Electric field of a dipole and behaviour in uniform field

Two equal and opposite charges, glued a tiny distance apart, are electrically silent from far away — their total charge is zero, so no observer measuring only "how much charge is out there" would notice them at all. And yet a water molecule, built almost exactly this way, is the reason water dissolves salt, sticks to itself in droplets, and carries life's chemistry. What a neutral pair does have, and a lone charge does not, is a dipole moment $\vec p$ — a built-in "handedness" pointing from the negative charge to the positive one. That single vector controls two separate stories this section tells. The first story is what the dipole's own field looks like to a distant observer. Stand directly in line with the pair and the field you feel is twice as strong as if you stood on the perpendicular bisector at the same distance — and both readings die away far faster than a single charge's field would, falling as $1/r^3$ instead of $1/r^2$ , because the near-equal-and-opposite charges almost cancel each other's effect. The second story flips the question: instead of asking what field the dipole makes, ask what an external uniform field does to the dipole. Place it at an angle and the field does not push it sideways at all — net force is zero — but it does twist it, like a compass needle nudged toward north, until $\vec p$ lines up with $\vec E$ . This is the physics behind polar molecules aligning in an electric field, the working principle sketch of microwave heating, and a favourite two-mark-then-derivation combo on CBSE boards. We build both formulas from Coulomb's law with nothing beyond algebra, then use them exactly as the board expects: quoted and applied, not re-derived from scratch every time.

Key takeaways

A dipole is a pair of equal and opposite charges $\pm q$ separated by $2a$ ; its dipole moment is $\vec p = q(2a)$ , pointing from $-q$ to $+q$ .
On the axial line (extension of the dipole axis), the field of a short dipole is $E_{\text{axial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$ , directed along $\vec p$ .
On the equatorial line (perpendicular bisector), the field is $E_{\text{equatorial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3}$ , directed opposite to $\vec p$ .
Both fall off as $1/r^3$ — one power of $r$ faster than a single point charge's $1/r^2$ — because the dipole is net-neutral; only the charge separation survives at large distance.
$E_{\text{axial}} = 2\,E_{\text{equatorial}}$ at the same distance $r$ from the dipole's centre: a fast sanity check on any answer.
In a uniform external field $\vec E$ , a dipole feels zero net force but a torque $\vec\tau = \vec p \times \vec E$ , magnitude $\tau = pE\sin\theta$ , which vanishes at $\theta = 0°, 180°$ and is maximum at $\theta = 90°$ .
Key idea: a short dipole's own field falls off as $1/r^3$ (axial twice equatorial at equal r), while an external uniform field exerts zero net force but a torque τ = pE sinθ that swings p toward E.

The core insight

A short dipole's field dies off as $1/r^3$ with the axial value twice the equatorial value at the same distance, and a uniform external field exerts zero net force on a dipole but a torque $\tau = pE\sin\theta$ that tries to align $\vec p$ with $\vec E$ .

Where it fits in the syllabus

This section sits inside "Electric Charges and Fields," directly after the electric dipole and dipole moment are introduced, in Unit I (Electrostatics) of CBSE Class 12 Physics. The 2025-26 prescribed scope names exactly two dipole-field results — axial and equatorial — plus the torque $\vec\tau = \vec p\times\vec E$ in a uniform field; these are meant to be applied, not formally re-derived from Coulomb's law in every answer, though the derivation is useful once to understand where the formulas come from. The field of a dipole at a general point (not on the axis or the equator) and the behaviour of a dipole in a non-uniform field are both explicitly out of scope for boards — this section stays within those two special points and the uniform-field case only.

The physical picture

Picture the dipole as two point charges, $+q$ at one end and $-q$ at the other, separated by $2a$ , with the midpoint called the dipole's centre. The dipole moment is a vector of magnitude $p = q(2a)$ pointing from $-q$ toward $+q$ ; it is the single quantity that survives when the dipole is viewed from far away ( $r \gg a$ ), because the individual charges $+q$ and $-q$ almost cancel and only their product with separation matters. For a point $P$ on the axial line — on the straight line through both charges, beyond one of them — superpose the field of $+q$ (pointing away from it, toward $P$ ) and the field of $-q$ (pointing toward it, i.e. also roughly toward $P$ 's direction along the axis, since $-q$ pulls). Because $P$ is closer to one charge than the other, the two fields do not fully cancel; working through the algebra for $r \gg a$ gives:

