Potential due to an electric dipole

Stand a bar magnet-shaped pair of charges on a table — a $+q$ and a $-q$ glued a hair's width apart — and ask a simple question: what does the electric potential look like all around it? A single point charge answers this in one clean line, $V \propto 1/r$ , spreading its influence equally in every direction. The dipole refuses to be so simple. Walk around it and the potential changes sign, collapses to exactly zero along one special line, and dies away twice as fast with distance as a lone charge would. Direction, not just distance, now matters. This is not a curiosity confined to physics problems. Every polar molecule — water foremost among them — behaves electrically like a tiny dipole, and it is precisely this direction-dependent, fast-fading potential that governs how one water molecule "feels" its neighbour, how a dipole aligns in an electric field, and why the equatorial plane of any dipole is a place of perfect electrical neutrality even though real charge sits on either side of it. The chapter "Electrostatic Potential and Capacitance" builds this result directly from the previous section's point-charge potential, using nothing more than the superposition principle — potentials add like ordinary numbers, not like vectors — and a little geometry. What comes out is one of the most quietly elegant formulas in the CBSE Class 12 syllabus: $V = \dfrac{p\cos\theta}{4\pi\varepsilon_0 r^2}$ , a single expression that swallows the axial case, the equatorial case, and every angle in between. This article rebuilds that formula from scratch, drills it against board-style and competitive-style numbers, and flags exactly where students bleed marks.

Key takeaways

A dipole's potential is the superposition of the potentials of its two point charges: $V = \dfrac{q}{4\pi\varepsilon_0}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$ — a plain scalar sum, no vector addition needed.
On the equatorial line (perpendicular bisector), $r_1 = r_2$ exactly, so the two terms cancel completely: $V_{\text{equatorial}} = 0$ , at every point on that line, not approximately but exactly.
On the axial line, for $r \gg a$ (short-dipole limit), the potential simplifies to $V_{\text{axial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2}$ — positive on the $+q$ side, negative on the $-q$ side.
The general formula, valid for any direction at $r \gg a$ , is $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$ , where $\theta$ is measured from the dipole axis; it reduces correctly to both special cases above.
Dipole potential falls off as $1/r^2$ , one power of $r$ faster than a single point charge's $1/r$ — a direct fingerprint of two nearly-cancelling charges.
This is a genuinely directional quantity: unlike a point charge, a dipole's potential is not spherically symmetric — it depends on angle as well as distance.
Key idea: a short dipole's potential is $V = \dfrac{p\cos\theta}{4\pi\varepsilon_0 r^2}$ , built by superposing the two point-charge potentials, exactly zero on the equatorial line, maximal (and $1/r^2$ ) on the axial line, and always one power of $r$ steeper than a single charge's potential.

The core insight

A short electric dipole's potential at distance $r \gg a$ and angle $\theta$ from its axis is $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$ , obtained by superposing the potentials of $+q$ and $-q$ , vanishing exactly on the equatorial line and falling off as $1/r^2$ — one power faster than the $1/r$ potential of a lone point charge.

Where it fits in the syllabus

This topic sits inside the CBSE Class 12 Physics chapter "Electrostatic Potential and Capacitance" (Unit II, Electrostatics), immediately after potential due to a point charge and immediately before potential due to a general system of charges — the natural bridge between the two. It leans on the dipole moment definition, $\vec p = q\cdot 2\vec a$ , directed from $-q$ to $+q$ , carried over from the previous chapter "Electric Charges and Fields." The 2025-26 CBSE prescription lists "potential due to a dipole" explicitly in scope, expects the short-dipole approximation $r \gg a$ throughout, and does not ask for the exact (non-approximated) potential at arbitrary $r$ — so this article stays inside that boundary and never derives beyond it.

