PhysicsCBSE Class 12 · Electric Charges and Fields

Electric Dipole Class 12: Dipole Moment, Field and Torque

Updated 15 October 2026 · Fact-checked

An electric dipole is a pair of equal and opposite charges ±q a distance 2a apart, with dipole moment p = q × 2a pointing from −q to +q. Far away its field falls as 1/r³: 2kp/r³ on the axis, kp/r³ on the equatorial line. A uniform field gives zero net force but a torque pE sin θ.

Understand Electric Dipole

Many molecules, such as water, carry no net charge, yet one end is slightly positive and the other slightly negative. Such a charge pair is the simplest example of an electric dipole: charges $+ q$ and $- q$ held a distance $2 a$ apart.

Dipole moment. The strength of a dipole is measured by its dipole moment $p = q \times 2 a \overset{p}{^}$ , a vector pointing **from $- q$ to $+ q$ **. Its SI unit is C m. The total charge is zero, but the two charges sit at different places, so their fields do not cancel fully.

Field on the axis (end-on position). Take a point P on the axis at distance $r$ from the centre, on the side of $+ q$ . The field of $+ q$ (distance $r - a$ ) points away from it, the field of $- q$ (distance $r + a$ ) points towards it, i.e. back towards the dipole. The first is larger, so $E_{a x ia l} = k q [\frac{1}{( r - a ) ^{2}} - \frac{1}{( r + a ) ^{2}}] = \frac{k \cdot 4 q a r}{( r ^{2} - a ^{2} ) ^{2}} = \frac{2 k p r}{( r ^{2} - a ^{2} ) ^{2}}$ , along $p$ . For $r ≫ a$ : $E_{a x ia l} = \frac{2 k p}{r ^{3}}$ .

Field on the equatorial line (broad-on position). At a point on the perpendicular bisector, at distance $r$ from the centre, both charges are $r^{2} + a^{2}$ away and give fields of equal size $\frac{k q}{r ^{2} + a ^{2}}$ . Their components perpendicular to the axis cancel, and the components parallel to the axis add, both pointing from $+ q$ towards $- q$ . Each parallel component is $\frac{k q}{r ^{2} + a ^{2}} \cdot \frac{a}{r ^{2} + a ^{2}}$ , so $E_{e q} = \frac{2 k q a}{( r ^{2} + a ^{2} ) ^{3/2}} = \frac{k p}{( r ^{2} + a ^{2} ) ^{3/2}}$ , opposite to $p$ . For $r ≫ a$ : $E_{e q} = - \frac{k p}{r ^{3}}$ .

Two results to remember: at the same large distance the axial field is twice the equatorial field, and a dipole field falls as $\frac{1}{r ^{3}}$ , faster than the $\frac{1}{r ^{2}}$ of a single charge, because the two opposite charges nearly cancel at large distances.

Dipole in a uniform field. In a uniform field $E$ , the forces $q E$ on $+ q$ and $- q E$ on $- q$ are equal and opposite, so the net force is zero. But they act at different points, forming a couple. With $θ$ the angle between $p$ and $E$ , the torque is $τ = q E \times 2 a sin θ = pE sin θ$ , or $τ = p \times E$ . It turns the dipole to line up with the field. The torque is largest ( $pE$ ) at $θ = 90°$ and zero at $0°$ and $180°$ .

Energy and equilibrium. Turning the dipole from angle $θ_{1}$ to $θ_{2}$ against this torque needs work $W = pE (cos θ_{1} - cos θ_{2})$ , and the potential energy is $U = - p \cdot E = - pE cos θ$ (you meet this again in Chapter 2). At $θ = 0°$ , $U = - pE$ is a minimum: stable equilibrium. At $θ = 180°$ , $U = + pE$ is a maximum: unstable equilibrium.

Non-uniform field. If the field is stronger at one end than the other, the two forces no longer cancel, so the dipole feels a net force as well as (in general) a torque. A dipole lined up with the field is pulled towards the stronger-field region. This is why a charged comb attracts neutral paper: the comb's field induces dipoles in the paper and the field is non-uniform.