E_{\text{axial}} = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{2p}{r^3}

where: $p$ is the dipole moment ( $\text{C}\,\cdot\,\text{m}$ ), $r$ is the distance from the dipole's centre to $P$ (with $r \gg a$ ), and $\varepsilon_0$ is the permittivity of free space. The direction is along the dipole axis, same sense as $\vec p$ (from $-q$ side toward and beyond $+q$ ). For a point $P$ on the equatorial line — on the perpendicular bisector of the dipole, equidistant from both charges — the two individual fields (each of magnitude $\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2+a^2}$ , since $P$ is a slant distance $\sqrt{r^2+a^2}$ from each charge) have equal magnitude but point in different directions; their components perpendicular to the axis cancel by symmetry, and only the components parallel to the axis (pointing from $+q$ 's side toward $-q$ 's side) survive and add. For $r \gg a$ this gives:

E_{\text{equatorial}} = \dfrac{1}{4\pi\varepsilon_0}\,\dfrac{p}{r^3}

where: same symbols as above; the direction is antiparallel to $\vec p$ — opposite to the dipole moment, because the net contribution is dominated by the nearer, opposite-sign charge pulling the field back the other way.

Figure

Axial field twice the equatorial field at equal distance from a short dipole, with field direction arrows shown

The second half of this section flips the question. Instead of the field the dipole produces, consider a dipole placed inside a pre-existing uniform external field $\vec E$ (imagine it between two large charged parallel plates). Each charge feels a force $qE$ : the $+q$ charge is pushed along $\vec E$ , the $-q$ charge is pushed opposite to $\vec E$ , with equal magnitude. Because the field is uniform, these two forces are exactly equal and opposite everywhere — they add to zero, so the dipole never accelerates bodily through the field. But if $\vec p$ is not aligned with $\vec E$ , the two equal-and-opposite forces act along two different, parallel lines separated by a perpendicular distance, and a pair of equal, opposite, non-collinear forces is precisely a couple — it produces a pure rotation, a torque:

\vec \tau = \vec p \times \vec E, \qquad \tau = pE\sin\theta

where: $\theta$ is the angle between $\vec p$ and $\vec E$ ; $\tau$ is measured in $\text{N}\,\cdot\,\text{m}$ . The torque direction (by the right-hand rule of the cross product) always rotates $\vec p$ toward $\vec E$ , so $\theta = 0$ (dipole aligned with the field) is the stable equilibrium, $\theta = 180°$ is an unstable equilibrium (zero torque, but any nudge grows), and $\theta = 90°$ gives the maximum possible torque, $\tau_{\max} = pE$ , for that $p$ and $E$ .

Derivation AField of a short dipole on the axial line

The goal is to combine the fields of the two point charges $+q$ and $-q$ , separated by $2a$ , at a point $P$ on their common axis, at distance $r$ from the dipole's centre, using nothing beyond Coulomb's law, vector addition along a single line, and the far-field approximation $r \gg a$ that turns the exact algebraic fraction into the clean $1/r^3$ result quoted above. This is the one CBSE-listed formula whose derivation is worth walking through once, since the equatorial and torque results follow the same superposition logic.

Step 1. Set up coordinates. Place the dipole's centre at the origin with $-q$ at $x=-a$ and $+q$ at $x=+a$ , and let $P$ be on the axis at $x = r$ , with $r > a$ . The distance from $+q$ to $P$ is $(r-a)$ , and the distance from $-q$ to $P$ is $(r+a)$ .

Step 2. Write the two individual fields at $P$ , both along the axis. The field of $+q$ points away from $+q$ , i.e. in the $+x$ direction (toward and beyond $P$ ): $E_{+} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r-a)^2}$ . The field of $-q$ points toward $-q$ , i.e. in the $-x$ direction: $E_{-} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{(r+a)^2}$ , directed opposite to $E_+$ .

Step 3. Combine them, since $E_+$ is larger in magnitude ( $P$ is closer to $+q$ than to $-q$ ): $E_{\text{axial}} = E_+ - E_- = \dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{(r-a)^2} - \dfrac{1}{(r+a)^2}\right]$ . Putting the bracket over a common denominator, $\dfrac{1}{(r-a)^2} - \dfrac{1}{(r+a)^2} = \dfrac{(r+a)^2-(r-a)^2}{(r-a)^2(r+a)^2} = \dfrac{4ar}{(r^2-a^2)^2}$ , using the identity $(r+a)^2-(r-a)^2 = 4ar$ .

Step 4. Substitute back: $E_{\text{axial}} = \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{4ar}{(r^2-a^2)^2} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2(q\cdot 2a)\,r}{(r^2-a^2)^2} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2p\,r}{(r^2-a^2)^2}$ , recognising $q(2a) = p$ , the dipole moment.

Step 5. Apply the short-dipole (far-field) approximation $r \gg a$ , so $a^2$ is negligible next to $r^2$ in the denominator: $(r^2-a^2)^2 \approx r^4$ . The expression collapses to $E_{\text{axial}} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2p\,r}{r^4} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$ — the quoted formula. The direction, since $E_+ > E_-$ , is along $+x$ , the same sense as $\vec p$ . Note: the equatorial result follows the identical superposition idea (add the two slant fields, keep only the axis-parallel components, apply $r\gg a$ ) but is not re-derived here in full — it is quoted, per the board's own treatment.

