Electric Dipole

So far, you have looked at the field produced by a single point charge and by collections of charges in general arrangements. But nature has a favourite charge configuration that shows up everywhere, from water molecules to radio antennas: a pair of equal and opposite charges sitting close together. This arrangement is called an electric dipole, and it produces a field with some beautifully distinctive features.

What Is an Electric Dipole?

Take a charge $+q$ and a charge $-q$ and place them a fixed distance $2a$ apart. That is an electric dipole. A few things to note right away:

• The line connecting the two charges defines a natural axis, called the dipole axis.
• By convention, the direction along this axis runs from $-q$ toward $+q$. This is the direction we will call $\hat{\mathbf{p}}$.
• The point exactly midway between the two charges is the centre of the dipole.
• The total charge is zero ($+q + (-q) = 0$), but this does not mean the field is zero. Because the charges are separated in space, their individual fields do not perfectly cancel at most locations.

At distances much larger than the separation ($r \gg 2a$), the partial cancellation becomes nearly complete, and the dipole field drops off faster than the $1/r^2$ field of a single charge. Let us work out the field explicitly at two special locations to see exactly how this plays out.

The Electric Field on the Dipole Axis

Fig 1.17(a): Electric field of a dipole at a point on the axis

Consider a point P on the dipole axis, at a distance $r$ from the centre, on the side of the positive charge. The positive charge $+q$ is at distance $(r - a)$ from P, while the negative charge $-q$ is at distance $(r + a)$ from P.

Setting Up the Individual Fields

The field at P due to $-q$ points toward $-q$, that is, opposite to $\hat{\mathbf{p}}$:

$\mathbf{E}_{-q} = -\frac{q}{4\pi\varepsilon_0(r+a)^2} \, \hat{\mathbf{p}} \qquad \text{[1.13(a)]}$

The field at P due to $+q$ points away from $+q$, which is along $\hat{\mathbf{p}}$:

$\mathbf{E}_{+q} = \frac{q}{4\pi\varepsilon_0(r-a)^2} \, \hat{\mathbf{p}} \qquad \text{[1.13(b)]}$

Notice that $\mathbf{E}_{+q}$ is stronger than $\mathbf{E}_{-q}$ because P is closer to $+q$ than to $-q$. So the net field will point along $\hat{\mathbf{p}}$.

Adding the Two Contributions

By superposition, the total field at P is:

$\mathbf{E} = \mathbf{E}_{+q} + \mathbf{E}_{-q} = \frac{q}{4\pi\varepsilon_0} \left[\frac{1}{(r-a)^2} - \frac{1}{(r+a)^2}\right] \hat{\mathbf{p}}$

To combine the two fractions, take a common denominator of $(r-a)^2(r+a)^2 = (r^2 - a^2)^2$. The numerator becomes:

$(r+a)^2 - (r-a)^2$

Expanding each square:

$(r^2 + 2ar + a^2) - (r^2 - 2ar + a^2) = 4ar$

So the exact result for any point on the axis is:

$\mathbf{E} = \frac{q}{4\pi\varepsilon_0} \frac{4ar}{(r^2 - a^2)^2} \, \hat{\mathbf{p}} \qquad \text{(1.14)}$

The Far-Field Approximation

When P is very far from the dipole ($r \gg a$), the term $a^2$ in the denominator becomes negligible compared to $r^2$. That means $(r^2 - a^2)^2 \approx r^4$, and the expression simplifies to:

$\mathbf{E} = \frac{4qa}{4\pi\varepsilon_0 r^3} \, \hat{\mathbf{p}} \qquad (r \gg a) \qquad \text{(1.15)}$

This is a clean, compact result. The field points along $\hat{\mathbf{p}}$ and falls off as $1/r^3$.

The Electric Field on the Equatorial Plane

Fig 1.17(b): Electric field of a dipole at a point on the equatorial plane

Now consider a point P on the equatorial plane, the plane that passes through the centre of the dipole and is perpendicular to the dipole axis. Let P be at distance $r$ from the centre.

Equal Distances, Equal Magnitudes

Both charges are at the same distance from P. Using the Pythagorean theorem, this distance is $\sqrt{r^2 + a^2}$. So the field magnitudes are:

$E_{+q} = \frac{q}{4\pi\varepsilon_0} \frac{1}{r^2 + a^2} \qquad \text{[1.16(a)]}$

$E_{-q} = \frac{q}{4\pi\varepsilon_0} \frac{1}{r^2 + a^2} \qquad \text{[1.16(b)]}$

The two magnitudes are equal. But the field vectors point in different directions: $\mathbf{E}_{+q}$ points away from $+q$ and $\mathbf{E}_{-q}$ points toward $-q$.

Resolving into Components

Break each field into two components: one along the dipole axis and one perpendicular to it.

By symmetry, the perpendicular components of $\mathbf{E}_{+q}$ and $\mathbf{E}_{-q}$ are equal in magnitude but opposite in direction. They cancel each other out completely.

The components along the dipole axis, however, both point in the same direction: from $+q$ toward $-q$, which is opposite to $\hat{\mathbf{p}}$. So these components add up.