Electric Dipole: formulas and key facts

Dipole moment: $p = q \times 2 a \overset{p}{^}$; Direction from $- q$ to $+ q$ ; SI unit C m.
Field on the axial line: $E_{a x ia l} = \frac{1}{4 π ε _{0}} \frac{2 p r}{( r ^{2} - a ^{2} ) ^{2}} \approx \frac{1}{4 π ε _{0}} \frac{2 p}{r ^{3}}$; Along $p$ . The approximation holds for $r ≫ a$ .
Field on the equatorial line: $E_{e q} = \frac{1}{4 π ε _{0}} \frac{p}{( r ^{2} + a ^{2} ) ^{3/2}} \approx \frac{1}{4 π ε _{0}} \frac{p}{r ^{3}}$; Opposite to $p$ .
Ratio at the same large distance: $E_{a x ia l} : E_{e q} = 2 : 1$; Both fall as $\frac{1}{r ^{3}}$ .
Torque in a uniform field: $τ = p \times E, τ = pE sin θ$; Net force is zero in a uniform field.
Work to rotate and potential energy: $W = pE (cos θ_{1} - cos θ_{2}), U = - p \cdot E = - pE cos θ$; Stable at $θ = 0°$ ( $U = - pE$ ), unstable at $θ = 180°$ ( $U = + pE$ ).

How to solve Electric Dipole questions

Use this method for dipole field, torque and energy questions.

1Find $p = q \times 2 a$ in C m; remember $2 a$ is the full separation, so do not halve it twice.
2Decide where the point is: on the axis, on the equatorial line, or (for torque and energy) what angle $θ$ the dipole makes with $E$ .
3Check whether $r ≫ a$ . If yes, use the short-dipole forms $\frac{2 k p}{r ^{3}}$ and $\frac{k p}{r ^{3}}$ ; if not, use the exact forms.
4Give the direction: axial field along $p$ , equatorial field opposite to $p$ .
5For torque, use $τ = pE sin θ$ ; for work, $W = pE (cos θ_{1} - cos θ_{2})$ with $θ$ measured from $E$ .
6In a derivation, draw the dipole, the point and both field arrows, write each field with its distance, then combine and approximate in the last line.

Electric Dipole 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 dipole

Suppose you build a model dipole: two tiny beads carrying +10 nC and −10 nC on the ends of a light insulating rod 4 cm long, pivoted at its middle between two charged plates that make a uniform field of $5 \times 1 0^{4}$ N/C.

(a) Find the dipole moment. (b) Find the torque when the rod is at 90° and at 60° to the field. (c) How much work must you do to turn it from lined up with the field to exactly opposite? (d) Which of these two positions will the rod stay in if nudged slightly?

Try it first, then show the solutionSolution

$p = q \times 2 a = 10 \times 1 0^{- 9} \times 0.04 = 4 \times 1 0^{- 10}$ C m.
At 90°: $τ = pE = 4 \times 1 0^{- 10} \times 5 \times 1 0^{4} = 2 \times 1 0^{- 5}$ N m. At 60°: $τ = pE sin 60° \approx 1.73 \times 1 0^{- 5}$ N m.
$W = pE (cos 0° - cos 180°) = 2 pE = 4 \times 1 0^{- 5}$ J.
Lined up ( $θ = 0°$ ) is stable: a nudge produces a torque that turns it back. Opposite ( $180°$ ) is unstable: a nudge makes it swing round.

Answer: (a) $4 \times 1 0^{- 10}$ C m; (b) $2 \times 1 0^{- 5}$ N m and about $1.73 \times 1 0^{- 5}$ N m; (c) $4 \times 1 0^{- 5}$ J; (d) the lined-up position.

Memory hook: A dipole is a compass needle for electric fields: it turns to point along E.

Polar molecules in a field

Imagine a single water molecule, a dipole of moment about $6.2 \times 1 0^{- 30}$ C m, placed in a strong uniform field of $1 0^{6}$ N/C.

(a) Find the largest torque on the molecule. (b) Find the work needed to turn it from lined up with the field to perpendicular to it. (c) Why does a uniform field not pull the molecule along?