Worked examples

Board level - axial and equatorial fields of a short dipole

Question: A short dipole of moment $p = 4\times10^{-9}\ \text{C}\,\cdot\,\text{m}$ is placed with its axis along the $x$ -direction. Find the electric field at a point $0.30\ \text{m}$ from its centre, (a) on the axial line, and (b) on the equatorial line.

Step 1. Axial field: $E_{\text{axial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3} = 9\times10^9\times\dfrac{2\times4\times10^{-9}}{(0.30)^3}$ .

Step 2. Evaluate the pieces: $(0.30)^3 = 0.027\ \text{m}^3$ , and the numerator $9\times10^9\times8\times10^{-9} = 72$ . So $E_{\text{axial}} = \dfrac{72}{0.027} \approx 2667\ \text{N/C}$ .

Step 3. Equatorial field: $E_{\text{equatorial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3} = 9\times10^9\times\dfrac{4\times10^{-9}}{0.027} = \dfrac{36}{0.027} \approx 1333\ \text{N/C}$ .

Step 4. Cross-check with the $2\times$ rule: $E_{\text{axial}}/E_{\text{equatorial}} = 2667/1333 \approx 2.0$ , exactly the factor the formulas predict — a fast way to catch an arithmetic slip on a board answer sheet.

Answer: $E_{\text{axial}} \approx 2.67\times10^3\ \text{N/C}$ directed along the dipole axis (direction of $\vec p$ ); $E_{\text{equatorial}} \approx 1.33\times10^3\ \text{N/C}$ , directed opposite to $\vec p$ .

Competitive level - torque, and the work to rotate a dipole from equilibrium to 90 degrees

Question: A dipole of moment $p = 2\times10^{-6}\ \text{C}\,\cdot\,\text{m}$ is placed at an angle of $30°$ to a uniform external field of magnitude $E = 5\times10^4\ \text{N/C}$ . (a) Find the torque on the dipole. (b) State the angle at which the torque would be maximum for the same $p$ and $E$ . (c) Find the work needed to rotate the dipole from its stable equilibrium orientation ( $\theta=0$ ) to $\theta = 30°$ , using $W = pE(1-\cos\theta)$ , a formula quoted for reference since its full derivation from $U=-\vec p\cdot\vec E$ belongs to the next chapter.

Step 1. Torque at $\theta = 30°$ : use $\tau = pE\sin\theta$ with $p = 2\times10^{-6}\ \text{C}\,\cdot\,\text{m}$ , $E = 5\times10^4\ \text{N/C}$ , $\sin30° = 0.5$ . So $\tau = (2\times10^{-6})(5\times10^4)(0.5)$ .

Step 2. Multiply in stages: $(2\times10^{-6})(5\times10^4) = 1\times10^{-1}$ ; then $(1\times10^{-1})(0.5) = 5\times10^{-2}\ \text{N}\,\cdot\,\text{m}$ .

Step 3. Maximum torque occurs where $\sin\theta$ is largest, i.e. $\sin\theta = 1$ at $\theta = 90°$ — the dipole moment perpendicular to the field. There, $\tau_{\max} = pE = (2\times10^{-6})(5\times10^4) = 1\times10^{-1}\ \text{N}\,\cdot\,\text{m} = 0.1\ \text{N}\,\cdot\,\text{m}$ .

Step 4. Work to rotate from $\theta=0$ to $\theta=30°$ : $W = pE(1-\cos\theta) = (1\times10^{-1})(1-\cos30°) = (0.1)(1-0.866) = (0.1)(0.134) \approx 1.34\times10^{-2}\ \text{J}$ .

Answer: $\tau \approx 5\times10^{-2}\ \text{N}\,\cdot\,\text{m}$ at $\theta = 30°$ ; torque is maximum, $\tau_{\max} = 0.1\ \text{N}\,\cdot\,\text{m}$ , at $\theta = 90°$ ; the work to rotate from stable equilibrium to $30°$ is about $1.34\times10^{-2}\ \text{J}$ . Timing note: on a competitive paper, compute $pE$ once as a single number (here $0.1\ \text{N}\,\cdot\,\text{m}$ ) and reuse it for both the torque-at-angle and work-to-rotate parts — recomputing $p\times E$ twice wastes 20-30 seconds you don't have in a timed section.

