The Dipole and the Field Point
An electric dipole consists of two charges $+q$ and $-q$ separated by a small distance $2a$. Its total charge is zero, so it cannot be summarised by a single Coulomb potential. Instead it is characterised by a dipole moment vector $\vec{p}$ of magnitude $p = q \times 2a$, pointing from $-q$ to $+q$ (NCERT 2.4). To find the potential at a field point $P$, we place the origin at the centre $O$ of the dipole and describe $P$ by its distance $r = OP$ and the angle $\theta$ that $\vec{r}$ makes with $\vec{p}$.
Because potential is a scalar tied to the work done by the field, it obeys the same superposition principle as the field itself. The dipole potential is therefore the plain algebraic sum of the potentials of $+q$ and $-q$ — no vector addition is required. This is precisely why the potential is easier to handle than the field, even though the geometry is identical.
Deriving the Dipole Potential
Let $r_1$ and $r_2$ be the distances of $P$ from $+q$ and $-q$. By superposition,
$V = \dfrac{q}{4\pi\varepsilon_0}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$
For a small dipole, geometry gives the two distances in terms of $r$ and $\theta$ (NCERT Eq. 2.10, NIOS Eq. derived from Fig. 16.4):
$r_1 \approx r - a\cos\theta, \qquad r_2 \approx r + a\cos\theta$
Retaining only first-order terms in $a/r$ via the binomial expansion, NCERT obtains $\dfrac{1}{r_1} \approx \dfrac{1}{r}\left(1 + \dfrac{a}{r}\cos\theta\right)$ and $\dfrac{1}{r_2} \approx \dfrac{1}{r}\left(1 - \dfrac{a}{r}\cos\theta\right)$. Substituting and using $p = 2qa$:
$V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{p}\cdot\hat{r}}{r^2}\qquad (r \gg a)$
This is NCERT Eq. (2.15), with $p\cos\theta = \vec{p}\cdot\hat{r}$ where $\hat{r}$ is the unit vector along $OP$. It is approximate, valid only when $r$ is large compared to the dipole size so that higher-order terms in $a/r$ are negligible. For an idealised point dipole ($2a \to 0$ at fixed $p$), the same expression is exact at every $r$.
Axial and Equatorial Special Cases
Two directions dominate the question bank. On the axial line the field point lies along $\vec{p}$, so $\theta = 0$, $\cos\theta = 1$, and the potential is maximal in magnitude. On the equatorial line — the perpendicular bisector of the dipole — $\theta = 90^\circ$, $\cos\theta = 0$, and the potential vanishes exactly. The equatorial result is purely geometric: every point on the bisector is equidistant from $+q$ and $-q$, so their contributions cancel.
| Location | Angle θ | cos θ | Potential V |
|---|---|---|---|
| Axial, +q side | $0^\circ$ | $+1$ | $+\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2} = +\dfrac{kp}{r^2}$ |
| Axial, −q side | $180^\circ$ | $-1$ | $-\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2} = -\dfrac{kp}{r^2}$ |
| Equatorial line | $90^\circ$ | $0$ | $0$ |
| General point | $\theta$ | $\cos\theta$ | $\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$ |
Here $k = \dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^9~\text{N m}^2\,\text{C}^{-2}$. The axial sign flips with the side: positive near $+q$ ($\theta = 0$), negative near $-q$ ($\theta = 180^\circ$), as in NCERT Eq. (2.16) and NIOS Eqs. (16.14)–(16.16).
The dipole result is just two Coulomb potentials added. Revise the single-charge case in Potential Due to a Point Charge before tackling the angular form.
Angular Dependence and the 1/r² Falloff
Unlike a point charge, whose potential depends only on $r$, the dipole potential depends on both $r$ and the angle $\theta$. As $\theta$ sweeps from $0$ to $180^\circ$, $\cos\theta$ runs from $+1$ through $0$ at $90^\circ$ to $-1$, so the potential is positive in the forward hemisphere, zero on the equatorial plane, and negative in the rear hemisphere. The dependence is, however, axially symmetric about $\vec{p}$: rotating $P$ around the axis at fixed $\theta$ traces a cone of constant potential (NCERT 2.4(i)).