Try it first, then show the solutionSolution

Largest torque at 90°: $τ_{ma x} = pE = 6.2 \times 1 0^{- 30} \times 1 0^{6} = 6.2 \times 1 0^{- 24}$ N m.
$W = pE (cos 0° - cos 90°) = pE = 6.2 \times 1 0^{- 24}$ J.
The forces on the positive and negative ends are equal and opposite in a uniform field, so the net force is zero; only a torque acts.

Answer: (a) $6.2 \times 1 0^{- 24}$ N m; (b) $6.2 \times 1 0^{- 24}$ J; (c) equal and opposite forces cancel in a uniform field.

Memory hook: Uniform field: twist but no pull. Non-uniform field: twist and pull.

Tricks to remember Electric Dipole

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

Minus to plus
The dipole moment arrow goes from the negative charge to the positive charge, the opposite of field lines between them.
Use it when: Getting the direction of p and of the fields right.
Axial is double, equatorial is opposite
End-on: 2kp/r³ along p. Broad-on: kp/r³ against p.
Use it when: Ratio and direction questions.
Cube for a couple
One charge: 1/r². A pair of opposite charges: 1/r³, one power faster, because the two nearly cancel.
Use it when: How the dipole field varies with distance.
Torque is sine, energy is cosine
τ = pE sin θ is biggest at 90°; U = −pE cos θ is lowest at 0°.
Use it when: Picking the right trig function.

Electric Dipole: solved examples

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

Example 1: Field at axial and equatorial points · 3 marks · Short answerBoard level

A dipole is made of charges ±2 nC separated by 4 mm. Find (a) the dipole moment, and the field at a distance of 20 cm from its centre (b) on the axis and (c) on the equatorial line. Give directions.

Try it first, then show the solutionSolution

(a) $p = 2 \times 1 0^{- 9} \times 4 \times 1 0^{- 3} = 8 \times 1 0^{- 12}$ C m, from −q to +q. (1)
(b) $r = 20$ cm $≫ a = 2$ mm, so $E_{a x ia l} = \frac{2 k p}{r ^{3}} = \frac{2 \times 9 \times 1 0 ^{9} \times 8 \times 1 0 ^{- 12}}{( 0.2 ) ^{3}} = 18$ N/C, along $p$ . (1)
(c) $E_{e q} = \frac{k p}{r ^{3}} = 9$ N/C, opposite to $p$ . (1)

Answer: (a) $8 \times 1 0^{- 12}$ C m; (b) 18 N/C along p; (c) 9 N/C opposite to p.

Example 2: Derivation: field on the axial line · 3 marks · Short answerBoard level

Derive an expression for the electric field at a point on the axis of an electric dipole at a distance r from its centre. Write the form for a short dipole.

Try it first, then show the solutionSolution

Draw −q at A and +q at B, with AB = 2a, centre O and point P on the axis beyond B, OP = r. Then BP = r − a and AP = r + a. (½)
Field of +q at P: $E_{+} = \frac{k q}{( r - a ) ^{2}}$ along $\overset{p}{^}$ . Field of −q at P: $E_{-} = \frac{k q}{( r + a ) ^{2}}$ along $- \overset{p}{^}$ . (1)
Net: $E = k q [\frac{( r + a ) ^{2} - ( r - a ) ^{2}}{( r ^{2} - a ^{2} ) ^{2}}] = \frac{4 k q a r}{( r ^{2} - a ^{2} ) ^{2}} = \frac{2 k p r}{( r ^{2} - a ^{2} ) ^{2}}$ , along $\overset{p}{^}$ . (1)
For $r ≫ a$ , $(r^{2} - a^{2})^{2} \approx r^{4}$ , so $E = \frac{1}{4 π ε _{0}} \frac{2 p}{r ^{3}}$ . (½)

Answer: $E_{a x ia l} = \frac{1}{4 π ε _{0}} \frac{2 p r}{( r ^{2} - a ^{2} ) ^{2}}$ along p; for a short dipole $\frac{1}{4 π ε _{0}} \frac{2 p}{r ^{3}}$ .