Common traps

Forgetting the factor of 2 between the axial and equatorial formulas — a very common slip is using the equatorial formula for an axial point, or dropping the 2 in the axial expression entirely.
Using the general point-charge $1/r^2$ falloff for a dipole's field instead of the correct $1/r^3$ falloff, which applies specifically to short dipoles viewed from far away.
Forgetting that the equatorial field is antiparallel to $\vec p$ , not parallel to it — a direction reversal relative to the axial field's direction.
Mixing up $r$ (distance from the dipole's centre) with the distance from one of the individual charges; the formulas are defined with $r$ measured from the centre, and only valid for $r \gg a$ .
On the torque question, quoting $\tau = pE$ as a blanket formula and forgetting the $\sin\theta$ factor — $pE$ is only the maximum torque, at $\theta = 90°$ , not the general expression.

Common trapTorque without force: a dipole in a uniform field never accelerates bodily, no matter how strong the torque

The single biggest misconception here is assuming that a nonzero torque means a nonzero net force. In a uniform field the two forces on $+q$ and $-q$ are always exactly equal in magnitude and exactly opposite in direction, so they always sum to zero — the dipole's centre of mass does not accelerate, however large $\tau = pE\sin\theta$ becomes. What a nonzero torque means is that those two equal-and-opposite forces act along two different lines (unless $\vec p \parallel \vec E$ ), and a pair of equal, opposite, non-collinear forces is a pure couple: it spins the object without translating it. This distinction — zero net force, nonzero torque — is exactly the situation of a compass needle in the Earth's magnetic field: the needle does not fly off, it only rotates. If the external field were non-uniform instead, the two forces would differ in magnitude too, and the dipole would then feel a net force as well as a torque — but that case is outside the CBSE syllabus for this topic and should not appear in a board answer.

You should be able to…

State and correctly apply the axial-field and equatorial-field formulas for a short dipole, including their directions relative to $\vec p$ .
Explain, in words, why a dipole's field falls off as $1/r^3$ rather than $1/r^2$ .
Derive the axial-field formula from Coulomb's law by superposing the two individual charge fields and applying $r \gg a$ .
Compute the torque on a dipole in a uniform external field using $\tau = pE\sin\theta$ , and identify the orientations of stable equilibrium, unstable equilibrium, and maximum torque.
Explain clearly why the net force on a dipole in a uniform field is zero even while a torque acts on it.

Ask the tutor: want to see the equatorial-field derivation worked out in the same step-by-step style as the axial one, or try a few more torque-and-equilibrium problems? Ask the AI tutor.

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

Because a dipole has zero total charge. A single point charge's field falls as $1/r^2$ because there is genuinely "more charge left over" at every distance — nothing cancels. A dipole's two charges are equal and opposite, so from far away they almost completely cancel each other's field; what survives is only the small residual effect caused by the tiny separation $2a$ between them, and that residual shrinks one extra power of $r$ faster, giving $1/r^3$ . The larger $r$ gets relative to $a$ , the more nearly the two charges look like they're sitting at the same point, and the weaker the leftover field becomes.

If the net force on a dipole in a uniform field is always zero, how does a polar molecule like water actually feel an electric field at all?

It feels the field through the torque, not through a net push. When you put a water molecule (which behaves like a tiny permanent dipole) between charged plates, it does not fly toward either plate — the equal and opposite forces on its slightly positive and slightly negative ends cancel exactly. What happens instead is that the torque $\tau = pE\sin\theta$ twists the molecule until its dipole moment lines up with the field. That alignment is exactly what happens (on a much larger, statistical scale, and with more subtlety) inside a dielectric placed in a capacitor, and it is the qualitative picture behind microwave ovens agitating water molecules with a rapidly flipping field.

Why is theta = 0 stable and theta = 180 degrees unstable, if the torque is zero at both?

Zero torque only tells you the dipole is momentarily not turning — it does not tell you whether a small nudge grows or shrinks. At $\theta = 0$ ( $\vec p$ parallel to $\vec E$ ), nudge the dipole slightly to $\theta = \epsilon$ and the torque $\tau = pE\sin\epsilon$ is small but points to rotate it back toward $\theta = 0$ — restoring, so it's stable. At $\theta = 180°$ ( $\vec p$ antiparallel to $\vec E$ ), nudge it slightly and the torque now points to rotate it further away from $180°$ , toward $0°$ — the nudge grows, so it's unstable, exactly like a pencil balanced on its tip versus one lying flat. This is the same stable-versus-unstable-equilibrium idea you meet with a pendulum at its lowest versus highest point.

Do I need to memorise the axial-field derivation for boards, or just the final formulas?

For this specific topic, CBSE expects you to be able to apply the axial and equatorial formulas and the torque formula fluently — that is what shows up as two- and three-mark numerical questions. The full algebraic derivation (the one worked through in this section using $(r-a)$ and $(r+a)$ ) is good to understand once so the $1/r^3$ result and the factor of 2 make sense to you, and it does occasionally appear as a derivation question, but it is not something you need to reproduce from memory for every short-answer question — know the results cold, and be able to sketch the reasoning (superposition plus $r \gg a$ ) if asked to justify them.