Electric Dipole

A pair of equal and opposite charges separated by a small distance is an electric dipole — the simplest non-trivial charge distribution and the prototype of polar molecules like $H_2O$.

Concept

A dipole consists of charges $+q$ and $-q$ separated by a vector $\vec{2a}$ from $-q$ to $+q$. The dipole moment is $\vec p = q\cdot \vec{2a},$ pointing from the negative to the positive charge. Unit: $C\cdot m$.

Axial line: along the line through both charges. Equatorial line: perpendicular bisector of the dipole axis.

For a point at distance $r$ from the centre (with $r \gg a$):

• Axial field: $E_\text{axial} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$, parallel to $\vec p$.
• Equatorial field: $E_\text{eq} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3}$, antiparallel to $\vec p$.

Note that both fall as $1/r^3$, faster than a point charge ($1/r^2$). The reason: at large $r$ the dipole looks almost neutral.

Derivation

Axial point. Take the dipole centred at origin, with $+q$ at $+a\hat x$ and $-q$ at $-a\hat x$. Field at point $(r, 0, 0)$ with $r>a$: $E = \frac{kq}{(r-a)^2} - \frac{kq}{(r+a)^2}.$ Combine over a common denominator: $E = kq\,\frac{(r+a)^2 - (r-a)^2}{(r^2-a^2)^2} = kq\,\frac{4ar}{(r^2-a^2)^2}.$ With $p = 2aq$ and using $r \gg a$: $E_\text{axial} \approx \frac{2kp}{r^3}.$ Direction: along $+\hat x$ (from $-q$ to $+q$), i.e. parallel to $\vec p$.

Equatorial point. At $(0, r, 0)$ each charge is at distance $\sqrt{r^2+a^2}$. By symmetry, the components along $\hat y$ cancel and only the $-\hat x$ components survive: $E = 2\cdot\frac{kq}{r^2+a^2}\cdot\frac{a}{\sqrt{r^2+a^2}} = \frac{2kqa}{(r^2+a^2)^{3/2}}.$ For $r\gg a$ this becomes $E_\text{eq} \approx \frac{kp}{r^3},$ antiparallel to $\vec p$.

Worked Example

A dipole has charges $\pm 2\,nC$ separated by $1\,cm$. Find the axial field at $10\,cm$ from the centre.

$p = q(2a) = 2\times 10^{-9} \times 0.01 = 2\times 10^{-11}\,C\cdot m$.

$E_\text{axial} = \dfrac{2\times 9\times 10^9 \times 2\times 10^{-11}}{(0.10)^3} = \dfrac{3.6\times 10^{-1}}{10^{-3}} = 360\,N/C$.

On the equatorial line at the same distance, $E_\text{eq} = 180\,N/C$ — exactly half.

Common Confusions

• $\vec p$ points from $-$ to $+$. This is a convention; many textbooks emphasize it because the torque formula depends on it.
• Axial field is twice the equatorial field for the same $r$. Memorize this — common in JEE/NEET.
• Far-field formulas assume $r \gg a$. For close-up problems, use the exact expressions.
• Dipole field is NOT a $1/r^2$ field. It is $1/r^3$ because of partial cancellation between the two charges.

Key Takeaways

• $\vec p = q\,\vec{2a}$, direction from $-q$ to $+q$.
• $E_\text{axial} = 2kp/r^3$ along $\vec p$.
• $E_\text{eq} = kp/r^3$ opposite to $\vec p$.
• Axial/equatorial ratio is $2:1$.
• Dipole field decays as $1/r^3$ — faster than a single charge.