Example 3: Derivation: equatorial field and torque · 5 marks · Long answerBoard level

(a) Derive the expression for the electric field at a point on the equatorial line of a dipole, at distance r from its centre. (3)
(b) Show that a dipole in a uniform electric field feels no net force, and derive the torque on it. (2)

Try it first, then show the solutionSolution

(a) P lies on the perpendicular bisector, OP = r; each charge is $r^{2} + a^{2}$ from P, so $∣ E_{+} ∣ = ∣ E_{-} ∣ = \frac{k q}{r ^{2} + a ^{2}}$ . (1)
Resolve each field along and perpendicular to the axis. The perpendicular components are equal and opposite and cancel; the components along the axis both point from +q towards −q and add. (1)
$E = 2 \frac{k q}{r ^{2} + a ^{2}} cos α$ with $cos α = \frac{a}{r ^{2} + a ^{2}}$ , so $E = \frac{k ( 2 q a )}{( r ^{2} + a ^{2} ) ^{3/2}} = \frac{k p}{( r ^{2} + a ^{2} ) ^{3/2}}$ , opposite to $p$ ; for $r ≫ a$ , $E = - \frac{k p}{r ^{3}}$ . (1)
(b) Force on +q is $q E$ and on −q is $- q E$ ; their sum is zero. (1)
The two forces have different lines of action, separated by $2 a sin θ$ , so they form a couple: $τ = q E \times 2 a sin θ = pE sin θ$ , i.e. $τ = p \times E$ , tending to align $p$ with $E$ . (1)

Answer: (a) $E_{e q} = \frac{1}{4 π ε _{0}} \frac{p}{( r ^{2} + a ^{2} ) ^{3/2}}$ opposite to p; (b) net force zero, torque $τ = p \times E$ .

Example 4: Torque, work and energy of a dipole · 3 marks · Short answerHarder board level

A dipole consists of charges ±5 μC separated by 2 cm and is placed in a uniform field of $2 \times 1 0^{5}$ N/C at 30° to the field.
(a) Find the torque on it. (1)
(b) Find its potential energy in this position. (1)
(c) How much work is needed to turn it from the stable position to the unstable one? (1)

Try it first, then show the solutionSolution

$p = 5 \times 1 0^{- 6} \times 0.02 = 1 0^{- 7}$ C m; $pE = 1 0^{- 7} \times 2 \times 1 0^{5} = 0.02$ N m.
(a) $τ = pE sin 30° = 0.02 \times 0.5 = 0.01$ N m. (1)
(b) $U = - pE cos 30° = - 0.02 \times 0.866 \approx - 0.0173$ J. (1)
(c) Stable is $θ = 0°$ , unstable is $180°$ : $W = pE (cos 0° - cos 180°) = 2 pE = 0.04$ J. (1)

Answer: (a) 0.01 N m; (b) about −0.0173 J; (c) 0.04 J.

Example 5: When is the short-dipole formula good enough? · 3 marks · Short answerHOTS / competency

A dipole has its charges 4 mm apart. Using the exact axial formula, find by what percentage the short-dipole formula $\frac{2 k p}{r ^{3}}$ underestimates the axial field at (a) r = 1 cm and (b) r = 10 cm from the centre. What do you conclude?

Try it first, then show the solutionSolution

$\frac{E _{e x a c t}}{E _{a pp r o x}} = \frac{2 k p r / ( r ^{2} - a ^{2} ) ^{2}}{2 k p / r ^{3}} = \frac{r ^{4}}{( r ^{2} - a ^{2} ) ^{2}}$ , with a = 2 mm. (1)
(a) r = 10 mm: $\frac{1 0 ^{4}}{( 100 - 4 ) ^{2}} = \frac{10000}{9216} \approx 1.085$ ; the approximation gives $\frac{9216}{10000}$ of the true value, about 7.8% too low. (1)
(b) r = 100 mm: ratio $= \frac{1}{( 1 - 0.0004 ) ^{2}}$ ; the approximation is about 0.08% too low. The short-dipole form is reliable once r is many times a. (1)

Answer: (a) about 7.8% low; (b) about 0.08% low; use the exact form when r is only a few times a.

Example 6: Dipole near a point charge · 4 marks · Case-basedHarder board level

Read the passage and answer.