The physical picture

Picture the dipole as two point charges, $+q$ at one end and $-q$ at the other, separated by a small distance $2a$ , with the dipole moment $p = q(2a)$ pointing from $-q$ to $+q$ . Electric potential is a scalar — it has no direction of its own — so unlike electric field, where you must add vectors carefully, the potentials of the two charges at any external point simply add as ordinary signed numbers:

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{r_1} - \frac{q}{r_2}\right)

where: $r_1$ is the distance of the field point from $+q$ , $r_2$ is its distance from $-q$ , and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ is the permittivity of free space, so that $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^{9}\ \text{N m}^2\text{C}^{-2}$ . The physical reading is simple: the positive charge contributes a positive potential that weakens with distance from it, the negative charge contributes a negative potential that weakens with distance from it, and the net value at any point is whichever contribution wins after the two are added. Two special directions immediately settle the extremes. On the equatorial line — the perpendicular bisector of the line joining the charges — every point is, by the definition of a perpendicular bisector, exactly equidistant from $+q$ and $-q$ , so $r_1 = r_2$ and the bracket in the formula above becomes zero identically, for any value of $r_1$ whatsoever:

V_{\text{equatorial}} = 0

On the axial line, extended through the dipole's two charges, a field point at distance $r$ from the centre (with the short-dipole condition $r \gg a$ enforced) sits closer to one charge and farther from the other by almost exactly $a$ , and the calculation (carried out step by step in the derivation below) collapses to:

V_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^2} \qquad (r \gg a)

The sign of this axial potential is positive on the side nearer $+q$ and negative on the side nearer $-q$ — it is not symmetric the way a point charge's potential is, because a dipole has no overall charge to set one universal sign. Generalising to a point at distance $r$ from the dipole's centre, making an angle $\theta$ with the dipole axis (again under $r \gg a$ ), the single formula that covers every direction is:

V = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}

where: $\theta$ is measured from the direction of $\vec p$ (i.e. from the $-q$ -to- $+q$ line) to the line joining the dipole's centre to the field point. At $\theta = 0°$ , $\cos\theta = 1$ and this reduces exactly to $V_{\text{axial}}$ ; at $\theta = 90°$ , $\cos\theta = 0$ and it reduces exactly to $V_{\text{equatorial}} = 0$ ; at $\theta = 180°$ , $\cos\theta = -1$ , giving the negative axial value on the $-q$ side. One formula, three checks, all consistent. Compare this with the dipole's electric field, which the previous chapter showed varies as $1/r^3$ on the axis — one power of $r$ steeper than the potential's $1/r^2$ . That same one-power relationship (field is potential's "steepness," $E = -dV/dr$ along a radial line) also links a point charge's $1/r^2$ field to its $1/r$ potential. Recognising this pattern is a fast way to sanity-check either formula from memory under exam pressure.

Figure

A dipole with positive charge on the left and negative charge on the right, showing potential contours that bulge positive near the positive charge and negative near the negative charge, with a vertical dashed equatorial line down the middle marked as the zero potential line

Derivation APotential on the axial line of a short dipole

The goal is to show that the potential of a dipole at a point on its own axis, at distance $r$ from the centre with $r \gg a$ , collapses from the exact two-term superposition expression to the clean short-dipole result $V_{\text{axial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2}$ , using only algebra and the binomial approximation $(1+x)^{-1}\approx 1-x$ for small $x$ — no calculus beyond what has already been used earlier in the chapter.

Step 1. Set up the geometry. Place the dipole with $-q$ and $+q$ separated by $2a$ , centre at the origin, and consider a point $P$ on the axis at distance $r$ from the centre, on the side of $+q$ . The distance from $P$ to $+q$ is $r_1 = r - a$ , and the distance from $P$ to $-q$ is $r_2 = r + a$ .

Step 2. Write the exact superposed potential. By the superposition of scalar potentials, $V = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q}{r-a} - \dfrac{q}{r+a}\right)$ . Combine the two fractions over a common denominator: $V = \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{(r+a) - (r-a)}{(r-a)(r+a)} = \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{2a}{r^2 - a^2}$ .

Step 3. Introduce the dipole moment. Since $p = q(2a)$ , the numerator $q\cdot 2a$ is exactly $p$ , giving the exact (not yet approximated) result $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2 - a^2}$ . This form is valid at any $r > a$ , but it still carries the awkward $a^2$ term.

