PhysicsISC Class 12 · Electric Charges and Fields

Electric Dipole: Dipole Moment, Field on the Axis and Equator, and Torque

Updated 15 October 2026 · Fact-checked

An electric dipole is a pair of equal and opposite charges ±q separated by 2l. Its dipole moment $\vec{p} = q(2\vec{l})$ points from −q to +q. For a short dipole, the field is $\frac{2kp}{r^3}$ on the axis and $\frac{kp}{r^3}$ on the equatorial line, and a uniform field exerts no net force but a torque $\vec{\tau} = \vec{p} \times \vec{E}$.

Axial field points along p and is twice as strong as the equatorial field, which points opposite to p.

Illustration drawn for QuestionBuddy.

Understand Electric Dipole: Field and Torque

What a dipole is. Two equal and opposite point charges $+ q$ and $- q$ a small distance $2 l$ apart form an electric dipole. Polar molecules such as water behave like dipoles, and so do atoms in an external field.

Dipole moment. $p = q \times 2 l$ , a vector from $- q$ to $+ q$ along the axis, with SI unit C m. The axis is the line through both charges; the equatorial line (plane) is the perpendicular bisector of the axis. The net charge of a dipole is zero, but its field is not, because the two charges are at different places.

Field on the axial line. Take a point P on the axis at distance $r$ from the centre O, on the side of $+ q$ . Distances: $r - l$ from $+ q$ and $r + l$ from $- q$ .
$E_{+} = \frac{k q}{( r - l ) ^{2}}$ (away from +q, along $p$ ), $E_{-} = \frac{k q}{( r + l ) ^{2}}$ (towards −q, opposite to $p$ ).
$E_{a x ia l} = k q [\frac{1}{( r - l ) ^{2}} - \frac{1}{( r + l ) ^{2}}] = k q \frac{4 r l}{( r ^{2} - l ^{2} ) ^{2}} = \frac{2 k p r}{( r ^{2} - l ^{2} ) ^{2}},$
directed along $p$ . For a short dipole ( $r ≫ l$ ): $E_{a x ia l} = \frac{2 k p}{r ^{3}}$ .

Field on the equatorial line. At P on the perpendicular bisector, distance $r$ from O, each charge is $r^{2} + l^{2}$ away, so
$∣ E_{+} ∣ = ∣ E_{-} ∣ = \frac{k q}{r ^{2} + l ^{2}} .$
Resolve each along and perpendicular to the axis. The components perpendicular to the axis cancel; the components parallel to the axis both point from $+ q$ towards $- q$ (opposite to $p$ ) and add. With $cos θ = \frac{l}{r ^{2} + l ^{2}}$ ,
$E_{e q} = 2 \frac{k q}{r ^{2} + l ^{2}} \cdot \frac{l}{r ^{2} + l ^{2}} = \frac{k p}{( r ^{2} + l ^{2} ) ^{3/2}},$
directed opposite to $p$ . Short dipole: $E_{e q} = \frac{k p}{r ^{3}}$ .

Comparing. At equal distances from a short dipole, $\frac{E _{a x ia l}}{E _{e q}} = 2$ , the 2025 ISC result. The field falls as $\frac{1}{r ^{3}}$ , faster than a point charge's $\frac{1}{r ^{2}}$ , because the two charges partly cancel each other at large distances. The "short dipole" formulas are good when $r$ is at least about 10 times $l$ ; closer than that, use the exact forms.

Zero potential. Every point on the equatorial plane is equidistant from $+ q$ and $- q$ , so the potential there is zero. The locus of zero potential is the equatorial plane (perpendicular bisector), even though the field there is not zero.

Dipole in a uniform field. In a uniform field $E$ the forces $+ q E$ and $- q E$ are equal and opposite: net force zero. They act along different lines, so they form a couple. If $p$ makes angle θ with $E$ , the perpendicular distance between the forces is $2 l sin θ$ , so
$τ = q E \cdot 2 l sin θ = pE sin θ, τ = p \times E .$
The torque turns $p$ towards $E$ . It is maximum ( $pE$ ) at θ = 90° and zero at 0° (stable equilibrium) and 180° (unstable equilibrium).