A small dipole is placed near a fixed positive point charge Q, with its centre on a radial line from Q and its dipole moment pointing directly away from Q.
(a) Which end of the dipole is nearer Q? (1)
(b) Compare the sizes of the forces on the two ends and give the direction of the net force. (1)
(c) Is there any torque? Give a reason. (1)
(d) Use this idea to explain why a charged comb attracts small bits of neutral paper. (1)

Try it first, then show the solutionSolution

(a) $p$ points from −q to +q and away from Q, so −q is nearer Q. (1)
(b) The field of Q is stronger closer to it, so the attraction on −q is larger than the repulsion on +q: net force towards Q. (1)
(c) No torque: both forces act along the same radial line through the dipole's axis, so they have no turning effect. (1)
(d) The comb's field induces dipoles in the paper, lined up with the field; the comb's field is non-uniform, so these dipoles feel a net pull towards the comb. (1)

Answer: (a) −q; (b) larger pull on −q, net force towards Q; (c) no torque, forces along the axis; (d) induced dipoles in a non-uniform field are pulled towards stronger field.

Where students lose marks in Electric Dipole

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

Mistake: Drawing the dipole moment from + to −.
Field lines go from + to −, and students copy that direction.
Wrong working
p points from +q to −q.
Right working
p points from −q to +q.
Do this: Write 'from −q to +q' next to the arrow in every diagram.
Usually costs: ½ to 1 mark, plus wrong field directions later.
Mistake: Using a in place of 2a in p.
Confusing the half-separation a with the separation 2a.
Wrong working
p = q × a = 2 × 10⁻⁹ × 2 × 10⁻³
Right working
p = q × 2a = 2 × 10⁻⁹ × 4 × 10⁻³
Do this: Read the question: 'separated by d' means 2a = d.
Usually costs: The numerical answer mark.
Mistake: Approximating too early in a derivation.
Writing $\frac{2 p}{r ^{3}}$ straight away hides the exact result the marking scheme looks for.
Do this: Get the exact expression first, then write 'for r ≫ a' and simplify in the last line.
Usually costs: 1 mark in a 3-mark derivation.
Mistake: Saying a dipole in a uniform field moves towards the field.
Mixing up torque and force.
Wrong working
The dipole is pulled along the field.
Right working
Net force is zero; the dipole only rotates to align with E.
Do this: Uniform field: twist, no pull. Non-uniform field: twist and pull.
Usually costs: 1 mark in an assertion-reason or 2-mark reasoning question.
Mistake: Measuring θ from the wrong line in τ = pE sin θ.
Some take θ from the plates or from the perpendicular.
Do this: θ is always the angle between $p$ and $E$ .
Usually costs: The final mark of the numerical.

Exam tips

Every derivation needs a labelled diagram: −q, +q, 2a, centre O, point P, distance r and both field arrows.
End each field derivation with the short-dipole form and its direction relative to $p$ .
For torque, write $τ = p \times E$ and state the direction in words: it tends to align p with E.
Board papers have asked the equatorial-field derivation (2023, 55/1/1) and the torque or energy of a dipole in a uniform field (2024 and 2026, 55/1/1) as 5-mark parts; practise both in full.
In MCQs, check whether the question says uniform or non-uniform field before choosing 'zero net force'.