Step 4. Apply the short-dipole approximation. When $r \gg a$ , the term $a^2$ is negligible next to $r^2$ , so $r^2 - a^2 \approx r^2$ . Substituting gives the final short-dipole axial result, $V_{\text{axial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2}$ , matching the formula quoted in "The physical picture" and consistent with putting $\theta = 0°$ into the general formula $V = \dfrac{p\cos\theta}{4\pi\varepsilon_0 r^2}$ .

Worked examples

Board level - axial potential of a small dipole

An electric dipole has dipole moment $p = 4\times10^{-9}\ \text{C}\,\cdot\,\text{m}$ . Find the electric potential at a point on its axial line, $20\ \text{cm}$ from the centre of the dipole, on the side of the positive charge. Take $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N m}^2\text{C}^{-2}$ .

Step 1. Identify the formula: since the point lies on the axial line and $r = 0.20\ \text{m}$ is certainly much larger than any realistic dipole separation, use $V_{\text{axial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2}$ .

Step 2. Square the distance: $r^2 = (0.20)^2 = 0.04\ \text{m}^2$ .

Step 3. Substitute the numbers: $V = (9\times10^9)\times\dfrac{4\times10^{-9}}{0.04} = (9\times10^9)\times(10^{-7}) = 900\ \text{V}$ .

Answer: the potential is $+900\ \text{V}$ , positive because the chosen point lies on the side of $+q$ .

Competitive level - potential at a general angle, then locate the zero

A dipole of moment $p = 4\times10^{-9}\ \text{C}\,\cdot\,\text{m}$ (the same dipole as above) is now considered at a point $P$ at the same distance $r = 0.20\ \text{m}$ from its centre, but at an angle $\theta = 60°$ to the dipole axis. (a) Find the potential at $P$ . (b) State, without recomputation, at what angle the potential from this same dipole, at the same distance, would be exactly zero.

Step 1. Use the general short-dipole formula, since $P$ is now off-axis: $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$ .

Step 2. Insert the known axial result from the previous example as a shortcut: at this $r$ , $\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2} = 900\ \text{V}$ (computed above), so $V = 900\times\cos 60° = 900\times 0.5 = 450\ \text{V}$ .

Step 3. For part (b), recall that the general formula vanishes whenever $\cos\theta = 0$ , which happens at $\theta = 90°$ , regardless of the value of $r$ or $p$ — this is exactly the equatorial line, so the zero-potential direction is $\theta = 90°$ from the axis, matching the second worked result in Pass-1 (equatorial potential is always zero).

Answer: $V(60°) = 450\ \text{V}$ , and the potential vanishes at $\theta = 90°$ (the equatorial line), for any $r$ and any $p$ . Timing note: once the axial value at a given $r$ is known (or computed first, as here), every other angle at the same $r$ is just that axial value multiplied by $\cos\theta$ — no need to recompute $\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2}$ from scratch. This shortcut turns a potential two-mark repeat calculation into a ten-second multiplication, valuable when a competitive question asks for potential at several angles at the same distance.

Common traps

Writing dipole potential as $1/r$ , copying the point-charge formula out of habit — it is $1/r^2$ , because the near-cancellation of the $+q$ and $-q$ contributions removes the leading $1/r$ term entirely (seen explicitly in Step 3 of the derivation, where the $2a$ in the numerator is what survives).
Treating equatorial potential as "small" rather than exactly zero — by the geometry of the perpendicular bisector, $r_1 = r_2$ always, so the two terms cancel completely, not approximately, at every point on that line.
Confusing the axial-only formula $V_{\text{axial}} = \dfrac{p}{4\pi\varepsilon_0 r^2}$ (implicitly $\cos\theta = 1$ ) with the general angle-dependent formula $V = \dfrac{p\cos\theta}{4\pi\varepsilon_0 r^2}$ — using the axial formula for an off-axis point silently drops the $\cos\theta$ factor and overstates the answer.
Dropping the short-dipole condition $r \gg a$ — the exact result from Step 3 of the derivation, $V = \dfrac{p}{4\pi\varepsilon_0(r^2-a^2)}$ , is only simplified to $V = \dfrac{p}{4\pi\varepsilon_0 r^2}$ once $a^2$ is negligible; at small $r$ this approximation is not valid.
Mixing up potential with electric field: the field falls off one power faster ( $1/r^3$ on axis) than the potential ( $1/r^2$ ) — stating the wrong exponent for the wrong quantity is a very common mark-loss.