Non-uniform field. If the field is stronger at one charge than the other, the forces no longer cancel and the dipole also feels a net force, pulling it (when aligned with the field) towards the stronger field. This is why a charged comb attracts neutral bits of paper: it first induces dipoles, then pulls them into its stronger field.

Field lines of a dipole start on $+ q$ and end on $- q$ , bending round; on the equatorial line they are parallel to the axis, pointing from + side to − side.

Electric Dipole: Field and Torque: formulas and key facts

Dipole moment: $p = q (2 l)$; From −q to +q; unit C m.
Axial field (exact): $E = \frac{2 k p r}{( r ^{2} - l ^{2} ) ^{2}}$; Along $p$ .
Equatorial field (exact): $E = \frac{k p}{( r ^{2} + l ^{2} ) ^{3/2}}$; Opposite to $p$ .
Short dipole: $E_{a x} = \frac{2 k p}{r ^{3}}, E_{e q} = \frac{k p}{r ^{3}}$; Ratio 2 : 1 at equal distances.
Torque: $τ = p \times E, τ = pE sin θ$; Maximum pE at 90°.
Net force (uniform field): $F = q E - q E = 0$; Non-zero only in a non-uniform field.

How to solve Electric Dipole: Field and Torque questions

Decide where the point is (axis or equator) and whether the dipole is short.

1Find $p = q \times 2 l$ in C m (convert nC, μC, mm, cm).
2Locate the point: on the axis (in line with the charges) or on the equatorial line (perpendicular bisector).
3Check $r ≫ l$ . If yes, use $\frac{2 k p}{r ^{3}}$ or $\frac{k p}{r ^{3}}$ ; if not, use the exact formula.
4State the direction: along $p$ on the axis, opposite to $p$ on the equator.
5For torque, use $τ = pE sin θ$ with θ between $p$ and $E$ .
6For a point affected by two dipoles, find each field as a vector and add.

Electric Dipole: Field and Torque in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Torque on a polar molecule

Suppose a water molecule, with dipole moment about $6.2 \times 1 0^{- 30}$ C m, sits in the oscillating electric field of a microwave oven whose amplitude is $5 \times 1 0^{4}$ N C⁻¹.

Find the largest torque the field can exert on the molecule. Why does the oven heat water?

Try it first, then show the solutionSolution

$τ_{ma x} = pE = 6.2 \times 1 0^{- 30} \times 5 \times 1 0^{4} = 3.1 \times 1 0^{- 25}$ N m.
The field keeps reversing, so the torque keeps twisting the molecules back and forth; their jostling raises the temperature.

Answer: $3.1 \times 1 0^{- 25}$ N m; the reversing torque makes molecules rotate to and fro, producing heat.

Memory hook: Torque turns p towards E; a reversing E keeps the dipole turning.

Inverse-cube fall-off

Imagine a field sensor that reads 800 N C⁻¹ at 10 cm along the axis of a small dipole.

What will it read at 20 cm along the same axis?

Try it first, then show the solutionSolution

$E \propto \frac{1}{r ^{3}}$ for a short dipole.
Doubling r divides E by $2^{3} = 8$ : $\frac{800}{8} = 100$ N C⁻¹.

Answer: 100 N C⁻¹.

Memory hook: Point charge: square law. Dipole: cube law.

Tricks to remember Electric Dipole: Field and Torque

Short memory aids. Say each one aloud once, then use it on a question the same day.

Axis is twice
Same distance, short dipole: axial field is twice the equatorial field.
Use it when: Ratio questions.
Axis along, equator against
Axial field points along $p$ ; equatorial field points against $p$ .
Use it when: Stating directions.
Minus for axis, plus for equator
Exact denominators: $(r^{2} - l^{2})^{2}$ on the axis, $(r^{2} + l^{2})^{3/2}$ on the equator.
Use it when: Exact formulas.
Torque is a cross
$τ = p \times E$ , same pattern as $τ = m \times B$ for a magnet.
Use it when: Torque and equilibrium.

