}$ makes a field of magnitude $E_0$ at distance $r \gg a$ on its axis. Find the field at the same distance $r$ on its perpendicular bisector.

✅ Correct! On the bisector

E_y = -p/(4\pi\varepsilon_0 r^3)

, half the axis value

2p/(4\pi\varepsilon_0 r^3)

and opposite in sign.

❌ Close, but check the direction. At

\theta = 90^\circ

3\cos^2\theta - 1 = -1

, so

E_y

is negative: the field points against

\vec{p}

❌ Not quite. Both fields fall off as

1/r^3

, but the angular factor differs:

2

on the axis,

-1

on the bisector.

❌ Not quite. The factor

1/8

comes from doubling the distance. Here the distance is the same and only the angle changes.

❌ Not quite. Evaluate

E_y = \frac{p}{4\pi\varepsilon_0 r^3}(3\cos^2\theta - 1)

\theta = 0^\circ

and at

\theta = 90^\circ

Show solution

On the axis, $\theta = 0^\circ$ : $\sin\theta = 0$ and $\cos^2\theta = 1$ , so

E_x = 0,\qquad E_y = \frac{p}{4\pi\varepsilon_0 r^3}(3 - 1) = \frac{2p}{4\pi\varepsilon_0 r^3} = E_0.

On the bisector, $\theta = 90^\circ$ : $\cos\theta = 0$ , so

E_x = 0,\qquad E_y = \frac{p}{4\pi\varepsilon_0 r^3}(0 - 1) = -\frac{p}{4\pi\varepsilon_0 r^3} = -\frac{E_0}{2}.

The bisector field has magnitude $E_0/2$ and points opposite $\vec{p}$ .

Problem 3 · Exact Field Versus the Point-Dipole Formula

Given: $+q$ at $y = +a$ and $-q$ at $y = -a$ with $q = 1.0\ \text{nC}$ and $a = 1.0\ \text{cm}$ . The field point is on the axis at $y = 5.0\ \text{cm}$ . Use $\frac{1}{4\pi\varepsilon_0} = 8.99\times10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$ . Find $E_y$ exactly, then from the point-dipole formula.

Exact $E_y$

Point-dipole $E_y$

✅ Correct! At

r = 5a

the point-dipole formula is already within about

8\%

of the exact field, and the gap shrinks as

r

grows.

❌ Not quite.

\vec{E}_-

points back toward

-q

, so it enters with a minus sign: subtract the two terms instead of adding them.

❌ Not quite. That is the point-dipole approximation. The exact field uses the two actual distances,

4.0\ \text{cm}

and

6.0\ \text{cm}

❌ Check the exact field. On the axis

x = 0

, so

E_y = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{(y-a)^2} - \frac{1}{(y+a)^2}\right)

❌ Not quite. On the axis the angular factor is

3\cos^2\theta - 1 = 2

, so

E_y = 2p/(4\pi\varepsilon_0 r^3)

❌ Not quite. The dipole moment is

p = 2qa

with

2a = 2.0\ \text{cm}

; this value uses twice that.

❌ Check the point-dipole value. Use

p = 2qa

and

E_y = 2p/(4\pi\varepsilon_0 r^3)

with

r = 5.0\ \text{cm}

Show solution

Exact. With $x = 0$ the distances are $r_+ = y - a = 0.040\ \text{m}$ and $r_- = y + a = 0.060\ \text{m}$ :

E_y = (8.99\times10^{9})(1.0\times10^{-9})\left(\frac{1}{0.040^2} - \frac{1}{0.060^2}\right) = 8.99\,(625 - 277.8) \approx 3.12\times10^{3}\ \text{N/C}

Point dipole. $p = 2qa = (1.0\times10^{-9})(0.020) = 2.0\times10^{-11}\ \text{C}\cdot\text{m}$ and $\theta = 0^\circ$ :

E_y = \frac{2p}{4\pi\varepsilon_0 r^3} = \frac{2(8.99\times10^{9})(2.0\times10^{-11})}{(0.050)^3} \approx 2.88\times10^{3}\ \text{N/C}

The approximation is about $8\%$ low at $r = 5a$ . Both point up, along $\vec{p}$ .

Problem 4 · A Dipole Along x, in 3D

Given: a small dipole $\vec{p} = p\,\hat{i}$ at the origin. Find $\vec{E}$ from the vector form at the point $(0, 0, d)$ on the $z$ axis and at the point $(d, 0, 0)$ on the $x$ axis, with $d$ much larger than the dipole.

At $(0, 0, d)$

At $(d, 0, 0)$

✅ Correct! The

z

axis is perpendicular to

\vec{p}

, so it plays the bisector's role, and the

x

axis is now the dipole axis.

❌ Check the sign. With

\vec{p}\cdot\vec{r} = 0

only

-\vec{p}/r^3

survives, and it points against

\vec{p}

❌ Not quite. The surviving term is

-\vec{p}/r^3

, which is parallel to

\vec{p} = p\,\hat{i}

, not to

\vec{r}

❌ Not quite.

\vec{p}\cdot\vec{r} = 0

removes only the second term. The

-\vec{p}/r^3

term remains.

❌ Check the field on the $z$ axis. Compute

\vec{p}\cdot\vec{r}

for

\vec{r} = d\,\hat{k}

first.

❌ Not quite. Keep both terms:

-\vec{p}/d^3 + 3p\,\hat{i}/d^3 = 2p\,\hat{i}/d^3

❌ Check the field on the $x$ axis. Here

\vec{p}\cdot\vec{r} = pd

and

\vec{r} = d\,\hat{i}

, so both terms point along

\hat{i}

Show solution

Use $\vec{E}(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\left(-\frac{\vec{p}}{r^3} + \frac{3(\vec{p}\cdot\vec{r})\,\vec{r}}{r^5}\right)$ with $\vec{p} = p\,\hat{i}$ .

At $(0, 0, d)$ : $\vec{r} = d\,\hat{k}$ , $r = d$ , and $\vec{p}\cdot\vec{r} = 0$ , so

\vec{E} = \frac{1}{4\pi\varepsilon_0}\left(-\frac{p\,\hat{i}}{d^3}\right) = -\frac{p}{4\pi\varepsilon_0 d^3}\,\hat{i}.

At $(d, 0, 0)$ : $\vec{r} = d\,\hat{i}$ and $\vec{p}\cdot\vec{r} = pd$ , so

\vec{E} = \frac{1}{4\pi\varepsilon_0}\left(-\frac{p\,\hat{i}}{d^3} + \frac{3(pd)(d\,\hat{i})}{d^5}\right) = \frac{2p}{4\pi\varepsilon_0 d^3}\,\hat{i}.

These are the axis and bisector results of the video, rotated: the formula never refers to a particular axis.

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