Each field vector makes an angle $\theta$ with the line from P perpendicular to the dipole axis. The component along the dipole axis for each field is $E \cos\theta$, where $\cos\theta = \frac{a}{\sqrt{r^2 + a^2}}$.

Combining the Result

The total field is the sum of the two axial components, directed opposite to $\hat{\mathbf{p}}$:

$\mathbf{E} = -(E_{+q} + E_{-q}) \cos\theta \, \hat{\mathbf{p}}$

Substituting the expressions for $E_{+q}$, $E_{-q}$, and $\cos\theta$:

$\mathbf{E} = -\frac{2q}{4\pi\varepsilon_0(r^2 + a^2)} \times \frac{a}{\sqrt{r^2 + a^2}} \, \hat{\mathbf{p}}$

$= -\frac{2qa}{4\pi\varepsilon_0(r^2 + a^2)^{3/2}} \, \hat{\mathbf{p}} \qquad \text{(1.17)}$

The Far-Field Approximation

At large distances ($r \gg a$), we have $(r^2 + a^2)^{3/2} \approx r^3$, so:

$\mathbf{E} = -\frac{2qa}{4\pi\varepsilon_0 r^3} \, \hat{\mathbf{p}} \qquad (r \gg a) \qquad \text{(1.18)}$

The negative sign tells us the field points opposite to $\hat{\mathbf{p}}$: it is directed from the positive charge toward the negative charge.

The Dipole Moment: A Single Quantity That Captures the Dipole

Look at the far-field expressions Eq. (1.15) and Eq. (1.18). Neither contains $q$ and $a$ separately. Both depend only on the product $qa$, or more precisely, $q \times 2a$. This suggests defining a single quantity that encapsulates the “strength” of the dipole.

The dipole moment vector $\mathbf{p}$ is defined as:

$\mathbf{p} = q \times 2a \, \hat{\mathbf{p}} \qquad \text{(1.19)}$

Its magnitude is the charge $q$ multiplied by the full separation $2a$, and its direction is from $-q$ to $+q$.

Using $\mathbf{p}$, the far-field results take on an elegant form:

On the dipole axis:

$\mathbf{E} = \frac{2\mathbf{p}}{4\pi\varepsilon_0 r^3} \qquad (r \gg a) \qquad \text{(1.20)}$

On the equatorial plane:

$\mathbf{E} = -\frac{\mathbf{p}}{4\pi\varepsilon_0 r^3} \qquad (r \gg a) \qquad \text{(1.21)}$

Key Features of the Dipole Field

Three important observations stand out from these results:

• Inverse-cube decay: The dipole field falls off as $1/r^3$, not $1/r^2$. This faster decay happens because the fields of $+q$ and $-q$ nearly cancel at large distances, leaving only a small residual that diminishes rapidly.
• Direction depends on position: On the axis, the field points along $\hat{\mathbf{p}}$ (from $-q$ to $+q$). On the equatorial plane, it points opposite to $\hat{\mathbf{p}}$ (from $+q$ to $-q$). At a general point, both the magnitude and direction depend on the angle between the position vector $\mathbf{r}$ and the dipole moment $\mathbf{p}$.
• Axial field is twice the equatorial field at the same distance: comparing Eq. (1.20) and Eq. (1.21), the axial field magnitude is $\frac{2p}{4\pi\varepsilon_0 r^3}$ while the equatorial field magnitude is $\frac{p}{4\pi\varepsilon_0 r^3}$.

The Point Dipole: An Idealised Limit

Imagine shrinking the dipole: let the separation $2a$ get smaller and smaller, while increasing the charge $q$ so that the product $p = q \times 2a$ stays the same. In the limit, $2a \to 0$ and $q \to \infty$ with $p$ held constant. This idealised configuration is called a point dipole.

For a point dipole, the condition $r \gg a$ is automatically satisfied for any finite distance $r$, because $a$ is effectively zero. So Eqs. (1.20) and (1.21) become exact at all distances, not just at far-away points.

Physical Significance: Why Dipoles Matter in the Real World

Electric dipoles are not just a textbook exercise. They show up naturally in the structure of molecules, and this has profound consequences for how materials behave in electric fields.

Non-Polar Molecules

In many molecules, the centre of positive charge (the average position of all the protons, weighted by their charges) coincides exactly with the centre of negative charge (the average position of all the electrons). The centre of a collection of positive point charges is defined in the same way as the centre of mass (replacing mass with charge): $\mathbf{r}_{\text{centre}} = \frac{\sum_i q_i \mathbf{r}_i}{\sum_i q_i}$.

When these two centres sit at the same point, the net dipole moment is zero. Molecules like $CO_2$ and $CH_4$ belong to this category. They are called non-polar molecules.

However, when you place a non-polar molecule in an external electric field, the positive and negative charge centres get pulled slightly apart. The molecule then develops an induced dipole moment that lasts only as long as the external field is present.

Polar Molecules

In some molecules, the geometry is such that the centres of positive and negative charge do not coincide, even without any external field. These molecules carry a permanent electric dipole moment and are called polar molecules.