Electric Dipole: Field and Torque: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: When is the torque greatest? · 1 mark · MCQBoard level

A dipole is placed in a uniform electric field. The torque on it is maximum when the angle between $p$ and $E$ is:
(a) 0°
(b) 45°
(c) 90°
(d) 180°

Try it first, then show the solutionSolution

$τ = pE sin θ$ is largest when $sin θ = 1$ .

Answer: (c) 90°.

Example 2: Axial and equatorial fields · 2 marks · Short answerBoard level

Charges of $\pm 5 μ$ C are 2 mm apart. Find the dipole moment and the field at 10 cm from the centre on (a) the equatorial line, (b) the axis.

Try it first, then show the solutionSolution

$p = 5 \times 1 0^{- 6} \times 2 \times 1 0^{- 3} = 1 \times 1 0^{- 8}$ C m; r = 10 cm ≫ 1 mm, so the dipole is short. (1)
$E_{e q} = \frac{k p}{r ^{3}} = \frac{9 \times 1 0 ^{9} \times 1 0 ^{- 8}}{1 0 ^{- 3}} = 9 \times 1 0^{4}$ N C⁻¹ (opposite to p); $E_{a x} = 1.8 \times 1 0^{5}$ N C⁻¹ (along p). (1)

Answer: $p = 1 0^{- 8}$ C m; (a) $9 \times 1 0^{4}$ N C⁻¹; (b) $1.8 \times 1 0^{5}$ N C⁻¹.

Example 3: Derivation: equatorial field · 5 marks · Long answerHarder board level

Derive an expression for the electric field at a point on the equatorial line of an electric dipole of length 2l, at distance r from its centre. Write the result for a short dipole.

Try it first, then show the solutionSolution

Each charge is at distance $r^{2} + l^{2}$ from P, so $∣ E_{+} ∣ = ∣ E_{-} ∣ = \frac{k q}{r ^{2} + l ^{2}}$ ; $E_{+}$ points away from +q and $E_{-}$ towards −q. (1)
Resolve each field along the axis and perpendicular to it. The perpendicular components are equal and opposite and cancel. (1)
The components along the axis are both directed from +q to −q and add: $E = 2 \frac{k q}{r ^{2} + l ^{2}} cos θ$ with $cos θ = \frac{l}{r ^{2} + l ^{2}}$ . (1)
$E = \frac{k q ( 2 l )}{( r ^{2} + l ^{2} ) ^{3/2}} = \frac{k p}{( r ^{2} + l ^{2} ) ^{3/2}}$ , directed opposite to $p$ . (1)
Short dipole ( $r ≫ l$ ): $E = \frac{k p}{r ^{3}} = \frac{p}{4 π ε _{0} r ^{3}}$ . (1)

Answer: $E = \frac{1}{4 π ε _{0}} \frac{p}{( r ^{2} + l ^{2} ) ^{3/2}}$ , antiparallel to $p$ ; $\frac{p}{4 π ε _{0} r ^{3}}$ for a short dipole.

Example 4: Torque and work · 3 marks · Short answerHarder board level

A dipole of moment $2 \times 1 0^{- 8}$ C m is in a uniform field of $5 \times 1 0^{3}$ N C⁻¹. Find (a) the maximum torque, (b) the torque when $p$ makes 30° with $E$ , (c) the work needed to turn it from the stable position to the unstable position.

Try it first, then show the solutionSolution

(a) $pE = 2 \times 1 0^{- 8} \times 5 \times 1 0^{3} = 1 \times 1 0^{- 4}$ N m. (1)
(b) $pE sin 30° = 5 \times 1 0^{- 5}$ N m. (1)
(c) $W = pE (cos 0° - cos 180°) = 2 pE = 2 \times 1 0^{- 4}$ J. (1)

Answer: (a) $1 0^{- 4}$ N m; (b) $5 \times 1 0^{- 5}$ N m; (c) $2 \times 1 0^{- 4}$ J.

Example 5: How short is short? · 3 marks · Short answerHOTS / competency

Charges of ±1 nC are 4 cm apart. Find the field on the axis 10 cm from the centre using the exact formula and the short-dipole formula, and comment.