You should be able to…

Derive the axial dipole potential from the superposition of the two point-charge potentials, including the algebra step where $r^2 - a^2$ simplifies to $r^2$ under $r \gg a$ .
State and correctly apply the general formula $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$ for any angle, and recover the axial and equatorial cases from it by substitution.
Explain physically, using the equidistance argument, why the equatorial potential is exactly zero rather than merely small.
Explain why dipole potential falls off as $1/r^2$ while a point charge's potential falls off as $1/r$ , and why the dipole's field is a further power of $r$ steeper still.
Compute potential at axial, equatorial, or general-angle points given the dipole moment and distance, including using an already-known axial value as a shortcut via $\cos\theta$ .

Ask the tutor: curious how the exact (non-approximated) potential formula behaves as $r$ approaches $a$ , or want the general-angle formula derived using the cosine rule for $r_1$ and $r_2$ instead of the axial shortcut — ask the AI tutor to work through it with you.

Why does the dipole potential fall off faster than a single charge's potential?

Because a dipole has zero net charge — the $+q$ and $-q$ contributions are opposite in sign and, at large distances, nearly equal in size, so they almost completely cancel. The derivation above shows this literally: the exact axial potential is $\dfrac{p}{4\pi\varepsilon_0(r^2-a^2)}$ , and the leading $1/r$ behaviour you would expect from either charge alone has already vanished by the time you combine the two fractions — what survives is a much smaller residual term controlled by the separation $2a$ , not by either charge individually. That residual falls off as $1/r^2$ rather than $1/r$ , precisely because it is a "leftover" of two near-equal, opposite-sign contributions, not a full-strength single-charge effect.

If the equatorial potential is exactly zero, does that mean there is no electric field there either?

No — this is a genuinely different question, and mixing the two up is a classic slip. The field on the equatorial line is not zero; it is directed antiparallel to the dipole moment and falls off as $1/r^3$ , as covered in the earlier chapter on dipole fields. Potential being zero only means that no work is needed to bring a unit positive charge from infinity to that specific point along any path — it says nothing about whether a force acts on a charge placed there. A useful way to remember it: zero potential at a point does not require zero field at that point, since potential is a cumulative, path-independent quantity while field is the local force per unit charge.

Can I use the general formula V = p cos theta over four pi epsilon naught r squared for any distance r, or only far away?

Only far away, in the sense of $r \gg a$ , where $a$ is the half-separation of the dipole's charges. This is called the short-dipole approximation, and it is baked into the derivation: the exact result before approximating is $V = \dfrac{p}{4\pi\varepsilon_0(r^2-a^2)}$ on the axis, and only when $a^2$ is negligible compared to $r^2$ does that simplify to the clean $p/r^2$ (or, more generally, $p\cos\theta/r^2$ ) form. For a molecular-scale dipole like a water molecule, "far" can mean just a few nanometres, since $a$ itself is sub-nanometre; for the textbook problems in this chapter, any distance many times the given separation comfortably satisfies $r \gg a$ .

Is this the same dipole potential used to describe how polar molecules like water interact?

Yes, in spirit — this is exactly the formula that explains why a water molecule (which behaves electrically like a small dipole, oxygen slightly negative and the hydrogens slightly positive) exerts a direction-dependent influence on its neighbours rather than a uniform, distance-only influence the way a stray ion would. The board syllabus does not ask you to apply this formula to molecular chemistry, and the exact treatment of real polar molecules also needs quantum mechanics for the charge distribution — but the qualitative shape of the result, potential strongest and positive along the dipole axis, exactly zero on the perpendicular bisector, and falling as $1/r^2$ , is the same physics you are learning here in its idealised point-charge form.