12 min readConcept sheetSection B short answer (3 marks) or long answer (5 marks), often with a choice: field on the axis or the equatorial line, or torque in a uniform field; also a NEET favourite

Electric dipole: its field and the torque on it

A dipole is ±q, 2l apart, with p = q × 2l. Its field falls as 1/r³; in a uniform field it feels torque pE sin θ, no net force.

Dipole momentp = q × 2l; unit C m; a vector pointing from −q to +q

End-on (axial) fieldE = (1/4πε₀) 2p/r³, along p (short dipole, r much larger than l)

Broad-on (equatorial) fieldE = (1/4πε₀) p/r³, opposite to p: half the end-on value

Torque in a uniform fieldτ = pE sin θ, or τ = p × E; the net force is zero

What is an electric dipole?

An electric dipole is a pair of equal and opposite charges, +q and −q, a small distance 2l apart. A water molecule behaves like one. The dipole moment is p = q × 2l. Its unit is the coulomb metre (C m). p is a vector along the axis, pointing from −q to +q. The total charge is zero, but the field is not, because the two charges sit at different places. Your NIOS lesson calls a point on the axis the end-on position and a point on the perpendicular bisector the broad-on position.

Derivation: field at an end-on (axial) point

−q is at A and +q at B, with AB = 2l and centre O. P is on the axis beyond B, with OP = r. Let k = 1/4πε₀.

1
Field at P due to +q (distance r − l): E₁ = kq/(r − l)², along AP, away from B.
2
Field at P due to −q (distance r + l): E₂ = kq/(r + l)², along PA, towards A.
3
E₁ is larger, as P is nearer +q. Net field: E = E₁ − E₂ = kq[1/(r − l)² − 1/(r + l)²].
4
Simplify: [(r + l)² − (r − l)²] = 4lr, so E = kq × 4lr/(r² − l²)² = k(2lq)(2r)/(r² − l²)².
5
With p = 2lq: E = (1/4πε₀) 2pr/(r² − l²)².
6★
For a short dipole (r much larger than l), drop l²: E = (1/4πε₀) 2p/r³, in the direction of p.

★ key step

Derivation: field at a broad-on (equatorial) point

Same dipole. P is on the perpendicular bisector of AB, with OP = r.

1
Each charge is √(r² + l²) from P, so the two fields are equal in size: E₁ = E₂ = kq/(r² + l²).
2
E₁ points along BP, away from +q. E₂ points along PA, towards −q.
3
Resolve each field along AB and at right angles to it. The parts at right angles to AB are equal and opposite, so they cancel.
4
The parts along AB add. Both point from B towards A, that is, opposite to p.
5
With cos θ = l/√(r² + l²): E = 2E₁ cos θ = 2kql/[(r² + l²)√(r² + l²)] = kp/[(r² + l²)√(r² + l²)].
6★
For a short dipole: E = (1/4πε₀) p/r³, opposite to p. This is half the end-on field at the same distance.

★ key step

Derivation: dipole in a uniform field

A dipole of charges ±q, 2l apart, sits in a uniform field E, with its axis at an angle θ to E.

1
Force on +q: qE, along E. Force on −q: qE, opposite to E.
2
The two forces are equal and opposite, so the net force is zero: the dipole does not move as a whole.
3
They act along different lines, so they form a couple that turns the dipole.
4
Torque = one force × arm of the couple = qE × 2l sin θ.
5
With p = 2lq: τ = pE sin θ, or in vector form τ = p × E.
6★
So, τ = 0 at θ = 0° (p along E) and τ = pE, the largest, at θ = 90°. The field tries to line the dipole up with itself. In a non-uniform field there is also a net force, so the dipole moves as well as turns.

★ key step

How to draw the dipole figures

Draw a horizontal line AB. Put −q at A and +q at B, mark the centre O and write 2l along AB. Draw the arrow p from A to B.
End-on: mark P on the line beyond B, with OP = r. At P draw a long arrow E₁ pointing away from B and a short arrow E₂ pointing back towards A.
Broad-on: mark P above O, with OP = r, and join AP and BP. Draw equal arrows E₁ along BP extended and E₂ along PA. Show their resultant parallel to BA.
Torque: draw parallel arrows for E. Draw the dipole tilted at θ, with qE along E at +q and qE opposite to E at −q. Label the arm 2l sin θ.

Worked example: field and torque of a dipole

Problem

Charges of +4 μC and −4 μC are 3 mm apart. Find (a) the dipole moment, (b) the field 20 cm from its centre in the end-on and broad-on positions, and (c) the torque on it at 30° to a uniform field of 5 × 10⁴ N C⁻¹.

Working

(a) p = q × 2l = 4 × 10⁻⁶ C × 3 × 10⁻³ m = 1.2 × 10⁻⁸ C m.
r = 0.20 m is far larger than l = 1.5 mm, so use the short-dipole results.
(b) End-on: E = 2kp/r³ = 2 × 9 × 10⁹ × 1.2 × 10⁻⁸ ÷ (0.20)³ = 216 ÷ 0.008 = 2.7 × 10⁴ N C⁻¹.
Broad-on: E = kp/r³ = 108 ÷ 0.008 = 1.35 × 10⁴ N C⁻¹.
(c) τ = pE sin θ = 1.2 × 10⁻⁸ × 5 × 10⁴ × sin 30° = 6 × 10⁻⁴ × 0.5 = 3 × 10⁻⁴ N m.

Answer

(a) 1.2 × 10⁻⁸ C m; (b) 2.7 × 10⁴ N C⁻¹ along p, and 1.35 × 10⁴ N C⁻¹ opposite to p; (c) 3 × 10⁻⁴ N m

In real life

A water molecule is a natural dipole. In a microwave oven the electric field keeps reversing, the water dipoles keep turning to follow it, and this jostling heats the food.

Exam angle

1 marks: “What is the SI unit of electric dipole moment?” Coulomb metre (C m).

2 marks: “A dipole is made of charges ±2 μC placed 1 cm apart. Find the largest torque on it in a uniform field of 1 × 10⁵ N C⁻¹.” p = q × 2l = 2 × 10⁻⁶ × 0.01 = 2 × 10⁻⁸ C m. The torque is largest at θ = 90°: τ = pE = 2 × 10⁻⁸ × 1 × 10⁵ = 2 × 10⁻³ N m.

3 marks: “Derive the field of a short dipole at a point on its equatorial line. Compare it with the end-on field at the same distance.” Equal fields kq/(r² + l²) from ±q; the parts across the axis cancel; the parts along it add to 2kql/[(r² + l²)√(r² + l²)]. For r much larger than l, E = kp/r³, opposite to p. The end-on field, 2kp/r³ along p, is twice as large.

Remember it

End-on is double, broad-on is backward: 2p/r³ along p on the axis, p/r³ against p on the bisector. Both fall as 1/r³.