Try it first, then show the solutionSolution

$p = 1 0^{- 9} \times 0.04 = 4 \times 1 0^{- 11}$ C m; l = 0.02 m, r = 0.10 m. (1)
Exact: $\frac{2 k p r}{( r ^{2} - l ^{2} ) ^{2}} = \frac{2 \times 9 \times 1 0 ^{9} \times 4 \times 1 0 ^{- 11} \times 0.1}{( 0.0096 ) ^{2}} \approx 781$ N C⁻¹. (1)
Short-dipole: $\frac{2 k p}{r ^{3}} = 720$ N C⁻¹, about 8% low, because r is only 5 times l. (1)

Answer: 781 N C⁻¹ (exact) and 720 N C⁻¹ (approximate); the short-dipole formula is not accurate here.

Example 6: A molecular dipole · 4 marks · Case-basedHOTS / competency

Suppose a molecule behaves as charges of $\pm 3.2 \times 1 0^{- 19}$ C separated by $1.0 \times 1 0^{- 10}$ m, and is placed in a uniform field of $2 \times 1 0^{5}$ N C⁻¹.
(i) Find its dipole moment. (1 mark)
(ii) Find the maximum torque on it. (1 mark)
(iii) What is the net force on it? (1 mark)
(iv) Find the work needed to turn it from alignment with the field to perpendicular to it. (1 mark)

Try it first, then show the solutionSolution

(i) $p = 3.2 \times 1 0^{- 19} \times 1 0^{- 10} = 3.2 \times 1 0^{- 29}$ C m.
(ii) $τ_{ma x} = pE = 6.4 \times 1 0^{- 24}$ N m.
(iii) Zero: the field is uniform, so the forces on the two charges cancel.
(iv) $W = pE (cos 0° - cos 90°) = pE = 6.4 \times 1 0^{- 24}$ J.

Answer: (i) $3.2 \times 1 0^{- 29}$ C m; (ii) $6.4 \times 1 0^{- 24}$ N m; (iii) zero; (iv) $6.4 \times 1 0^{- 24}$ J.

Where students lose marks in Electric Dipole: Field and Torque

The slips examiners see most often, and what to write instead.

Mistake: Calling the locus an "equipotential surface"
The ISC 2025 examiners noted candidates who wrote "equipotential surface" instead of naming the equatorial plane, or confused a locus with a point.
Do this: Name it: the perpendicular bisector (equatorial plane) of the dipole.
Usually costs: 1 mark.
Mistake: Not using short-dipole formulas
Also in 2025: candidates did not use $\frac{2 k p}{r ^{3}}$ and $\frac{k p}{r ^{3}}$ for the ratio.
Do this: Write both formulas, then divide.
Usually costs: 1 mark.
Mistake: Broadside-on vs equatorial
Candidates mixed the magnet term "broadside-on" with the dipole's equatorial position.
Do this: For electric dipoles say "axial line" and "equatorial line".
Usually costs: ½ to 1 mark.
Mistake: Wrong direction of the equatorial field
Students say it is along $p$ .
Wrong working
$E_{e q}$ along $p$ .
Right working
$E_{e q} = - \frac{k p}{r ^{3}}$ .
Do this: Draw the two fields and add their components.
Usually costs: 1 mark in derivations.

Exam tips

Draw the diagram for every derivation, marking +q, −q, P, r, l and the field vectors.
State the short-dipole condition $r ≫ l$ before simplifying.
Keep p in C m; 1 nC × 1 mm = $1 0^{- 12}$ C m.
In torque questions, check whether the angle given is with the field or with the plates.