Water ($H_2O$) is a classic example. Its bent shape means the oxygen end carries a partial negative charge while the hydrogen end carries a partial positive charge, giving the molecule a permanent dipole moment. This permanent polarity is responsible for many of water’s remarkable properties, from its high boiling point to its ability to dissolve salts.

The behaviour of various materials in the presence or absence of electric fields, and the interesting phenomena that arise, are deeply connected to whether their constituent molecules are polar or non-polar.

Worked Example: Computing the Dipole Field at Two Points

Example 1.9: Two charges $\pm 10$ $\mu$C are placed 5.0 mm apart. Find the electric field at (a) a point P on the axis, 15 cm from the centre on the side of the positive charge, and (b) a point Q on the equatorial plane, 15 cm from the centre.

Fig 1.18: Dipole with observation points P (on axis) and Q (on equatorial plane)

Part (a): Field at P on the Axis

Given: $q = 10 \, \mu\text{C} = 10^{-5}$ C, separation $2a = 5.0$ mm so $a = 2.5$ mm $= 0.25$ cm, distance $r = 15$ cm.

Step 1: Field due to $+q$ at P.

The positive charge is at distance $(r - a) = (15 - 0.25) = 14.75$ cm from P. The field points away from $+q$, i.e., along BP (to the right):

$E_{+q} = \frac{10^{-5}}{4\pi \times 8.854 \times 10^{-12}} \times \frac{1}{(14.75 \times 10^{-2})^2}$

$= 4.13 \times 10^6 \text{ N C}^{-1} \quad \text{along BP}$

Step 2: Field due to $-q$ at P.

The negative charge is at distance $(r + a) = (15 + 0.25) = 15.25$ cm from P. The field points toward $-q$, i.e., along PA (to the left):

$E_{-q} = \frac{10^{-5}}{4\pi \times 8.854 \times 10^{-12}} \times \frac{1}{(15.25 \times 10^{-2})^2}$

$= 3.86 \times 10^6 \text{ N C}^{-1} \quad \text{along PA}$

Step 3: Net field.

The two fields point in opposite directions along the axis. The net field is their difference, directed along BP (since $E_{+q} > E_{-q}$):

$E = 4.13 \times 10^6 - 3.86 \times 10^6 = 2.7 \times 10^5 \text{ N C}^{-1} \quad \text{along BP}$

Step 4: Verification using the far-field formula.

The ratio $r/a = 15/0.25 = 60$ is very large, so the far-field formula should give a good approximation. First, compute the dipole moment:

$p = q \times 2a = 10^{-5} \times 5 \times 10^{-3} = 5 \times 10^{-8} \text{ C m}$

Then apply Eq. (1.20):

$E = \frac{2p}{4\pi\varepsilon_0 r^3} = \frac{2 \times 5 \times 10^{-8}}{4\pi \times 8.854 \times 10^{-12} \times (0.15)^3}$

$= 2.6 \times 10^5 \text{ N C}^{-1}$

This is along the dipole moment direction (from A to B), which agrees well with the exact calculation. The small difference (about 4%) comes from the approximation $r \gg a$, which is not perfectly satisfied even at $r/a = 60$.

Part (b): Field at Q on the Equatorial Plane

Step 1: Field due to $+q$ at Q.

The distance from $+q$ to Q is $\sqrt{r^2 + a^2} = \sqrt{15^2 + 0.25^2} \approx 15.002$ cm (essentially 15 cm since $a$ is tiny):

$E_{+q} = \frac{10^{-5}}{4\pi \times 8.854 \times 10^{-12}} \times \frac{1}{(15^2 + 0.25^2) \times 10^{-4}}$

$= 3.99 \times 10^6 \text{ N C}^{-1} \quad \text{along BQ}$

Step 2: Field due to $-q$ at Q.

By symmetry, this has the same magnitude:

$E_{-q} = 3.99 \times 10^6 \text{ N C}^{-1} \quad \text{along QA}$

Step 3: Resolve and add.

The components perpendicular to the dipole axis (along OQ) cancel. The components along the dipole axis both point from B toward A (opposite to $\hat{\mathbf{p}}$) and add up. Each axial component is $E \cos\theta$, where $\cos\theta = \frac{a}{\sqrt{r^2 + a^2}} = \frac{0.25}{\sqrt{15^2 + 0.25^2}}$:

$E_{\text{net}} = 2 \times \frac{0.25}{\sqrt{15^2 + 0.25^2}} \times 3.99 \times 10^6$

$= 1.33 \times 10^5 \text{ N C}^{-1} \quad \text{along BA}$

Step 4: Verification using the far-field formula.

Using Eq. (1.21):

$E = \frac{p}{4\pi\varepsilon_0 r^3} = \frac{5 \times 10^{-8}}{4\pi \times 8.854 \times 10^{-12} \times (0.15)^3}$

$= 1.33 \times 10^5 \text{ N C}^{-1}$

The direction is opposite to $\hat{\mathbf{p}}$, that is, from B to A. The far-field formula matches the exact calculation perfectly at this distance.

Comparing the two results: The axial field ($2.6 \times 10^5$ N/C) is almost exactly twice the equatorial field ($1.33 \times 10^5$ N/C), just as the far-field formulas predict.