Practice questions on Electric Dipole

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 direction of the electric field at a point on the equatorial line of a dipole is
(a) along the dipole moment
(b) opposite to the dipole moment
(c) perpendicular to the dipole moment
(d) zero, as the fields cancel
Hint
Draw both field arrows and add their components.
Answer
(b) opposite to the dipole moment.
Q2 · 2 marks · Very short answerBoard level
Define electric dipole moment. Write its SI unit and its direction.
Hint
Product of one charge and the separation.
Answer
Dipole moment is the product of the magnitude of either charge and the separation between the charges, $p = q \times 2 a$ . SI unit: C m. Direction: from −q to +q along the axis.
Q3 · 2 marks · Very short answerHarder board level
A short dipole produces a field E at a point on its axis at distance r. Find the field (size and direction) at a point on its equatorial line at distance 2r.
Hint
Write both in terms of kp/r³.
Answer
$E = \frac{2 k p}{r ^{3}}$ . At 2r on the equator: $\frac{k p}{( 2 r ) ^{3}} = \frac{k p}{8 r ^{3}} = \frac{E}{16}$ , directed opposite to $p$ .
Q4 · 2 marks · Very short answerHarder board level
A molecule behaves as a dipole and feels a maximum torque of $9 \times 1 0^{- 26}$ N m in a uniform field of $1 0^{4}$ N/C. (a) Find its dipole moment. (b) If the charges are ±e, find their separation.
Hint
Maximum torque is pE.
Answer
(a) $p = \frac{τ _{ma x}}{E} = \frac{9 \times 1 0 ^{- 26}}{1 0 ^{4}} = 9 \times 1 0^{- 30}$ C m. (b) $d = \frac{p}{e} = \frac{9 \times 1 0 ^{- 30}}{1.6 \times 1 0 ^{- 19}} \approx 5.6 \times 1 0^{- 11}$ m.
Q5 · 3 marks · Short answerHarder board level
A dipole of moment $2 \times 1 0^{- 8}$ C m is in a uniform field of $3 \times 1 0^{4}$ N/C, at 60° to the field. (a) Find the torque. (b) Find the work done in turning it to 180°. (c) In which position is its potential energy least?
Hint
pE = 6 × 10⁻⁴ J.
Answer
(a) $τ = pE sin 60° = 6 \times 1 0^{- 4} \times 0.866 \approx 5.2 \times 1 0^{- 4}$ N m. (b) $W = pE (cos 60° - cos 180°) = 6 \times 1 0^{- 4} \times 1.5 = 9 \times 1 0^{- 4}$ J. (c) At θ = 0°, lined up with the field, where U = −pE.
Q6 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): A dipole placed in a non-uniform electric field may feel a net force.
Reason (R): In a non-uniform field, the forces on the two charges of a dipole are not equal in magnitude.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Net force = qE₁ − qE₂.
Answer
(a). With $E_{1} \neq = E_{2}$ at the two ends, $q E_{1} - q E_{2} \neq = 0$ , so there is a net force.
Q7 · 3 marks · Short answerHOTS / competency
Two short dipoles, each of moment p, are placed at the origin: one along the x-axis and one along the y-axis. Find the net field at a point (r, 0), r being large, and its angle with the x-axis.
Hint
The point is axial for one dipole and equatorial for the other.
Answer
Dipole along x: the point is on its axis, $\frac{2 k p}{r ^{3}}$ along +x. Dipole along y: the point is on its equatorial line, $\frac{k p}{r ^{3}}$ along −y. Net $E = \frac{k p}{r ^{3}} 4 + 1 = \frac{5 k p}{r ^{3}}$ , at $tan^{- 1} (1/2) \approx 26.6°$ below the +x-axis.

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.What is the direction of the dipole moment?Show answer
From −q to +q along the axis.
2.How does the field of a short dipole vary with distance?Show answer
As 1/r³.
3.What is the ratio of axial to equatorial field at the same large distance?Show answer
2 : 1.
4.What is the net force on a dipole in a uniform field?Show answer
Zero.
5.Write the torque on a dipole in a uniform field.Show answer
τ = pE sin θ, or $τ = p \times E$ .
6.In which orientation is a dipole in stable equilibrium?Show answer
When p is parallel to E (θ = 0°), with U = −pE.

Electric Dipole: frequently asked questions

Why does the dipole field fall off as 1/r³ and not 1/r²?

The two charges are equal and opposite. Far away, their fields almost cancel, and what is left depends on the small difference in distance to the two charges. That leftover falls one power of r faster, as 1/r³.

Is the field zero anywhere on the equatorial line of a dipole?

No. At every point on the equatorial line the components along the axis add up, giving a field opposite to the dipole moment. It only tends to zero very far away.

What is an ideal dipole?

A dipole in which the separation 2a tends to zero and the charge q tends to infinity in such a way that p = q × 2a stays finite. For such a dipole the short-dipole formulas are exact at every distance.

Does a dipole in a uniform field have potential energy?

Yes. U = −pE cos θ, taking the zero of energy at θ = 90°. It is least (−pE) when the dipole lines up with the field and greatest (+pE) when it points opposite.

Why are neutral pieces of paper attracted by a charged comb?

The comb's field makes each bit of paper a small dipole (induced polarisation). The comb's field is non-uniform, being stronger nearer the comb, so the induced dipoles feel a net force towards the comb.