Practice questions on Electric Dipole: Field and Torque

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The SI unit of electric dipole moment is:
(a) C m⁻¹
(b) C m
(c) N C⁻¹
(d) C m²
Hint
Charge × distance.
Answer
(b).
Q2 · 2 marks · Short answerBoard level
A short dipole produces a field E at a point on its equatorial line at distance r from its centre. (i) What is the potential at that point? (ii) What is the field at distance 2r on its axis, in terms of E? (A variation on ISC 2025.)
Hint
Equidistant from both charges; $E_{a x ia l} = \frac{2 k p}{r ^{3}}$ .
Answer
(i) Zero: the point lies on the equatorial plane, the locus of zero potential. (ii) $\frac{2 k p}{( 2 r ) ^{3}} = \frac{1}{4} \cdot \frac{k p}{r ^{3}}$ , so $\frac{E}{4}$ .
Q3 · 2 marks · Short answerHarder board level
At what distance on the axis of a short dipole is the field equal to the field at 20 cm on its equatorial line?
Hint
$\frac{2}{r ^{3}} = \frac{1}{0. 2 ^{3}}$ .
Answer
$r = 0.2 \times 2^{1/3} \approx 25.2$ cm.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A dipole in a uniform electric field experiences no net force.
Reason (R): The forces on its two charges are equal in magnitude and opposite in direction.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Torque can still act.
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
A dipole made of ±2 μC charges experiences a maximum torque of $8 \times 1 0^{- 3}$ N m in a field of $2 \times 1 0^{5}$ N C⁻¹. Find the separation of the charges.
Hint
$p = \frac{τ _{ma x}}{E}$ .
Answer
$p = 4 \times 1 0^{- 8}$ C m; $2 l = \frac{p}{q} = 0.02$ m = 2 cm.
Q6 · 3 marks · Short answerHOTS / competency
Two identical short dipoles of moment p are placed at the same point at right angles. A point P is at distance r on the axis of the first and on the equatorial line of the second. Find the resultant field at P.
Hint
The two fields are perpendicular.
Answer
$\frac{2 k p}{r ^{3}}$ and $\frac{k p}{r ^{3}}$ at right angles: $E = \frac{5 k p}{r ^{3}}$ , at $tan^{- 1} \frac{1}{2} \approx 26.6°$ to the first dipole's axis.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose charges of ±4 nC are fixed 3 mm apart, and the dipole is placed in a uniform field of $3 \times 1 0^{4}$ N C⁻¹ at 45° to it.
(i) Find the dipole moment. (1 mark)
(ii) Find the field due to the dipole 15 cm away on its axis and on its equatorial line. (2 marks)
(iii) Find the torque exerted on it by the external field. (1 mark)
Hint
r = 15 cm ≫ 1.5 mm.
Answer
(i) $1.2 \times 1 0^{- 11}$ C m. (ii) Axial $\frac{2 k p}{r ^{3}} = 64$ N C⁻¹; equatorial 32 N C⁻¹. (iii) $pE sin 45° \approx 2.5 \times 1 0^{- 7}$ N m.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Direction of the dipole moment?Show answer
From −q to +q.
2.Field on the axis of a short dipole?Show answer
$\frac{2 k p}{r ^{3}}$ , along $p$ .
3.Torque on a dipole in a uniform field?Show answer
$τ = p \times E$ .
4.p = 10⁻⁹ C m in E = 10⁴ N C⁻¹ at 90°: torque?Show answer
$1 0^{- 5}$ N m.
5.When does a dipole feel a net force?Show answer
In a non-uniform field.

Electric Dipole: Field and Torque: frequently asked questions

Why does the dipole field fall as 1/r³?

At large distances the fields of +q and −q nearly cancel. What remains depends on their small separation divided by an extra power of r, giving \frac{1}{r^3} instead of \frac{1}{r^2}.

Is the field zero where the potential is zero?

No. On the equatorial plane the potential is zero but the field is \frac{kp}{r^3}, directed opposite to \vec{p}.

What is the stable position of a dipole?

With \vec{p} along \vec{E} (θ = 0°): the torque is zero and the potential energy is minimum. At 180° the torque is also zero but the equilibrium is unstable.

Do I need to derive both axial and equatorial fields?

Yes. The ISC 2027 syllabus asks for the derivation of E at a point on the axis and on the equatorial line, and for the torque on a dipole in a uniform field.

How was this tested recently?

In 2025 a 2-mark question asked for the locus of zero potential and the ratio of axial to equatorial field of a short dipole.