The second feature is the falloff. At fixed direction, $V \propto 1/r^2$, faster than the $1/r$ of a single charge. The reason is cancellation — the $+q$ and $-q$ contributions nearly annul, leaving only a residue proportional to $a/r$, which multiplies the underlying $1/r$ into $1/r^2$ (NCERT 2.4(ii)).
Equatorial potential is zero — but the field is not
Students confuse "zero potential" with "zero field". On the equatorial line $V = 0$, yet the electric field there is decidedly non-zero (it points anti-parallel to $\vec{p}$). Potential vanishing does not force the field to vanish — that requires $\dfrac{dV}{dr} = 0$, a different condition.
Equatorial line: $V = 0$ but $\vec{E} \neq 0$. The two are independent claims.
Field 1/r³ versus Potential 1/r²
The single most tested contrast in this subtopic is the difference in falloff between the dipole field and the dipole potential. They are not the same power, and mixing them is a guaranteed error. The field is the spatial gradient of the potential, $E = -\dfrac{dV}{dr}$; differentiating a $1/r^2$ potential introduces one extra power of $r$ in the denominator, yielding $1/r^3$.
| Source | Potential V | Field E |
|---|---|---|
| Point charge | $\propto \dfrac{1}{r}$ | $\propto \dfrac{1}{r^2}$ |
| Dipole | $\propto \dfrac{1}{r^2}$ | $\propto \dfrac{1}{r^3}$ |
The pattern is consistent: in every row the field falls off one power faster than the potential, because $E$ is the derivative of $V$. Memorising "$1/r^2$ for dipole potential, $1/r^3$ for dipole field" closes the most common single-mark loss in this chapter.
Do not write the dipole potential as 1/r
A point-charge potential goes as $1/r$; a dipole potential goes as $1/r^2$. The extra power comes from the near-cancellation of $+q$ and $-q$. Carrying the point-charge $1/r$ habit into a dipole problem gives a wrong distance-scaling and a wrong numerical answer.
Dipole: $V \propto 1/r^2$ (not $1/r$); $E \propto 1/r^3$ (not $1/r^2$).
Worked Example
A short dipole has moment $p = 4 \times 10^{-6}~\text{C m}$. Find the potential at an axial point $2~\text{m}$ from its centre. Take $\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^9$ SI units.
On the axis, $\theta = 0$ so $\cos\theta = 1$ and $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2}$.
$V = 9\times10^9 \times \dfrac{4\times10^{-6}}{(2)^2} = 9\times10^9 \times \dfrac{4\times10^{-6}}{4} = 9\times10^{3}~\text{V}$.
So the axial potential is $\pm 9 \times 10^{3}~\text{V}$ (sign set by which end of the axis $P$ lies on). Note the formula used the $1/r^2$ dependence, not $1/r$, and the axial $\cos\theta = 1$.
Potential Due to an Electric Dipole
- For $r \gg a$: $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{p}\cdot\hat{r}}{r^2}$, with $\theta$ measured from the axis along $\vec{p}$.
- Axial line: $V = \pm \dfrac{kp}{r^2}$ ($+$ near $+q$ at $\theta=0$, $-$ near $-q$ at $\theta=180^\circ$).
- Equatorial line ($\theta = 90^\circ$): $V = 0$, yet $\vec{E} \neq 0$ there.
- Dipole potential $\propto 1/r^2$ (faster than a point charge's $1/r$) because the $\pm$ charges nearly cancel.
- Dipole field $\propto 1/r^3$; the field always falls one power faster than the potential.
- The formula is approximate for a real dipole ($r\gg a$) and exact for an ideal point dipole.